It was more due to him wanting to keep the lesson simple and if it were done with 7, it would have been complicated and requires this lesson to be complete before they could do the math for that. And the other student suggested by itself not just 5.
This guy is such a great teacher, you can see how he engages with the class to be a guide and helper in a journey of discovery. Not some remote pedagogue, he gets alongside them. He loves his subject, and wants us to love it too. Exceptional.
It's a nice demonstration but he does the discovery instead of the students. Not this spectacular, but more beneficial to let the students do the thinking.
Why do kids think addition when we talk about exponents??? "What number, when multiplied by itself, gives you 100?" "50" 🤦♂️ "Multiply. Multiply. Multiply!"
Every math teacher ever: "Let's just use whatever number I think of... Could be anything... Anything at all." *inserts very specific number to illustrate the lesson*
Which is a useful skill in higher math, but that's a bad way to teach it: One of the most useful tools when dealing with complicated algebra is the knowledge that as long as you multiply both sides of the equation by the same thing, you can multiply them by *anything*. So, just like the teacher in the example, you pick something *useful*. This concept is tangentially useful later with substitutions as well. Of course, the trick is being able to tell what is going to be useful to you in a given problem...
@@treyslider6954 to multiply both side by 0 mean u are dividing 0/0. The reason u can multiply both side by any number its because they would divide by themselve. But u cant divide 0/0
@@IStMl I wouldn't be so condescending if I were you, considering the hypocrisy of your entire statement... you just said "ironical", which isn't even a word... plus I don't think you know the meaning of ironic lmao. The word you're looking for that describes the joke is "hyperbolic"... so I don't think you're one to ridicule others' academic abilities.
But by the method he got to =7 I.e. by multiplying the exponents, you do get 10. That’s what I don’t get. You get two different answers from two different methods.
@@Praneettigga No, you must have made a mistake with your arithmetic. The exponents are 1/10 and 10, and they are being multiplied. 1/10·10 = 1, and the base is 7, so you get 7^1 = 7, which is indeed correct.
I really enjoy your videos! Keep up the great work! Love your enthusiasm and your style of breaking down and trying to show things in a simple manner. This is how most math teachers should be.
I love the personality and energy, and the wording is great for students who "get" maths, but there are clearly students that haven't grasped previous concepts (such as roots that are anything but square roots). Also, it's never specifically stated that a^1/X is the x root of a, and by the end some students have clearly missed that and still think it's magic.
@@amosw766 I'm not saying that he was wrong for asking, I'm just pointing out the fact that if you know about the topic it's gonna make you cringe. Also, he wasn't so wrong since the 3 fives technically came from the 1/3 so it was actually kind of linked together anyways... Being real here those were the types of questions I'd ask in my math class ;) except I'd reflect a bit more before asking ( Which isn't the right way to do it since there isn't a right or a wrong way to ask a question ! )
Most people learn this stuff in the early years of HS, but I like Eddie because he explains the origins and concepts of everything he teaches. Like, he tells us *why* things are the way they are, which is the main way I remember things so it's super helpful (I find it harder to cram stuff).
I'm going into an engineering degree and I've done real well at calculus. Not sure why I'm watching but this is a good teacher in action Also at about 6:50 it is written that 10root7=7, it's nit picking but it'll be counted as wrong in an exam
I love how often you say "Get your calculator out". Back in my day they'd say "No calculators, you need to know how to do this in your head, You won't always have a calculator with you"
@@jeremiahbarron2158 Allow me to be the party pooper here, this is not an okay boomer moment for 2 main reasons: 1- It is not a boomer thing, it happened to Millenials and to some Gen Z's and is still happening in many countries to Gen Z's in school....sooooooo....yeah, not a boomer thing, he is not talking about the time people had to go several miles to make a phone call or something. 2- okay boomer doesn't apply here cause the guy is not criticizing anyone nor glorifying "boomer's days"...so the whole foundation and context of using the "okay boomer" card are non-existent. PS: waiting for your reply to this with "okay boomer"
isnt it funny how simple and logical this is but no teacher could ever make me realize it up until now :/ i wonder if there is a playlist with these simple but essential rules. would love to see em all through.
This is exactly what I was looking for, thanks! the 1/x exponents I actually learnt about in economics for growth rates but I was curious about other fractional exponents, not presented here.
Might I add, you can have the roots/exponents in any order you want. They are both basically the same operation like division is backwards multiplication so a*b/c=a/c*b.
"2/3 is really 2 * 1/3" - This isn't true with powers, though what you continued on with was. EDIT: I misinterpreted what Iqtidar said, so this is wrong and he was right, I apologise. x^(1/3)^2 = x^(2/3), but x^2 * x^(1/3) != x^(2/3) (which was how I originally interpreted it).
Root by the denominator, power by the numerator (you can remember this because numerator sounds more like generator than denominator and generators generate power) 2^ 2/3 = 2^2 then take that, and cube root. Equation: x^(y/z) = x^y root z'ed
You explain very well how it works when raising to the power of a fraction when the numerator is 1, How about adding further explanations on the implications and effect when you are raising to a fractional power and the numerator is not 1.
I think he was just doing basics for the sake of the students so they can understand this first before getting into the more complex stuff but it's basically the numerator will be a power and the denominator will be a root so in your example it would be the 7th root of 3^5
This is actually starting to come back to me. I also remember learning a version where the square roots were two different fractions, and I think it's also possible for the square root to be fraction. I was hoping he'd get into those more complex aspects.
Yeah, if you have a fractional root, it's like this: n'th root of x = x^(1/n) so, one-half ' th root of x = x^(1 / 1 / 2) and 1 / 1 / 2 = 2 (because there are two halves in one whole. In general, dividing by a fraction is the same as multiplying by the flipped fraction. Like, dividing by 2/3 is the same as multiplying by 3/2.) So, one-half ' th root of x = x^2. This also works for general fractions: (a/b)'th root of x = x^(1 / a / b) = x^(b/a) Additionally, with some further math, x^(b/a) can be written as the a'th root of x^b. So, (a/b) ' th root of x = a'th root of x^b . Example: (3/2) ' th root of x = 3rd root of x^2 (or "cube root of x^2) Another example: (3/2) ' th root of 8 = the 3rd root of 8^2 = the cube root of 64 = 4.
@@captaincaseyvids8579 Basically, you have the bth root of said number to the power of a. Or vice versa, (cube root (5))^3 = 5 = cube root (5^3). You just apply the same reasoning here. Say for simplicity 2^(3/4). We don't now what it is, but we know how to manipulate it based on the "rules" for exponents (they're on the left side of the board if you need a refresher, and technically they're just short cuts based on taking exponents and turning them into sequential multiplications). So, we already know that 2^(1/4) = 4th root of 2, and need to turn that 1 into a 3. By the rules, either we have (2^3)^(1/4), or (2^(1/4)) ^3, either way it is 2^(3 * 1/4) = 2^(3/4). So 4th√(2^3) or (4th√(2))^3. Or generally, for X^(a/b) = bth√(X^a) = (bth√(X))^a
When he said _let's put some other crazy number here like ..._ I thought ... he's going to say _i_ ! He's actually going to show us what the _ith_ root of a number means! 🤣
Been knowing you for a while now, but I've never actually watched your videos (College student right here). Now I'm watching them because it's possible I'll become a high school math teacher pretty soon... and I want to learn how to teach :) Learning from the best! Very good approach. Thank you a lot sir!
A conceptual step for fractional indices is; The fraction of the (identical) factors. So for a = 3 x 3 x 3 x 3 x 3, then a^(4/5) means 4/5 of the factors, ie 3 x 3 x 3 x 3
At 6:20 Eddie makes a mistake by equating the ten's root of seven to seven. It is understood he did not mean that, but the students must be very confused after that.
wow, I really needed a refresh. I haven't had any math class since the start of learning virtually and totally forgot how fractional exponents worked. I appreciate it :) thanks
there's actually a shortcut here... you multiply the numerator of the exponent to the whole number, while the Denominator will serve as the root of the product of the numerator and the whole number if I'm wrong, by all means, correct me.
6:47 "true because we said so" was how I was though at high-school. This is the first time I seen it explained so intuitively so I'll never have to remember where does what go in m/n over X. Once you understand something you don't have to rely on memory because you'll always be able to fall back on logic. The more logic you accumulate the easier is gets. Got one anecdote from when I was around 8 years old and we learned about division with leftover. They never explained what "leftover" part was: the stuff left in your hand or the stuff you are missing to hand out so everyone is satisfied. For example 33/7 is 4(+5) but for me it was 5(+2). I need two more so there's no leftover to hand out. Lol I got the lowest grade on the test in class but it was the most valuable grade I got in my life. Because from then on I learned the importance of understanding rather than memorizing. Since then I was top guy in math class throughout my education and loved math most of the time. It's all about the teacher. Until, many many years later, I learned from Numberphile that 1+2+3+... = -1/12. Yeah, fek math after that :)
That's a pretty cool understanding, actually! "I need two more" is the same as saying "I have negative two left over." You could perfectly say 5 with remainder -2. I'd give credit for that if I were the teacher grading. Also, the -1/12 thing isn't a true equality. It's just a cool kind of thing you can work out with algebra, but it's not an actual value. It's sorta like a "glitch in the system"; it doesn't really matter, it's just cool. So I wouldn't let that deter you from learning interesting math! I'm glad you liked the video :)
@@VikeingBlade The answer does come useful in a few situations with physics where summing an infinite number of possibilities sums to a negative possibility less than 1 which you can approach using the same sort of technique, but it's actually for a very different reason to their abuse of a divergent series, in physics the possibilities interfere so multiple similar but out of phase possibilities can be less likely than any single possibility in the sum.
Great explanations, great to follow along. I am puzzled by the audience you teach it to though. What is the square root of 100? 50!, what is 4x4x4? 54 ! 🤷♂️ it would appear the stuff you teach is way beyond their current level of comprehension. Did you walk into the wrong classroom?😉
Tas That one is doable but a little harder. It comes down to the fact that irrational numbers, at their core, are defined by which rational numbers are to their left and which to their right. So you can just say what numbers your result is between: 2^pi is more than 2^(31/10) but less than 2^(32/10). And you can do that until the indices get infinitely close, and then you know where your result is.
it depends If you are just interested in a numerical result you could also use logarithms to "bring down" the exponent. a ^ x is the same as eg. e ^ ( x * ln ( a ) ) you can calculate that for a numerical result or if this happens in some mathematical derivation continue with that. But of course: if you haven't learned about logairthms yet, there still is something waiting to be learned.
Wut how else do you do your multiplication? That’s like the only way to do it. Unless I’m really dumb and don’t know some other way, pretty sure the only way is breaking it down
Btw, the "break it down and add parts) is really just (a+b) * (c + d) = ac + ad + bc + bd (in any order, ofc) but with the occasional trivial steps removed, and some turbo nesting simplified. The above rule is itself just x(c+d) = xc + xd, applied recursively (x = a + b -> xc + xd = (a+b)c + (a+b)d = ac + bc + ad + bd. For example, 46 * 4 = (40 + 6) * (4 + 0) = we skip this step excluding the trivials [40*4 + 40*0 + 6 * 4 + 6 * 0] = 40*4+6*4. When we have multiple digits on both, it is just the longer version (23 * 16) = (20 + 3) * (10 + 6) = 6 * 3 + 6 * 20 + 10 * 3 + 10 * 20 [this is literally the same order you learn to do in school when you put the 16 under the 23 and do the "shift 1 to the left" trick which is really making the writing more clear ignoring the zeroes. When you get to something with more digits, you're just doing the longer x(c+d+e+...) where x=(a+b+...) version of it in a simplified way, but it is still the same thing.
@@louisvictor3473Here are two examples that show how I was shown: 46*4 = 40*4 + 6*4 = 160 + 24 = 184 Or 46 * 32 = 40*32 + 6*32 = 1280 + 192 = 1472 Just seems simpler the way I was shown to make math easier to do. Just break up one of the numbers so it’s a multiple of ten, multiply that and the part you took off by the other number and add them back together. Nothing fancy. Cheers
@@Ender7j Exactly. When you break one of those apart, you're basically doing x(a+b), but faster. Just pointing out where those methods come from, and why they're mathematically sound.
Couple days ago I was watching Taylor Series, Euler's identity, linear differential equations, and matrices; Today I ended up here with Fractional indices.
Not fair,, he should be a teacher in some nerd school..because I think the students in his class should discuss genuine ideas which will make it more interesting. And believe me, in India we lack these kind of teachers. There are, but not many...
At 4:00 (and just after 10:00), "three times" should have been "twice" (or "two times", if you prefer). Such a good teacher deserves far smarter students, but I suppose such slow students need a really good teacher. I strongly suspect that if Mr. Woo were teaching in Malaysia, his students would shout out in unison the correct answers to his questions.
@@crystallizard5867 Did you listen to the answers to his questions? Did they strike you as being the product of quick thinking and astute reasoning? Did you notice how few students were even offering answers? It may seem mean-spirited to point out that the class, as a whole, was not particularly focused or interested, but that doesn't make it any less true.
Everytime I watch one of his videos, I have difficulties understanding NOT what he teaches, but the age of the ones he's teaching to. They all have 16/17 y/o voices, but can't even make 3+3
The square root of 100 is also negative 10 (-10). Okay,......he brings that out and the class is lost. Imagine asking the class to write down 3 to the power of two-thirds (2/3).
If there isn't already, there needs to be a Part 2 to this, to examine fractional indices where the numerator isn't 1. For example, what does 5 to the power 2/3rds mean? I know the answer, just saying that it would be useful for those learning to be able to explore the next step. :-)
@@bayon8291 So if we're at the point where we understand what 5^(1/3) is, we can just use the exponent rules again. As we know, multiplying two numbers with the same base involves adding the powers, but this can equally be applied "in reverse". We can think of 5^(2/3) as being 5^(1/3 + 1/3). Applying the "addition of powers" rule, we can see that this must be the same as 5^(1/3) x 5^(1/3). Now we have something (5^(1/3)) multiplied by itself, in other words, we have 5^(1/3) all squared. In the video, around the 3:00 mark, Eddie looks at a similar situation but in that case, he uses it to show that 5^(1/3) is essentially the cube root of 5. That's because in that example, 5^(1/3) is being multiplied by itself not once but twice, so there is a "chain of three" and we are cubing it. That is why 5^(1/3) must be the cube root. In our case we're multiplying 5^(1/3) by itself just the once, i.e. squaring it. So 5^(2/3) can be thought of as the square of the cube root of 5. However, there's another interesting application of the exponent rules here. One of the rules says that if we raise a power to another power, we multiply the powers. So, (x^y)^z = x^ (yz). But multiplication is commutative, so (yz) = (zy). We can arrive at the number (2/3) by multiplying 2 and 1/3, but the order doesn't matter. 2 x 1/3 is the same as 1/3 x 2. Therefore, we could get to a power of 2/3 by having something raised to the power 1/3, and then squaring it (in which case the powers multiplied would be 1/3 x 2) or, we could just as well square something first, then take its cube root (in which case multiplying the powers gives 2 x 1/3). The take away here is that both approaches give the same result. So, whilst a power of 2/3 can be thought of as the square of the cube root (as we saw above), it can just as equally be thought of as the cube root of the square. That's a beautiful result, and it means that we have at least two ways of approaching how to evaluate (if that's what we want to do). For example, let's look at 64^(2/3). We can either work out the square of 64, and then take its cube root, or we can take its cube root, then square that. We should get the same result. Now I don't know about you but I'm not particularly keen on working out 64 squared. It's not *too* difficult but it's a bigger number than I need to work with, and then I have to think about how to work out a cube root of that big number. Not impossible but not too obvious. Instead, I may know, or can reasonably easily work out that the cube root of 64 is 4 (since 4 x 4 x 4 = 64). Now, I can simply square that to get my answer, 16. If I *had* done it the first way, I would have had 64 x 64, equalling 4096. Then I'd have to work out the cube root of 4096, which is indeed 16 (since 16 x 16 x 16 = 4096). If our numbers are nice and friendly then trial and error may work, but again there's no simple way to do this without a calculator or using some other processes so its much easier to calculate the cube root of 64 instead of 4096. In summary, if you have a number x raised to a fractional power (y/z), the result is equivalent to (i) raising x to the "yth" power, then taking the "zth" root of the result; or (ii) taking the "zth" root of x, then raising the result to the "yth" power. Hope that helps!
I know this is very late but think of it like the other rule where (a^m)n=a^mn but go in reverse so picture 2/3 as the mn. You can split 2/3 as 2×1/3 and and that basically makes it (5^2)1/3 which is 25^1/3 so 5^2/3 =25^1/3. So basically you raise it by the numerator and then root it by the denominator.
How does he get kids to care and participate? I had good teachers and still 80% of the class were clowns that NO teacher would ever get motivated to learn and participate.
Why do kids think addition when we talk about exponents??? "What number, when multiplied by itself, gives you 100?" "50" 🤦♂️ "Multiply. Multiply. Multiply!"
One interesting fact that students should definitely be careful of is that while (a^m)^n = a^(m·n) is always true when m and n are integers, this is not so when they are rational non-integers, and so the order in which you exponentiate matters. For example, (x^2)^(1/2) = |x|, because x^(1/2) is typically defined in such a way that it is never negative. Therefore, even if x is negative, (x^2)^(1/2) is not. So the exponents will cancel one another, but the sign of x is changed if x < 0, because ^(1/2) can never return a negative value by definition. However, if you change the order of the exponents, you get [x^(1/2)]^2 = x, which is a different result than |x|. Once you begin to deal with exercises where x is an arbitrary complex number, it only becomes more complicated and the difference becomes more noticeable. This is why one should be careful when explaining fractional exponents. The convention is to define x^(m/n) as [x^(1/n)]^m and not as (x^m)^(1/n) because it leads to more consistency.
![KAYTRANADA - Witchy (feat. Childish Gambino) [Official Video]](http://i.ytimg.com/vi/jGK3YVmGZ3Y/mqdefault.jpg)
Eddie: what should i multiply this by?
Student: 7
Eddie: how bout no
Another Student: 5
Eddie: YES!
7 is not a creative number.
I've noticed the same guy always saying 7.
It was more due to him wanting to keep the lesson simple and if it were done with 7, it would have been complicated and requires this lesson to be complete before they could do the math for that. And the other student suggested by itself not just 5.
OmniscientWarrior ll
refunct and portal 2 gamer gang
I wish my math teacher was this enthusiastic
same
Mine's just depressed
lmao mines a heavy drinker
I’m lucky enough to have a pretty good one
i have decent math teacher who is really calm, helpful but it ends up with people underestimating him
I really want to know what age group he is teaching
@@saahilmehta143 yea bro, I'm from India too
@@saahilmehta143 bhai jaisa in bachon ka haal hai lg nhi rha hai inko roots aur cubes ka pta hai. Aur india mei square wagera 6-7 mein kara te hain.
@Saahil Mehta ye here in the uk its taught in year 9- 10. They sound like maybe college or uni
In my school we are learning this when we are 13
Saahil Mehta in iraq we taught that at. 7th grade
You know he means business when keeps his calculator in a case.
To be fair most scientific calculators come in a case. Though granted not as fancy as his.
Asian teacher with a calculator, that's disappointing
@@asphalt325 yikes dude
he cares about his most powerful tool yo
@@asphalt325 bruh
He makes these lessons so interesting that although I know the stuff he teaches I still watch it for entertainment...
I feel that, I’m cutting into my geography revision because I’m watching this. Productive Procrastination™️
Trademark that, just in case.
Me too 😅, damn so much for a revision
@@mofumofu512 Why procrastinate with entertainment when you can procrastinate productively?
This guy is such a great teacher, you can see how he engages with the class to be a guide and helper in a journey of discovery. Not some remote pedagogue, he gets alongside them. He loves his subject, and wants us to love it too. Exceptional.
It's a nice demonstration but he does the discovery instead of the students. Not this spectacular, but more beneficial to let the students do the thinking.
@@kalmanbekesi5980 The spectacular fact is that he is minority.
Why do kids think addition when we talk about exponents???
"What number, when multiplied by itself, gives you 100?"
"50" 🤦♂️
"Multiply. Multiply. Multiply!"
Likely because of the fact that if you have the same base, you just add the exponents to each other, eg: 2^2×2^4 = 2^(2+4) = 2^6
im sorry mr terence tao
I actually knew a kid who tried to convince me that 5q - q = 5. Yes, really. You had to say 1q, even though 1q is the same as q and we don't say 1.
I imagine it's because they are learning something new
That happens when you are not used to exponents yet
7:19 fifty.
Teacher: MULTIPLY
Every math teacher ever: "Let's just use whatever number I think of... Could be anything... Anything at all." *inserts very specific number to illustrate the lesson*
Which is a useful skill in higher math, but that's a bad way to teach it:
One of the most useful tools when dealing with complicated algebra is the knowledge that as long as you multiply both sides of the equation by the same thing, you can multiply them by *anything*. So, just like the teacher in the example, you pick something *useful*. This concept is tangentially useful later with substitutions as well.
Of course, the trick is being able to tell what is going to be useful to you in a given problem...
@@treyslider6954 Maybe don’t multiply both sides by 0 but yeah.
@@EpicWayWay I mean, it's still valid. 0*A = 0*B for all A & B (but yeah that one probably isn't useful...)
@@treyslider6954 to multiply both side by 0 mean u are dividing 0/0.
The reason u can multiply both side by any number its because they would divide by themselve. But u cant divide 0/0
How about we use the complex number (π + ei)
8:35
I really was surprised, that they got 4*4 correct on the first try
Ikr
I StM I huh why
@@brandonm7952 It's ironical. It's a very basic thing, but those students look really bad and stupid. So he was making fun of them.
@@IStMl rude much? but what am i to expect on yt comments section?
@@IStMl I wouldn't be so condescending if I were you, considering the hypocrisy of your entire statement... you just said "ironical", which isn't even a word... plus I don't think you know the meaning of ironic lmao. The word you're looking for that describes the joke is "hyperbolic"... so I don't think you're one to ridicule others' academic abilities.
Why are all the best teachers in RUclips and not in actual classrooms?
He is in an actual classroom.
Oh he is in an actual classroom but just not yours or mine..
And you only see those 10min videos when your in the mood so you're automatically more likely to listen
I'm in a classroom and I'm awesome.
Tookoko he’s enthusiastic and makes it sort of fun, there is a difference.
I couldn't concentrate knowing that he forgot ^10 on the 7^(1/10)
I could, thanks to ADHD, but I also never forgot that it was a thing and had focus on it as well, also thanks to ADHD.
Me neither
He didn't though, he wrote it out correctly
Edit: never mind I see it now, had to watch it like 5 times to see it
But by the method he got to =7 I.e. by multiplying the exponents, you do get 10. That’s what I don’t get. You get two different answers from two different methods.
@@Praneettigga No, you must have made a mistake with your arithmetic. The exponents are 1/10 and 10, and they are being multiplied. 1/10·10 = 1, and the base is 7, so you get 7^1 = 7, which is indeed correct.
I really enjoy your videos! Keep up the great work! Love your enthusiasm and your style of breaking down and trying to show things in a simple manner. This is how most math teachers should be.
I love how you teach them I learned so much just by watching a few of your videos
The fact that I understand all of this with almost no effort even after my grades in maths 15 years ago were rubbish is absolutely astounding
This isn’t hard at all
For what grade is this?
@@munadoesstuff3989 year 7 or 8. This is pretty basic math but its part of the fundamentals you need to know for more advanced topics
No, this stuff is very easy. I understood it even before fully watching the video as I could figure ithe rest out myself in a few seconds.
I love the personality and energy, and the wording is great for students who "get" maths, but there are clearly students that haven't grasped previous concepts (such as roots that are anything but square roots). Also, it's never specifically stated that a^1/X is the x root of a, and by the end some students have clearly missed that and still think it's magic.
if only my teachers had 1% of this guy ability and pleasure to teach
Needed to brush up on this exact thing for reasons. This guy is a really great teacher.
“Is the 3 there because there’s three fives” that killed me
I literally wanted to leave the video lol
@@Osirion16 crazy that people aren't born knowing this shit. So rude that a 13 year old asks a question because the topic is unfamiliar!
@@amosw766 I'm not saying that he was wrong for asking, I'm just pointing out the fact that if you know about the topic it's gonna make you cringe. Also, he wasn't so wrong since the 3 fives technically came from the 1/3 so it was actually kind of linked together anyways...
Being real here those were the types of questions I'd ask in my math class ;) except I'd reflect a bit more before asking ( Which isn't the right way to do it since there isn't a right or a wrong way to ask a question ! )
I mean you probably were once in that state. It’s always good to ask if you don’t understand
this is so amazing how simply this guy answers questions i have been thinking of for a longer while
Whait what kind of place do you teach in to have people not knowing the square root of 100 ?????
Pre school
Third trimester
The square root of 100 is 10.
Any year 8 class.
hey! it keeps happening no matter how old u are! sometimes 2x3 seems 5 LOL!
usually i get bored in maths class but this guy got me invested in watching the whole video in full screen-
Most people learn this stuff in the early years of HS, but I like Eddie because he explains the origins and concepts of everything he teaches. Like, he tells us *why* things are the way they are, which is the main way I remember things so it's super helpful (I find it harder to cram stuff).
I think everyone finds cramming stuff more difficult. Understanding is a great aid to memory.
Thanks Eddie, needed some help visualizing rational exponents :)
I'm going into an engineering degree and I've done real well at calculus. Not sure why I'm watching but this is a good teacher in action
Also at about 6:50 it is written that
10root7=7, it's nit picking but it'll be counted as wrong in an exam
yea i think he meant to put (10root7)^10=7
This was exactly the thing that kept making me stumble but oh my you explained it so easily
I love how often you say "Get your calculator out". Back in my day they'd say "No calculators, you need to know how to do this in your head, You won't always have a calculator with you"
Michael Lanman ok boomer
Phones: Let me introduce my self!
@@jeremiahbarron2158 damn you just slayed him
@@jeremiahbarron2158 Allow me to be the party pooper here, this is not an okay boomer moment for 2 main reasons:
1- It is not a boomer thing, it happened to Millenials and to some Gen Z's and is still happening in many countries to Gen Z's in school....sooooooo....yeah, not a boomer thing, he is not talking about the time people had to go several miles to make a phone call or something.
2- okay boomer doesn't apply here cause the guy is not criticizing anyone nor glorifying "boomer's days"...so the whole foundation and context of using the "okay boomer" card are non-existent.
PS: waiting for your reply to this with "okay boomer"
But notice that he only does that to show that the math adds up and remove the probability of error. Otherwise, they have to do this in their heads.
Sir You teach in an excellent way
Thanks alot Eddie!! You are one of my best teachers in life
isnt it funny how simple and logical this is but no teacher could ever make me realize it up until now :/ i wonder if there is a playlist with these simple but essential rules. would love to see em all through.
8:25 Always find a way to bring old Ramanujan into the class room!
love your lessons, making math very interesting
Which means:
X^(a/b) = b root of X^a
10^(3/4) = 4th root of 10^3
hypercube root
This is exactly what I was looking for, thanks! the 1/x exponents I actually learnt about in economics for growth rates but I was curious about other fractional exponents, not presented here.
@@AdrianAbdel The general case was missing from the video :-( So i wrote it for everyone... :-)
Bedloe Did you really need someone to extrapolate the general case for you ?
Might I add, you can have the roots/exponents in any order you want. They are both basically the same operation like division is backwards multiplication so a*b/c=a/c*b.
This man is actually amazing 😘
Wait I learned this in 7th grade
I mean, this looks to be the very first introduction to Exponential / logarithmic function
s, so it's normal to do his stuff again in 10th grade
An amazing math teacher that I've ever seen.
Just a question, how do you do numbers to fractions like 2/3
Like how would you do 3 to the power of 2/3
3^2/3 would be equal to the cube root of 3^2, the denominator is the root and the numerator is the power.
2/3 is really 2 * 1/3
so it would be cube root squared
((x)^1/3)^2
"2/3 is really 2 * 1/3" - This isn't true with powers, though what you continued on with was.
EDIT: I misinterpreted what Iqtidar said, so this is wrong and he was right, I apologise. x^(1/3)^2 = x^(2/3), but x^2 * x^(1/3) != x^(2/3) (which was how I originally interpreted it).
The denominator (in this case, 3) is the root. The numerator (in this case, 2) is the power! So this will be cube root of 3, and then square it
Root by the denominator, power by the numerator
(you can remember this because numerator sounds more like generator than denominator and generators generate power) 2^ 2/3 = 2^2 then take that, and cube root.
Equation: x^(y/z) = x^y root z'ed
Excellent teacher with enthusiasm to teach
Thank you so much Eddie I'm so thankful for this video. My maths Teacher is horrible at teaching that. Keep it up man. Thanks heaps, I get it now 👍🏽
You explain very well how it works when raising to the power of a fraction when the numerator is 1, How about adding further explanations on the implications and effect when you are raising to a fractional power and the numerator is not 1.
You have the discriminate be raised to the power of the numerator. In the case of the numerator being 1, it’s implied to raise it to the 1st
I was hoping he would do a crazy fraction too like 3 to the 5/7 power
I think he was just doing basics for the sake of the students so they can understand this first before getting into the more complex stuff but it's basically the numerator will be a power and the denominator will be a root so in your example it would be the 7th root of 3^5
everyone gangsta till eddie woo pulls up without monetizing his vids
when he realized no one in the class could do an elementary cube root in their heads, he must have been like "wow I just wasted my day..."
Teachers aren’t like that
I feel bad for the guy who asked why Eddie is multiplying the cube root of 5, 3 times. He's so lost.
Where was Mr Woo when I was at school? I would have been a Maths genius by now
I've been wondering this exact thing for 10 years, wish they'd taught this at my high school.
My school would become heaven if i see him at my school..
Love from India
That teacher is teaching where?
Edit: Cherrybrook Technology High School
I love that you hear the class talking at the start (normal high school), and then it eventually becomes raptured silence. He makes learning a joy.
This is actually starting to come back to me. I also remember learning a version where the square roots were two different fractions, and I think it's also possible for the square root to be fraction. I was hoping he'd get into those more complex aspects.
Yeah, if you have a fractional root, it's like this:
n'th root of x = x^(1/n)
so,
one-half ' th root of x = x^(1 / 1 / 2)
and 1 / 1 / 2 = 2 (because there are two halves in one whole. In general, dividing by a fraction is the same as multiplying by the flipped fraction. Like, dividing by 2/3 is the same as multiplying by 3/2.)
So, one-half ' th root of x = x^2.
This also works for general fractions:
(a/b)'th root of x = x^(1 / a / b) = x^(b/a)
Additionally, with some further math, x^(b/a) can be written as the a'th root of x^b.
So,
(a/b) ' th root of x = a'th root of x^b .
Example:
(3/2) ' th root of x = 3rd root of x^2 (or "cube root of x^2)
Another example:
(3/2) ' th root of 8 = the 3rd root of 8^2 = the cube root of 64 = 4.
Do you mind explaining what happens if we have a/b as a power as opposed to 1/n?
@@captaincaseyvids8579 Basically, you have the bth root of said number to the power of a. Or vice versa, (cube root (5))^3 = 5 = cube root (5^3). You just apply the same reasoning here. Say for simplicity 2^(3/4). We don't now what it is, but we know how to manipulate it based on the "rules" for exponents (they're on the left side of the board if you need a refresher, and technically they're just short cuts based on taking exponents and turning them into sequential multiplications).
So, we already know that 2^(1/4) = 4th root of 2, and need to turn that 1 into a 3. By the rules, either we have (2^3)^(1/4), or (2^(1/4)) ^3, either way it is 2^(3 * 1/4) = 2^(3/4). So 4th√(2^3) or (4th√(2))^3. Or generally, for X^(a/b) = bth√(X^a) = (bth√(X))^a
@@louisvictor3473 Much Thanks!
School made me remember the rules. This guy shows where the rules come from. I wish I had a teacher half as good as him.
I was always sarcastic around teachers, but this guy I'd be so interested I wouldn't try anything with him.
You were a terrible person hahaha
At time index 6:18 10th root of 7 is equal 7 . Small mistake but in general very good lesson.... thanks
What grade are you teaching? I would love to be in your class. Amazing tutoring bro!
probably like 6th
When he said _let's put some other crazy number here like ..._
I thought ... he's going to say _i_ ! He's actually going to show us what the _ith_ root of a number means! 🤣
Been knowing you for a while now, but I've never actually watched your videos (College student right here). Now I'm watching them because it's possible I'll become a high school math teacher pretty soon... and I want to learn how to teach :) Learning from the best! Very good approach. Thank you a lot sir!
It's interesting how these less intuitive laws are derived as a consequence of fundamental laws.
I already knew this...thought he might go more in depth conceptually. Oh well, better luck next time.
me too. I know it's the same as the roots, but I was interested in knowing how roots are calculated
A conceptual step for fractional indices is; The fraction of the (identical) factors. So for a = 3 x 3 x 3 x 3 x 3, then a^(4/5) means 4/5 of the factors, ie 3 x 3 x 3 x 3
Basically i guess u take the root of that number which is in denominator and keep the number in the space on which it has the power as a fraction
I wish I had this guy as a math teacher.
How hold are they ?? I've learned that when I was 14
13*
Too old.
@@ayushi26 Why are you correcting him? If he learned it at 14 he learned it at 14
@@Fillster im pretty sure it was a joke
At 6:20 Eddie makes a mistake by equating the ten's root of seven to seven. It is understood he did not mean that, but the students must be very confused after that.
wow, I really needed a refresh. I haven't had any math class since the start of learning virtually and totally forgot how fractional exponents worked. I appreciate it :) thanks
there's actually a shortcut here...
you multiply the numerator of the exponent to the whole number, while the Denominator will serve as the root of the product of the numerator and the whole number
if I'm wrong, by all means, correct me.
6:47 "true because we said so" was how I was though at high-school. This is the first time I seen it explained so intuitively so I'll never have to remember where does what go in m/n over X. Once you understand something you don't have to rely on memory because you'll always be able to fall back on logic. The more logic you accumulate the easier is gets.
Got one anecdote from when I was around 8 years old and we learned about division with leftover. They never explained what "leftover" part was: the stuff left in your hand or the stuff you are missing to hand out so everyone is satisfied. For example 33/7 is 4(+5) but for me it was 5(+2). I need two more so there's no leftover to hand out. Lol I got the lowest grade on the test in class but it was the most valuable grade I got in my life. Because from then on I learned the importance of understanding rather than memorizing. Since then I was top guy in math class throughout my education and loved math most of the time. It's all about the teacher.
Until, many many years later, I learned from Numberphile that 1+2+3+... = -1/12. Yeah, fek math after that :)
That's a pretty cool understanding, actually! "I need two more" is the same as saying "I have negative two left over." You could perfectly say 5 with remainder -2. I'd give credit for that if I were the teacher grading.
Also, the -1/12 thing isn't a true equality. It's just a cool kind of thing you can work out with algebra, but it's not an actual value. It's sorta like a "glitch in the system"; it doesn't really matter, it's just cool. So I wouldn't let that deter you from learning interesting math! I'm glad you liked the video :)
@@VikeingBlade
The answer does come useful in a few situations with physics where summing an infinite number of possibilities sums to a negative possibility less than 1 which you can approach using the same sort of technique, but it's actually for a very different reason to their abuse of a divergent series, in physics the possibilities interfere so multiple similar but out of phase possibilities can be less likely than any single possibility in the sum.
Great explanations, great to follow along. I am puzzled by the audience you teach it to though. What is the square root of 100? 50!, what is 4x4x4? 54 ! 🤷♂️ it would appear the stuff you teach is way beyond their current level of comprehension. Did you walk into the wrong classroom?😉
What about irrational indices?
Tas That one is doable but a little harder. It comes down to the fact that irrational numbers, at their core, are defined by which rational numbers are to their left and which to their right. So you can just say what numbers your result is between: 2^pi is more than 2^(31/10) but less than 2^(32/10). And you can do that until the indices get infinitely close, and then you know where your result is.
If an irrational number can be written as an infinite series then you could do:
2^e=2^(2 + 1/2 + 1/6 + ... ) = 2^2 * 2^(1/2) * 2^(1/6) * ...
it depends
If you are just interested in a numerical result you could also use logarithms to "bring down" the exponent.
a ^ x is the same as eg. e ^ ( x * ln ( a ) )
you can calculate that for a numerical result or if this happens in some mathematical derivation continue with that.
But of course: if you haven't learned about logairthms yet, there still is something waiting to be learned.
So 3 methods out of the comments:
1. Approximate (frame it)
2. Use a series if possible
3. Give the exact result using logs
Who wishes Eddie Woo was your math teacher
the 10th root of 7 is not equal to 7
it is (10th root of 7)^10 is equal to 7
Nerd
Which is exactly what he taught
Such a complicated way to learn, just type the formula: x^a/b= b sqr(x^a) .
I haven’t seen someone do multiplication like that in a long time. That was how I was taught to do it, break it down and add the parts. Nice
Wut how else do you do your multiplication? That’s like the only way to do it. Unless I’m really dumb and don’t know some other way, pretty sure the only way is breaking it down
Btw, the "break it down and add parts) is really just (a+b) * (c + d) = ac + ad + bc + bd (in any order, ofc) but with the occasional trivial steps removed, and some turbo nesting simplified. The above rule is itself just x(c+d) = xc + xd, applied recursively (x = a + b -> xc + xd = (a+b)c + (a+b)d = ac + bc + ad + bd.
For example, 46 * 4 = (40 + 6) * (4 + 0) = we skip this step excluding the trivials [40*4 + 40*0 + 6 * 4 + 6 * 0] = 40*4+6*4. When we have multiple digits on both, it is just the longer version (23 * 16) = (20 + 3) * (10 + 6) = 6 * 3 + 6 * 20 + 10 * 3 + 10 * 20 [this is literally the same order you learn to do in school when you put the 16 under the 23 and do the "shift 1 to the left" trick which is really making the writing more clear ignoring the zeroes. When you get to something with more digits, you're just doing the longer x(c+d+e+...) where x=(a+b+...) version of it in a simplified way, but it is still the same thing.
@@louisvictor3473Here are two examples that show how I was shown:
46*4 = 40*4 + 6*4 = 160 + 24 = 184
Or
46 * 32 = 40*32 + 6*32 = 1280 + 192 = 1472
Just seems simpler the way I was shown to make math easier to do. Just break up one of the numbers so it’s a multiple of ten, multiply that and the part you took off by the other number and add them back together. Nothing fancy. Cheers
@@Ender7j Exactly. When you break one of those apart, you're basically doing x(a+b), but faster. Just pointing out where those methods come from, and why they're mathematically sound.
Couple days ago I was watching Taylor Series, Euler's identity, linear differential equations, and matrices; Today I ended up here with Fractional indices.
Btw, the 10th root of 7 is not equal to 7 :D
It says: [ 7^(1/10) ]^10
You multiply the exponents, and the fraction cancels out, and you're just left with 7^1, which is just 7.
theheroinfather yeah, he just forgot to add the ^10 part at the end
Not fair,, he should be a teacher in some nerd school..because I think the students in his class should discuss genuine ideas which will make it more interesting.
And believe me, in India we lack these kind of teachers. There are, but not many...
At 4:00 (and just after 10:00), "three times" should have been "twice" (or "two times", if you prefer). Such a good teacher deserves far smarter students, but I suppose such slow students need a really good teacher. I strongly suspect that if Mr. Woo were teaching in Malaysia, his students would shout out in unison the correct answers to his questions.
Do your parents love you? Do you love yourself?The correct answer is no, because nobody should love you after what you have just said.
@@crystallizard5867 Did you listen to the answers to his questions? Did they strike you as being the product of quick thinking and astute reasoning? Did you notice how few students were even offering answers? It may seem mean-spirited to point out that the class, as a whole, was not particularly focused or interested, but that doesn't make it any less true.
@@AnonimityAssured i'd say they were rather engaged but you're right, not the brightest. eg "whats the square root of 100"...
"50" ????
i remember doing this in year 8 lol
These students should stop talking nonsense and start listening.
Exactly, such a good teacher for such bad students
@@IStMl
Omfg get your head out of your ass
At least they're participating. I've never heard such an active class before.
At least they are not falling asleep or zoning out
@@inigo8740 Yes but they don't have the decency to quit the talking when the interesting teacher talks.
What I want to know, is what happens when it's 5^(2/3) or 7^(7/10) as examples. Basically, instead of 1/x, what happens when it's y/x?
3:10 seven
My best youtube teacher
Just realised I watched a whole video explaining what a cube root is... I finished A level maths what the hell am I doing here? lol
me 2 bruh.....
Everytime I watch one of his videos, I have difficulties understanding NOT what he teaches, but the age of the ones he's teaching to. They all have 16/17 y/o voices, but can't even make 3+3
6:20 - 10th root of 7 = 7? The bastard didn't fix that.
Yeah that bugged me too😂
Square root of 100 is +/- 10
I came here to say the exact same thing.
what year class is that
I also wanna know. Someone said sq. root of 100 is 50
Looks really dumb class tho
Ahsan Waleed Exactly
11 I think
The square root of 100 is also negative 10 (-10).
Okay,......he brings that out and the class is lost.
Imagine asking the class to write down 3 to the power of two-thirds (2/3).
The students seen to be atleast 15-16 and if this is how they are at that age, no wonder Asians seem legend to them.
They can't be 15-16.... Look at their silly answers
They don't know 64 cube root... They aren't 16
eddie:square root of 100?
sutdent: hmmmmmmmmmmmm 50?
eddie: ;-;
If there isn't already, there needs to be a Part 2 to this, to examine fractional indices where the numerator isn't 1. For example, what does 5 to the power 2/3rds mean? I know the answer, just saying that it would be useful for those learning to be able to explore the next step. :-)
So what is 5^2/3
@@bayon8291 So if we're at the point where we understand what 5^(1/3) is, we can just use the exponent rules again.
As we know, multiplying two numbers with the same base involves adding the powers, but this can equally be applied "in reverse".
We can think of 5^(2/3) as being 5^(1/3 + 1/3). Applying the "addition of powers" rule, we can see that this must be the same as 5^(1/3) x 5^(1/3).
Now we have something (5^(1/3)) multiplied by itself, in other words, we have 5^(1/3) all squared. In the video, around the 3:00 mark, Eddie looks at a similar situation but in that case, he uses it to show that 5^(1/3) is essentially the cube root of 5. That's because in that example, 5^(1/3) is being multiplied by itself not once but twice, so there is a "chain of three" and we are cubing it. That is why 5^(1/3) must be the cube root.
In our case we're multiplying 5^(1/3) by itself just the once, i.e. squaring it. So 5^(2/3) can be thought of as the square of the cube root of 5.
However, there's another interesting application of the exponent rules here. One of the rules says that if we raise a power to another power, we multiply the powers. So, (x^y)^z = x^ (yz).
But multiplication is commutative, so (yz) = (zy). We can arrive at the number (2/3) by multiplying 2 and 1/3, but the order doesn't matter. 2 x 1/3 is the same as 1/3 x 2.
Therefore, we could get to a power of 2/3 by having something raised to the power 1/3, and then squaring it (in which case the powers multiplied would be 1/3 x 2) or, we could just as well square something first, then take its cube root (in which case multiplying the powers gives 2 x 1/3).
The take away here is that both approaches give the same result. So, whilst a power of 2/3 can be thought of as the square of the cube root (as we saw above), it can just as equally be thought of as the cube root of the square. That's a beautiful result, and it means that we have at least two ways of approaching how to evaluate (if that's what we want to do).
For example, let's look at 64^(2/3). We can either work out the square of 64, and then take its cube root, or we can take its cube root, then square that. We should get the same result. Now I don't know about you but I'm not particularly keen on working out 64 squared. It's not *too* difficult but it's a bigger number than I need to work with, and then I have to think about how to work out a cube root of that big number. Not impossible but not too obvious.
Instead, I may know, or can reasonably easily work out that the cube root of 64 is 4 (since 4 x 4 x 4 = 64). Now, I can simply square that to get my answer, 16.
If I *had* done it the first way, I would have had 64 x 64, equalling 4096. Then I'd have to work out the cube root of 4096, which is indeed 16 (since 16 x 16 x 16 = 4096). If our numbers are nice and friendly then trial and error may work, but again there's no simple way to do this without a calculator or using some other processes so its much easier to calculate the cube root of 64 instead of 4096.
In summary, if you have a number x raised to a fractional power (y/z), the result is equivalent to
(i) raising x to the "yth" power, then taking the "zth" root of the result; or
(ii) taking the "zth" root of x, then raising the result to the "yth" power.
Hope that helps!
I know this is very late but think of it like the other rule where (a^m)n=a^mn but go in reverse so picture 2/3 as the mn. You can split 2/3 as 2×1/3 and and that basically makes it (5^2)1/3 which is 25^1/3 so 5^2/3 =25^1/3. So basically you raise it by the numerator and then root it by the denominator.
6:20 mistake. its 7^10 inside the square root symbol.
How does he get kids to care and participate?
I had good teachers and still 80% of the class were clowns that NO teacher would ever get motivated to learn and participate.
intuitive
why do his videos always end so abruptly? >_
Why do kids think addition when we talk about exponents???
"What number, when multiplied by itself, gives you 100?"
"50" 🤦♂️
"Multiply. Multiply. Multiply!"
(4/3)^-(4/3)^-(4/3) = ?
Solve details please
I can't believe I'm watching this right after I had my test on this.
8:56 hold on let me write it down:)
"whenever you're unsure, get your calculator out." - Eddie Woo
One interesting fact that students should definitely be careful of is that while (a^m)^n = a^(m·n) is always true when m and n are integers, this is not so when they are rational non-integers, and so the order in which you exponentiate matters. For example, (x^2)^(1/2) = |x|, because x^(1/2) is typically defined in such a way that it is never negative. Therefore, even if x is negative, (x^2)^(1/2) is not. So the exponents will cancel one another, but the sign of x is changed if x < 0, because ^(1/2) can never return a negative value by definition. However, if you change the order of the exponents, you get [x^(1/2)]^2 = x, which is a different result than |x|. Once you begin to deal with exercises where x is an arbitrary complex number, it only becomes more complicated and the difference becomes more noticeable. This is why one should be careful when explaining fractional exponents. The convention is to define x^(m/n) as [x^(1/n)]^m and not as (x^m)^(1/n) because it leads to more consistency.
I'm currently finishing up my diploma, could not have told you those were called indices.
I really hope his students appreciate how lucky they are