I got my degree in math about forty years ago and I love this channel. He goes over a lot of stuff that I took for granted for decades and I sit here and go "Wow! I never knew that." For example, I've known about sines and cosines since 10th grade trig BUT no one ever explained that cosine is shorthand for the complement of sine. Also, I now see the relationship between sine and sinuous (I'm fascinated by words and their origins.).
I like that if BX is negative, you can imagine X is inside the circle. Therefore the line PQ cuts the circle in two places, so it isn't a tangent... which is a contradiction because we defined it as such in the set up.
The fact still stands that BX is distance, not displacement. Since it is a scalar, not a vector, it has no direction. Without direction, it can't possibly have a negative magnitude.
I had a friend who was very interested in art, and one of the things he did almost every day was practice drawing circles, straight lines, squares etc. Not for hours, mind you, but just 5-10 minutes each day. He obviously got very good at drawing these basic shapes without a straight edge, ruler, or compass. Point is, if you want to have a skill, you have to practice it.
@@TheFirstNamelessOne It’s not false or true by absolute measures, it’s false in relation to the hypothesis. Example: Probably a bad one as I commented on his a while ago: hypotheses is that my car is red. Proof by contradiction starts with the assumption that my car has no color. We then show that my car must have color and further discover that that color could be red. That’s usually good enough to make the needed point. My car has a color. Red is an option. Then you just need a picture.
Outstanding video. I knew you would assume that it wasn't perpendicular, but I couldn't come up with how to formulate the contradiction from that. My favorite historical proof by contradiction is Euclid's proof of the infinitude of primes.
(I will use exactly the same figure as shown in the video) Instead of saying that Angle OAX < Angle OXA, we should make use of the properties of a Right Angled Triangle. We know that the side opposite to the right angle is the Hypotenuse (The Longest Side) So, OA > OB. The rest of the procedure is just the same. Is it correct to prove this theorem like this?
Yes you could but the topic is of circles. So using of radii instead of sides of a triangles would be appropriate. Nevertheless your method is right too.
You can, Except there is one hole here. The triangle is OAX. Therefore, according to the hypotenuse is longest side property, it should be OA>OX*, so you basically still arrive at the same point he arrived. Only he gave them in even more detail cos it's easier for them to understand, and he's covering all questions that some curious minds among them will have 5 years later.
Great video. Thought he was going to use Pythagoras Theorem after labelling points X and B. OA^2 = OX^2 + AX^2 OA^2 = (OB + BX)^2 + AX^2, but OB = OA therefore OA^2 = (OA + BX)^2 + AX^2 OA^2 = OA^2 + 2(OA)(OB) + OB^2 + AX^2 0= OA^2 + 2(OA)(OB) + OB^2 + AX^2 -2(OA)(OB) = OB^2 + AX^2 This is impossible because the sum of two squared number cannot be negative hence initial assumption must be wrong. Therefore tangent is always perpendicular to the radius of a circle it subtends. Question: If I started out by saying, "By Pythagoras Theorem, OA^2 is IDENTICAL to OX^2 + AX^2 (sorry, I can't use the three line equal sign) would line 3 be sufficient to support the contradiction ?
Lets see if it can be written as a fraction, that is assume: sqrt(2)=p/q where p and q are whole numbers with no common factors Squaring both sides: 2=p^2/q^2 2q^2=p^2 So p^2 is divisible by 2. But any even square number must have a square root that is also even. This implies p is also divisible by 2 and can be written: p=2k where k is a whole number Now going back a few steps we can substitute p=2k for p: 2q^2=(2k)^2 2q^2=4k^2 q^2=2k^2 And we see q^2 is divisible by 2 implying q is divisible by 2 using the same reasoning as above. But then p and q each have a factor of 2 which is in contradiction with our assumption. So the square root of 2 can not be written as a fraction, so it is not rational. It must be irrational. QED
@@shadow-cz3vr We want p/q in its simplest form. If p/q has common factors it is not in its simplest form. In fact, we could then simplify p/q into a simpler fraction by dividing top and bottom by the greatest common factor (gcf) possible. If p and q have common factors then p=ba and q=ca (where a,b,c are natural numbers and a is the gcf) then ba/ca=b/c. We could then do the proof with b/c instead of p/q to achieve the same conclusion. If we don't state this, the proof doesn't work because the common factor we find may have been there to begin with. I should also add that the proof works by assuming something and then showing that the assumption creates an impossibility. We can assume whatever we want and in this case we are assuming p/q is in simplest form so that when we find that there's a common factor it is in contradiction with our assumption.
Sir can't we simply write if angle OxA = 90 the side opposite to this angle is hypotnuse that shoud be longer than other two sides... But it is clear from figure OA is less than Ox..it is only possible when x comes at A
I continued with the supposition that BX can be negative which means that X is inside the circle but we know that X and A are in PQ and in the circle which means PQ isnt a tangent line because it intersects two points in a circle
To explain these things you need a simple example. So , say you need to turn a ship around in the harbor, and for the distance you move forward into your turn, given the room available, how much of a change in direction or sharpness of the turn must you do to get the boat turned in the given space without crashing into the other side of the harbor. So for every yard you drive the boat, you need an angular turn that succeeds in getting the boat turned around, or 180 degrees, the direction is changed. So then that is why they used the sine and cosine, the perpendicular and the tangent, to navigate a ship in the water to turn about or bow , making it curve and turn, by having a way to calculate the steering , before you begin to turn the boat around.
1 question, shouldn't be tan 90 must be undetermined since you see it would be infinity/0 (Since perpendicular will be infinite and base would be 0 at tan 90 degree)
I am confused with the description of the angle: for me angle OAX is the right angle. I have always known that the letter in the middle is the angle or is there another convention? Thanks.
We say angle OXA is is 90 degrees, based on the assumption that OAX is smaller than 90 degrees. The drawing disagrees, so you should not look there, but instead follow the logic.
Whizzer191 Thank you. However it is obvious that it is a label mistake ; Mr Woo writes that OX is perpendicular to PQ so A and X should be reversed on the drawing. And writing angle MNP means angle N and nothing else. Best regards.
Yves, there are no mistakes in this video. Eddie is deliberately starting with a false situation in order to prove that it's false. We know that angle OAX is 90 degrees but only because we are told so. Eddie is proving that this is true by showing that it can't be any other size of angle. If you're still confused then keep watching the video until the penny drops. I'm a maths lecturer and this video is much better than other proofs I've seen on this particular topic.
Hi Mike, thank you for your answer. Wow, I made a real fool of myself here. Of course the video is correct. I listened to it the other day whilst doing something else and basically focused on the contradiction, ignoring the first six minutes. Eddie's videos are brilliant and inspiring. Cheers, Yves
This is identical to another "proof" I just watched which is false/incomplete. Please identify the assumption which has been contradicted? It is not enough to draw a picture with the length Bx > 0, and then say because Bx < 0 we have a contradiction. What is the assumption which justifies the assertion that Bx > 0 as it appears in the picture?
But you didn't explain how the tangent trig function relates to the tangent of a circle. Your didn't explain why the tangent function is called the tangent.
I think you omitted one fact that is needed for the formal proof: that a tangent to a circle is never inside the circle (at any other point). Without this X could be between B and O, instead of B being between O and X.
Amazing , but I have an even easier explanation It takes into account the fact that the altitude is always the shortest path between a point and a given line on a plane. Here we take the tangent as the line, the radius as the point. As the tangent is just touching the circle, the shortest path is the radius (easy to see) hence it's perpendicular.
You can skip the bit about "smaller angles opposite smaller sides" just by saying that, since it's a right angle triangle, Pythagoras's theorem applies. Therefore c (the hypotenuse which is in this case OA) squared is equal to a squared plus b squared. Take a to be OX and b to be AX (although it doesn't matter which one is which) since you are adding b^2 to a^2 to make c^2. a (OX) must be less than c (OA). But this contradicts as OX = OB + BX and OB is equal to OA as they are both radii.
Wow wow wow the bigger an angle, the bigger is his opposite side ? That is a very costly assumption, I'm gonna need more than "it is a triangles thing" ! By the way, a simplest demo is that the circle is symmetrical. So OAP = QAO, but OAP = 180° - QAO, so OAP = QAO = 90°
Dont know if anyone replied to you about this... but how it goes is: by definition, Sine of an angle is equal to the ratio of "opposite side divided by the hypotenuse". Hence, sine Q would be b/c. by definition, Cosine of an angle is equal to the ratio of "adjacent side divided by the hypotenuse". Hence, cosine Q would be a/c. you can remember these ratios by using the mnemonic "SOHCAHTOA" which stands for "Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent"
tan is short for tangent. Tangent comes from Latin, tangere- to touch. A straight line can cut a circle at at two points of the circumference, In this case the line is called a secant ( from the Latin secare - to cut), or it can just touch at one point (tangent), or like the vast majority of lines miss altogether.
You can also say that OA is in front of 90 deg so its the largest line in the riangle, but OX is outside the circle meanong its larger than the radius, and we have acontradiction again
Drawing straight line is just easy. They have mastered it like if you do it over and over again, you are also approaching to perfection but no one is perfect, just approaching.Assymptote? lol 😂😂 MATH IS ❤❤ Proving by Contradiction is kind'a hard. You always assume correct as incorrect. Therefore you imagine that incorrrect premise and interpret it in a logical way like its correct and still doesn't make sense at the end 😂😂.
Hey everyone, there is a basic mistake he did in his derivation, if anyone of you could find out it then i will get satisfied that I'm not only intelligent person in this comment section. Pls satisfy me.
I can feel the love you have for learning and teaching maths. Keep it up sir
Man! He drew a STRAIT line from O to x intersecting the circle at b without a ruler! That's talent!
Actually, he drew a straight line.
tell me if you can make a line O to x without intersecting the circle though...
@@JTtheking134 hum, the point is that the line is straight, no one cares about the fact that it is intersecting the circle 😆
@@goldwinger5434 and Mike is in _dire_ straits.
@@warplanner8852 One could say he is a tangent 'Brother in Arms'...
Just look how straight that Ox line is with no ruler
Finally a math teacher that doesn’t start a lesson by saying: “Ok, this is going to be boring...”
No shit math ain’t boring
Ikr they need passion in it
true not boring
I got my degree in math about forty years ago and I love this channel. He goes over a lot of stuff that I took for granted for decades and I sit here and go "Wow! I never knew that." For example, I've known about sines and cosines since 10th grade trig BUT no one ever explained that cosine is shorthand for the complement of sine. Also, I now see the relationship between sine and sinuous (I'm fascinated by words and their origins.).
@Rusty Never really gave it any thought but origin of words was never covered.
You have a talent for teaching. Keep up the good job.
Outstanding teacher.
I like that if BX is negative, you can imagine X is inside the circle. Therefore the line PQ cuts the circle in two places, so it isn't a tangent... which is a contradiction because we defined it as such in the set up.
Genius
That is what I was thinking
The fact still stands that BX is distance, not displacement. Since it is a scalar, not a vector, it has no direction. Without direction, it can't possibly have a negative magnitude.
He answered every question I had in mind except how he drew such a perfect circle
Brilliantly explained!
5:37 just pointing it out
I wish I had this teacher so captivating and willing to explain.
I had a friend who was very interested in art, and one of the things he did almost every day was practice drawing circles, straight lines, squares etc. Not for hours, mind you, but just 5-10 minutes each day. He obviously got very good at drawing these basic shapes without a straight edge, ruler, or compass.
Point is, if you want to have a skill, you have to practice it.
I wish I had this guy teaching me maths in school I swear to god!
Kid coughs in 2015: carries on*
Kid coughs in 2020: everyone, get out get out!
First time, I am interested in Mathematics .
Super class
omg, he can make everything simple and easy, that's incredible!!!! congrats
This guy will get an OAM at some point in the future, I'm calling it now....
lovely explanation ...... this man is a maestro
Correction: In proof by contradiction, you don't start with a false premise. You start with a premise you are testing.
No you don’t!
@@HansWeberHimself Let's assume that you do
@@MadaraUchihaSecondRikudo Funny, but no. 😉
@@HansWeberHimself If you knew it was false, what's the point?
@@TheFirstNamelessOne It’s not false or true by absolute measures, it’s false in relation to the hypothesis.
Example: Probably a bad one as I commented on his a while ago: hypotheses is that my car is red. Proof by contradiction starts with the assumption that my car has no color. We then show that my car must have color and further discover that that color could be red. That’s usually good enough to make the needed point. My car has a color. Red is an option. Then you just need a picture.
Outstanding video. I knew you would assume that it wasn't perpendicular, but I couldn't come up with how to formulate the contradiction from that.
My favorite historical proof by contradiction is Euclid's proof of the infinitude of primes.
I gotta subscribe. I love your accent. You sound so fancy.
I don't know if you're Australian but your accent, sir, is impeccable. As a British I am intrigued :D Sounds much better than the American accent.
(I will use exactly the same figure as shown in the video)
Instead of saying that Angle OAX < Angle OXA, we should make use of the properties of a Right Angled Triangle.
We know that the side opposite to the right angle is the Hypotenuse (The Longest Side)
So, OA > OB.
The rest of the procedure is just the same.
Is it correct to prove this theorem like this?
Hmm interesting, I may agree with you.
Siddharth Varshney yeah and honestly its better. You could also use the fact that sin,cos < 1
Yes you could but the topic is of circles. So using of radii instead of sides of a triangles would be appropriate. Nevertheless your method is right too.
You can, Except there is one hole here. The triangle is OAX. Therefore, according to the hypotenuse is longest side property, it should be OA>OX*, so you basically still arrive at the same point he arrived.
Only he gave them in even more detail cos it's easier for them to understand, and he's covering all questions that some curious minds among them will have 5 years later.
what math book do you suggest using that has challenging problems ?
Great video. Thought he was going to use Pythagoras Theorem after labelling points X and B.
OA^2 = OX^2 + AX^2
OA^2 = (OB + BX)^2 + AX^2, but OB = OA therefore
OA^2 = (OA + BX)^2 + AX^2
OA^2 = OA^2 + 2(OA)(OB) + OB^2 + AX^2
0= OA^2 + 2(OA)(OB) + OB^2 + AX^2
-2(OA)(OB) = OB^2 + AX^2
This is impossible because the sum of two squared number cannot be negative hence initial assumption must be wrong. Therefore tangent is always perpendicular to the radius of a circle it subtends.
Question: If I started out by saying, "By Pythagoras Theorem, OA^2 is IDENTICAL to OX^2 + AX^2 (sorry, I can't use the three line equal sign) would line 3 be sufficient to support the contradiction ?
you are the best teacher ever, thank you
Man, ox is one straight line
11:38
Can someone pls explain how he get Bx
Since the radius OB has an equal length to the radius OA, for OB + Bx to be less than OA, that would mean Bx would have to be less than zero.
I don't think it's O, I think he wrote Bx < 0
Second line you drew looked as if you used a ruler 👌
Oh boy! I’m impressed of his straight line drawing ability
Amazing. Now what’s the square root of two?
Lets see if it can be written as a fraction, that is assume:
sqrt(2)=p/q where p and q are whole numbers with no common factors
Squaring both sides:
2=p^2/q^2
2q^2=p^2
So p^2 is divisible by 2. But any even square number must have a square root that is also even.
This implies p is also divisible by 2 and can be written:
p=2k where k is a whole number
Now going back a few steps we can substitute p=2k for p:
2q^2=(2k)^2
2q^2=4k^2
q^2=2k^2
And we see q^2 is divisible by 2 implying q is divisible by 2 using the same reasoning as above.
But then p and q each have a factor of 2 which is in contradiction with our assumption.
So the square root of 2 can not be written as a fraction, so it is not rational. It must be irrational.
QED
@@THEGLORYRISING Why can't p and q have common factors?
@@shadow-cz3vr We want p/q in its simplest form. If p/q has common factors it is not in its simplest form. In fact, we could then simplify p/q into a simpler fraction by dividing top and bottom by the greatest common factor (gcf) possible.
If p and q have common factors then p=ba and q=ca (where a,b,c are natural numbers and a is the gcf) then ba/ca=b/c. We could then do the proof with b/c instead of p/q to achieve the same conclusion.
If we don't state this, the proof doesn't work because the common factor we find may have been there to begin with.
I should also add that the proof works by assuming something and then showing that the assumption creates an impossibility. We can assume whatever we want and in this case we are assuming p/q is in simplest form so that when we find that there's a common factor it is in contradiction with our assumption.
a surd
1.414 🤷🏻
Sir can't we simply write if angle OxA = 90 the side opposite to this angle is hypotnuse that shoud be longer than other two sides...
But it is clear from figure OA is less than Ox..it is only possible when x comes at A
Thanks ! This has helped me a lot.
This man is good. Life made easy
You are wonderfully well, sir.
However, I need a tutorial on domain and range of trigonometry functiond
sir very nice iam big fan of you from india
I continued with the supposition that BX can be negative which means that X is inside the circle but we know that X and A are in PQ and in the circle which means PQ isnt a tangent line because it intersects two points in a circle
To explain these things you need a simple example. So , say you need to turn a ship around in the harbor, and for the distance you move forward into your turn, given the room available, how much of a change in direction or sharpness of the turn must you do to get the boat turned in the given space without crashing into the other side of the harbor. So for every yard you drive the boat, you need an angular turn that succeeds in getting the boat turned around, or 180 degrees, the direction is changed. So then that is why they used the sine and cosine, the perpendicular and the tangent, to navigate a ship in the water to turn about or bow , making it curve and turn, by having a way to calculate the steering , before you begin to turn the boat around.
what a BANGER video man
1 question, shouldn't be tan 90 must be undetermined since you see it would be infinity/0 (Since perpendicular will be infinite and base would be 0 at tan 90 degree)
I am confused with the description of the angle: for me angle OAX is the right angle. I have always known that the letter in the middle is the angle or is there another convention? Thanks.
We say angle OXA is is 90 degrees, based on the assumption that OAX is smaller than 90 degrees. The drawing disagrees, so you should not look there, but instead follow the logic.
Whizzer191 Thank you. However it is obvious that it is a label mistake ; Mr Woo writes that OX is perpendicular to PQ so A and X should be reversed on the drawing. And writing angle MNP means angle N and nothing else. Best regards.
Yves, there are no mistakes in this video. Eddie is deliberately starting with a false situation in order to prove that it's false. We know that angle OAX is 90 degrees but only because we are told so. Eddie is proving that this is true by showing that it can't be any other size of angle. If you're still confused then keep watching the video until the penny drops. I'm a maths lecturer and this video is much better than other proofs I've seen on this particular topic.
Hi Mike, thank you for your answer. Wow, I made a real fool of myself here. Of course the video is correct. I listened to it the other day whilst doing something else and basically focused on the contradiction, ignoring the first six minutes. Eddie's videos are brilliant and inspiring. Cheers, Yves
This is identical to another "proof" I just watched which is false/incomplete. Please identify the assumption which has been contradicted? It is not enough to draw a picture with the length Bx > 0, and then say because Bx < 0 we have a contradiction. What is the assumption which justifies the assertion that Bx > 0 as it appears in the picture?
But you didn't explain how the tangent trig function relates to the tangent of a circle. Your didn't explain why the tangent function is called the tangent.
David Adams
The Tangent cuts the Circle...I thought that was obvious?
That's the next video. He broke it up into two parts, and this was only part 1.
homie circles are pretty impressive
What class is this? If they've been learning maths for 12 years, what age are they?
jumpingjflash
They are either 17 and Genii or he meant a smaller number!
I think you omitted one fact that is needed for the formal proof: that a tangent to a circle is never inside the circle (at any other point). Without this X could be between B and O, instead of B being between O and X.
What grade are your students in?
D T mural
Amazing , but I have an even easier explanation
It takes into account the fact that the altitude is always the shortest path between a point and a given line on a plane.
Here we take the tangent as the line, the radius as the point. As the tangent is just touching the circle, the shortest path is the radius (easy to see) hence it's perpendicular.
You can skip the bit about "smaller angles opposite smaller sides" just by saying that, since it's a right angle triangle, Pythagoras's theorem applies. Therefore c (the hypotenuse which is in this case OA) squared is equal to a squared plus b squared. Take a to be OX and b to be AX (although it doesn't matter which one is which) since you are adding b^2 to a^2 to make c^2. a (OX) must be less than c (OA). But this contradicts as OX = OB + BX and OB is equal to OA as they are both radii.
Wow wow wow the bigger an angle, the bigger is his opposite side ? That is a very costly assumption, I'm gonna need more than "it is a triangles thing" !
By the way, a simplest demo is that the circle is symmetrical. So OAP = QAO, but OAP = 180° - QAO, so OAP = QAO = 90°
8:32 This itself is a proof by contradiction
Excellent!
Can someone explain why sin Q (theta) = b/c and not b/a or c/a ? and same goes for cos (90-Q) /
Dont know if anyone replied to you about this... but how it goes is:
by definition, Sine of an angle is equal to the ratio of "opposite side divided by the hypotenuse". Hence, sine Q would be b/c.
by definition, Cosine of an angle is equal to the ratio of "adjacent side divided by the hypotenuse". Hence, cosine Q would be a/c.
you can remember these ratios by using the mnemonic "SOHCAHTOA" which stands for "Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent"
Sos Cas Toa where s is hypothenusa, o is opposite side, a is near by side and S is Sinus, C is Cosinus, and T is Tangus soscastoa
Lol, previous commenter wrote it in English, didn't know it works in English too this little pony trick 😸😸
If you started watching this video at 6:50 you’d think this bloke was off his trolley 😅
So basically we could have just looked at line BX and automatically said that’s impossible
michael jordan nice proof
thank you sir
Wait so why is tan called tan? 😭
From tangent
tan is short for tangent. Tangent comes from Latin, tangere- to touch. A straight line can cut a circle at at two points of the circumference, In this case the line is called a secant ( from the Latin secare - to cut), or it can just touch at one point (tangent), or like the vast majority of lines miss altogether.
Very good!!!
You can also say that OA is in front of 90 deg so its the largest line in the riangle, but OX is outside the circle meanong its larger than the radius, and we have acontradiction again
•˚• = therefor?
It's in Unicode. U+2234 ∴ Therefore.
unicode-table.com/en/2234/
Drawing straight line is just easy. They have mastered it like if you do it over and over again, you are also approaching to perfection but no one is perfect, just approaching.Assymptote? lol 😂😂
MATH IS ❤❤
Proving by Contradiction is kind'a hard. You always assume correct as incorrect. Therefore you imagine that incorrrect premise and interpret it in a logical way like its correct and still doesn't make sense at the end 😂😂.
Very detailed in explaining... But pace is way too slow for me. I m a quick and impatient learner ;>. Nonetheless you are a great teacher.
FBI open the door!!!....Too smart
I don't understand why he says OAX is a cute angle. I actually find it rather unappealing...
It’s simply because you have a shallow understanding
wowwwwwww
7:11 done
QED.
Skip the acute angle stuff and just go straight to the hypotenuse of a right angle triangle.
we must clone him....
Will u teach me pleaseeeeeee
I like
180 degrees in a triangle? Yeah, if it is a flat triangle. Try drawing that triangle on the surface of a ball and then measure the angles.
is he writing on a spherical desk?
lol of course we are assuming the triangle is drawn on flat euclidean geometry
Hey everyone, there is a basic mistake he did in his derivation, if anyone of you could find out it then i will get satisfied that I'm not only intelligent person in this comment section. Pls satisfy me.
Are you ok ?
Most annoying thing ive ever watched