but i don't see how, when seeing a problem like this for the first time, you'd know to replace n^(2) by 2n + 1. I don't see how you would know to do that.
+IffyProjects I agree. I mean, logically, it makes sense, but I automatically assume I CANNOT deviate from the problem due to getting it wrong in school. In real life, yes I could maybe make the assumption, but I wouldn't want to assume anything outside the direction of the problem when quizzes or tests come into play. I'd like an answer as well. "HOW did you know to do this step?"
+IffyProjects The only reason he did it the way he did was to make the proof look neater. You can absolutely start with the original inequality without replacing anything, and you'll eventually complete the proof. As OP pointed out in the first couple of minutes of the video, you'll arrive at a problem and have to take a step back solving the problem first.
In induction, you check first what you have to proof. Like 2^{k+1}>(k+1)²=k²+2k+1 Now to proof 2^k+2^k>k²+2k+1 Reduces down(by hypothesis) 2^k>2k+1 If you proof this, you're done! It has been 5 years since you posted. I hope so you have grown and could find it by yourself!
@@TechToppers I have been working on this problem for the past hour or two and you explained it so so well in one small comment. Thanks a ton just had my Eureka moment have a nice day
@@FSHnegativ Thanks a lot. But please remember, that the I wrote was just for intuition. Backtracking in mathematics is generally hard. You have to be very sure that your assumptions are correct. In which grade you are?
@@TechToppers im currently in the first semester of uni majoring in data science. Haven't had induction proofs in high school so I'm doing my best learning it as fast as possible. I actually just got done with the whole proof (n^2
If anybody was wondering why 2^n > n^2 becomes 2^n * 2 > n^2 * 2 is because in order to get 2^n+1 you need to multiply 2^n by 2 (hence 2^n * 2). Since it's an inequality n^2 becomes n^2 * 2.
You can go like a heat seaking missile right to the end. No need for ”expeditions” here and there. 1) Show that it is true for a smallest value of p. Bla bla 2) Assume the statement to be true for p. 3) Show that under the above assumption it is also true for p + 1. That is (p + 1)^2 LE 2^(p + 1) expand on both sides p^2 + 2p + 1 LE 2 * 2^p that is p^2 + 2p + 1 LE 2^p + 2^p Make use of our assumption under 2). That is, it now suffices to show that 2p + 1 LE 2^p. But 2p + 1 LE p^2 since after we subtract 2p and add 1 we get 2 LE p^2 - 2p +1 LE (p - 1)^2, which is true for p = 4, 5, 6,… And again by our assumption under 2) 2p + 1 LE p^2 LE 2^p Q. e. d. Comment: What we are as a whole to prove, and our assumption is like a balance leaning over to the right. We make use of our assumption and take away on both sides. The balance must not flip over to the other side! That is what suffices to prove. (And to do that we use our assumption once again.) Hope you see what I mean with a balance. The kind Mdm Justitia uses.
Sorry. Typofix. Should have been But 2p + 1 LE p^2 since after we subtract 2p and add 1 we get 2 LE p^2 - 2p +1 = (p - 1)^2, which is true for p = 4, 5, 6,…
Nice video! One question though. Isn't the whole first part a bit unnecessary if you could simply prove that (2^k-(2k+1)) >= 0 for k>=4, which it is. Or am I not allowed to just insert 4 like that?
Thank you so much! But I don't understand why you can't you do it directly. ie. assume true for n=k so 2^k is greater than k^2 then rtp: n= k+1 so 2^(k+1) is greater than (k+1)^2 LHS= 2 x 2^k then sub in the assumed stuff and becomes 2x2k^2 RHS=k^2+(2k+1) LHS is greater than RHS because k^2 is greater than 2k+1 for k>4 why can't you do it this way?
In my opinion, the flowchart at the beginning is more important to those who want to teach the subject whereas for students it may add another layer of difficulty since these videos are primarily watched by people who need reinforcement on the title/topic.
Your explanation was very helpful. I have had this question before but could not just figure out the induction step . I did understand what you did in both the first case and second but I have a small question. I understand that 2 to the power k+1 =2 to the power n times 2 but I don not seem understand how and why you had to write 2 to the power n+ 2 to the power n. May you please help me on that. Thank you..
***** Hello,you can find that part at 13:00 to about 13:10 . And it is 2 to the power k + 2 to the power k. Sorry ,I used n because I am always using n in my induction step too.Thank you.
Hi, I don't know if you still do videos or not. I found your explanations to be very simple and clear and wondered if you could show me how to do the following proofs: (1) let lcm(a,b) = l so l=pa and l=qb Prove that gcd(p,q) = 1 (2) if gcd(a,b) = 1 prove that gcd(a+b, a-b) is either 1 or 2. (3) Prove that if a | (bc) and gcd(a,b)=1 then a|c (4) If d=gcd(a,b) and f is any other common divisor of a and b, prove that f | d Any help with any of them is appreciated. If you don't do this anymore no problem. Thanks for all your videos.
Hey, I don't know if this would still be helpful but: Question 1: I'm going to do a proof by contradiction: p = l/a q = l/b ... (Given) Let's assume that gcd(p, q) ≠ 1 ... *(1)* Therefore, l/a and l/b have a common factor, say F Since, it's a factor, F is a positive integer ≠ 1 => l/a = mF; l/b = nF; where n and m are positive integers ≠ 1 => am = l/F; bn = l/F => l/F is a common multiple of both 'a' and 'b' (Since, m and n are integers > 1) Since F>1, this implies that l > l/F => lcm(a, b) = l/F ≠ l But, we know that lcm(a, b) = l Therefore, our assumption *(1)* was wrong. Therefore, gcd(p, q) = 1
Stian Sapiens I think u were not paying attention to the first 3 min of the video,. where he says he will do step 2 first and then step 1 and finally step 3.
The one thing I don't get about induction: why just assume something is true for n, then show it's true for n+1, if what you wanted to prove in the first place is that it's true for n?
actually you didnt have to demonstrate past 6:04 when you write consider as we multiplied by 2 which is a positive number which won't change the inequality right ?
Ek wil graag weet hoekom is 2 tot die mag ' n' groter en gelykaan ( 2n +1)? En by die einde van die bewys hoekom is k groter en gelykaan 4??(induction inequality example 6).
Hi Eddie, I find your videos very helpful, I kind of wonder the same thing like some other viewers here, can you explain a little more in detail when you swap n^2 by 2n+1, and how I can get to this step in general, for instance if I have to prove 2^(n-1)bigger and equal to n^2. Thanks very much!
Ik it's been 6 years but here's how I think about it; If A < B, we can edit this inequality so that it still becoles true. If A is less then B, then A is less than any number that's greater than B So if we swap B by a greater number, the RHS would become larger, hence keep being larger than A. It's all about making the hand side larger or smaller (by swapping for bigger or smaller numbers) so that the inequality remains true.
Hi eddie, i'm currently working on an assignment whereby, my assignment ask for to prove by induction that 2^n > n^2 for n>=5. is it possible for me to prove it ur way as ur question is quite similar?
That is same actually. n≥5 means all n greater than 5 including 5. n>4 means same... I know you have done your assignment but I thought to explain 😂 Btw, you could have replaced the given condition by condition in the video as they both are same... That's a 5 years ago...
In 3:38, you said that LHS is greater than or equal to RHS when you have only shown that LHS is greater than RHS, for n = 4. You have not necessarily convinced me about equality of the two sides.
Hi Eddie! Your video helped my alot! But I still have one question. Why did you change the letter n to k (n=k) and didn't continue using the letter n? Thank you !
I finally understand why you have proved that 2^n≥2n+1 is true at first, thank you so much!! Now I have another question, which is familiar with this one, that is: to prove that 3^n>n^3,{n=4,5,6...} Thank you.
I'm already getting the feeling that this video is gonna be directly responsible for a love affair I' going to have with mathematics for the next year or so.
Ok anyone watching this video: before you go and comment about not understanding something, take the time to carefully watch the ENTIRE thing. It will all makes sense :D
@@anitaojwani so I was hoping you have found the answer because your comment was posted 6 months ago haha is anyone else interested ? Give some answers people !
+Lopang Wayne Moalosi because 2^k is greater than k^2 by assumption, you replace 2^k by k^2 in the next row and replace sing = with greater than :D 13:27 check again :)
+Angela Fawn Leach continued from the previous steps... 2(2k+1)-2k-3 ..multiply 2 to get rid of parens making 4k+2-2k-3 then combine like terms... 4k-2k+2-3 making 2k-1
@ 8:46 7 is >= 0 why can you make 2^(k+1) - (2k+3) >=0 does it mean that it can also be =1? since it's >=0? but it should also be >=7 right? I can somehow understand the whole video but I want to fully grasp the concepts. I think there's something wrong somewhere in my understanding. Enlighten me please.
How can it affect, or does the EXACT number at the LHS really affect the whole equation? or just the category it belongs to matters (negative, positive,etc)?
If you're attempting this proof by yourself, you would have started with what he did towards the end of the video. And then realised you had to prove the lhs was greater than or equal to 2n+1
thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you reeaaaaaalllllyyyyyyyy muuuuuuch!!!!!!
![Induction Inequality Proof Example 6: [2^(2n)]*(n!)^2 ≥ (2n)!](http://i.ytimg.com/vi/6xdbKjZkjcg/mqdefault.jpg)
![Induction Inequality Proof Example 6: [2^(2n)]*(n!)^2 ≥ (2n)!](/img/tr.png)
but i don't see how, when seeing a problem like this for the first time, you'd know to replace n^(2) by 2n + 1. I don't see how you would know to do that.
+IffyProjects exactly...i was wondering the same thing
+IffyProjects I agree. I mean, logically, it makes sense, but I automatically assume I CANNOT deviate from the problem due to getting it wrong in school. In real life, yes I could maybe make the assumption, but I wouldn't want to assume anything outside the direction of the problem when quizzes or tests come into play. I'd like an answer as well. "HOW did you know to do this step?"
+IffyProjects ...I agree with you!!
+IffyProjects The only reason he did it the way he did was to make the proof look neater. You can absolutely start with the original inequality without replacing anything, and you'll eventually complete the proof. As OP pointed out in the first couple of minutes of the video, you'll arrive at a problem and have to take a step back solving the problem first.
it takes practice
hi! its very helpful, but I'm wondering how can you come up with the 2n+1 in the first place?
In induction, you check first what you have to proof.
Like 2^{k+1}>(k+1)²=k²+2k+1
Now to proof
2^k+2^k>k²+2k+1
Reduces down(by hypothesis)
2^k>2k+1
If you proof this, you're done!
It has been 5 years since you posted. I hope so you have grown and could find it by yourself!
@@TechToppers I have been working on this problem for the past hour or two and you explained it so so well in one small comment. Thanks a ton just had my Eureka moment have a nice day
@@FSHnegativ
Thanks a lot. But please remember, that the I wrote was just for intuition. Backtracking in mathematics is generally hard. You have to be very sure that your assumptions are correct.
In which grade you are?
@@TechToppers im currently in the first semester of uni majoring in data science. Haven't had induction proofs in high school so I'm doing my best learning it as fast as possible. I actually just got done with the whole proof (n^2
@@TechToppers How does the reduction work? Because you're halving one side and square-rooting the other.
If anybody was wondering why 2^n > n^2 becomes 2^n * 2 > n^2 * 2 is because in order to get 2^n+1 you need to multiply 2^n by 2 (hence 2^n * 2). Since it's an inequality n^2 becomes n^2 * 2.
8:36 I don’t get how you can say 7 is greater than/equal to 0...
this proofs are left me speechless😭
Why not use a chain for the first one?
For example:
assume: 2n + 1
great video, it really helped me out! :)
The only problem i have is this: why do you assume that 2^n is greater than or equal to 2n+1?
You can go like a heat seaking missile right to the end. No need for ”expeditions” here and there.
1) Show that it is true for a smallest value of p. Bla bla
2) Assume the statement to be true for p.
3) Show that under the above assumption it is also true for p + 1. That is
(p + 1)^2 LE 2^(p + 1) expand on both sides p^2 + 2p + 1 LE 2 * 2^p that is p^2 + 2p + 1 LE 2^p + 2^p
Make use of our assumption under 2). That is, it now suffices to show that
2p + 1 LE 2^p.
But 2p + 1 LE p^2 since after we subtract 2p and add 1 we get
2 LE p^2 - 2p +1 LE (p - 1)^2,
which is true for p = 4, 5, 6,…
And again by our assumption under 2)
2p + 1 LE p^2 LE 2^p
Q. e. d.
Comment:
What we are as a whole to prove, and our assumption is like a balance leaning over to the right. We make use of our assumption and take away on both sides. The balance must not flip over to the other side! That is what suffices to prove. (And to do that we use our assumption once again.) Hope you see what I mean with a balance. The kind Mdm Justitia uses.
Sorry. Typofix. Should have been
But 2p + 1 LE p^2 since after we subtract 2p and add 1 we get 2 LE p^2 - 2p +1 = (p - 1)^2, which is true for p = 4, 5, 6,…
Nice video! One question though. Isn't the whole first part a bit unnecessary if you could simply prove that (2^k-(2k+1)) >= 0 for k>=4, which it is. Or am I not allowed to just insert 4 like that?
How does one know the correlary piece they need to prove before hand?
But how is 7 greater or equal to 0? How can we make that assumption, when 7 would never equal 0?
7 has to be greater (which is the case) OR equal to 0.
Otherwise a>=b would be true only if a=b=0.
Thank you so much! But I don't understand why you can't you do it directly.
ie.
assume true for n=k
so 2^k is greater than k^2
then rtp: n= k+1
so 2^(k+1) is greater than (k+1)^2
LHS= 2 x 2^k then sub in the assumed stuff and becomes 2x2k^2
RHS=k^2+(2k+1)
LHS is greater than RHS because k^2 is greater than 2k+1 for k>4
why can't you do it this way?
+Charlene Chau 2x2k^2 IN lANE 7 shouLD BE 2xk^2
Would you consider the first proof the lemma that you need for the current proof or proof in consideration? Good job:
In my opinion, the flowchart at the beginning is more important to those who want to teach the subject whereas for students it may add another layer of difficulty since these videos are primarily watched by people who need reinforcement on the title/topic.
You solved exactly same thing I got for homework, heh :D
Now I understand it completely.
BTW, RTP means required to prove
Assume 2^k > k^2 for k > 4. 2^(k+1) = 2(2^k) = 2k^2 > 2k^2 = (k+1)^2 + k^2-2x-1 > (k+1)^2, since k^-2k-1 = (k-1)^2-2 > 0 for k>4
Your explanation was very helpful. I have had this question before but could not just figure out the induction step . I did understand what you did in both the first case and second but I have a small question. I understand that 2 to the power k+1 =2 to the power n times 2 but I don not seem understand how and why you had to write 2 to the power n+ 2 to the power n. May you please help me on that. Thank you..
***** Hello,you can find that part at 13:00 to about 13:10 . And it is 2 to the power k + 2 to the power k. Sorry ,I used n because I am always using n in my induction step too.Thank you.
I do it this way, hope it's correct.
2.2^k>=2.k^2
2^(k+1)>=(k+1)^2
To prove 2.k^>=(k+1)^2
2.k^2>=k^2+2k+1
k^2-2k-1>=o.......(1)
True if k>2.4...
Can we do it by putting the equation on one side and zero on another side for every inequality question??????
I would just use 2^(k+1)>=2k^2 from the assumption, then it's a simple matter of proving 2k^2>=(k+1)^2 for each k>=4
@ 8:39, it should be > 0 not > or = 0 correct? if something is > or = 7, it is > 0 not > or = 0 right?
why did you start with 2^(n)>and equal to 2n+1
Hi, I don't know if you still do videos or not. I found your
explanations to be very simple and clear and wondered if you could show
me how to do the following proofs:
(1) let lcm(a,b) = l so l=pa and l=qb
Prove that gcd(p,q) = 1
(2) if gcd(a,b) = 1 prove that gcd(a+b, a-b) is either 1 or 2.
(3) Prove that if a | (bc) and gcd(a,b)=1 then a|c
(4) If d=gcd(a,b) and f is any other common divisor of a and b, prove that f | d
Any help with any of them is appreciated. If you don't do this anymore no problem. Thanks for all your videos.
Hey, I don't know if this would still be helpful but:
Question 1: I'm going to do a proof by contradiction:
p = l/a
q = l/b ... (Given)
Let's assume that gcd(p, q) ≠ 1 ... *(1)*
Therefore, l/a and l/b have a common factor, say F
Since, it's a factor, F is a positive integer ≠ 1
=> l/a = mF; l/b = nF; where n and m are positive integers ≠ 1
=> am = l/F; bn = l/F
=> l/F is a common multiple of both 'a' and 'b' (Since, m and n are integers > 1)
Since F>1, this implies that l > l/F
=> lcm(a, b) = l/F ≠ l
But, we know that lcm(a, b) = l
Therefore, our assumption *(1)* was wrong.
Therefore, gcd(p, q) = 1
Everything is standard Number Theory plug in some numbers and try to understand it's behaviour and then formalize.
It has been 4 years but still😂
I can give you intuition if you want...
Sir, why m√a^n = a^n/m please explain
Thank you soo much for ur explanation,but I have a question why do assume that 2^n is greater or equal to 2n+1. Thank you
+Eddie Woo how do you know that you'll be needing that proof later on ?
Stian Sapiens I think u were not paying attention to the first 3 min of the video,. where he says he will do step 2 first and then step 1 and finally step 3.
The one thing I don't get about induction: why just assume something is true for n, then show it's true for n+1, if what you wanted to prove in the first place is that it's true for n?
Good question I know the answer. Should I tell?
2*2^(k)-(2k+1)>=k^2-2k+1-2=(k-1)^2-2 ,since k>=4,(k-1)^2>=9 -->k^2-2k-1>=9-2=7>=0
so i don't think we need to prove n^2>=2n+1
actually you didnt have to demonstrate past 6:04 when you write consider as we multiplied by 2 which is a positive number which won't change the inequality right ?
Ek wil graag weet hoekom is 2 tot die mag ' n' groter en gelykaan ( 2n +1)? En by die einde van die bewys hoekom is k groter en gelykaan 4??(induction inequality example 6).
Thank You soo much for this, but could you also prove n
2.2power k
How convert 2 power k + 2 power z
Tell me
Thank you this video was very helpful. ..but am a little bit confused. ...how come 2^k+1 =2^k+2^k. ...help
2^k+1 is equal to 2^k×2(Indices rule that is bases same so add powers)...then 2^k×2 means 2^k is written two times that is 2^k+2^k.
I don't get how you have two 2^k in 2.2^k. I'm confused
really late but that's not the case, 2^(k+1) is just 2(2^k) he wrote it as 2.2^k which is really 2*2^k
@@haloshiroe that's like from 4 years ago 😭 but thank you nonetheless. Your video helped a great deal🌟
@@snethembamsomi9390 here i am now currently confused😭
Hi Eddie, I find your videos very helpful, I kind of wonder the same thing like some other viewers here, can you explain a little more in detail when you swap n^2 by 2n+1, and how I can get to this step in general, for instance if I have to prove 2^(n-1)bigger and equal to n^2. Thanks very much!
Ik it's been 6 years but here's how I think about it;
If A < B, we can edit this inequality so that it still becoles true.
If A is less then B, then A is less than any number that's greater than B
So if we swap B by a greater number, the RHS would become larger, hence keep being larger than A.
It's all about making the hand side larger or smaller (by swapping for bigger or smaller numbers) so that the inequality remains true.
is it possible to start at step 2 if you had never worked this problem? Not likely
You are an excellent teacher, and that would be an understatement.
Hi eddie, i'm currently working on an assignment whereby, my assignment ask for to prove by induction that 2^n > n^2 for n>=5.
is it possible for me to prove it ur way as ur question is quite similar?
That is same actually.
n≥5 means all n greater than 5 including 5.
n>4 means same...
I know you have done your assignment but I thought to explain 😂
Btw, you could have replaced the given condition by condition in the video as they both are same...
That's a 5 years ago...
In 3:38, you said that LHS is greater than or equal to RHS when you have only shown that LHS is greater than RHS, for n = 4. You have not necessarily convinced me about equality of the two sides.
Hi Eddie! Your video helped my alot! But I still have one question.
Why did you change the letter n to k (n=k) and didn't continue using the letter n?
Thank you !
I finally understand why you have proved that 2^n≥2n+1 is true at first, thank you so much!!
Now I have another question, which is familiar with this one, that is: to prove that 3^n>n^3,{n=4,5,6...}
Thank you.
please solve 2n less than n! for, n greater or equal to 4
does this solution work for 2^n > n^2 ?? > not >=
I'm already getting the feeling that this video is gonna be directly responsible for a love affair I' going to have with mathematics for the next year or so.
im curious, how did your love affair go?
How do you prove this using minimum counterexample?
Wait a minute! This is the same guy in the video about parity bits that I watched! Man, this guy is all everywhere.
Thanks! Best explanation I've found. Finally got it!
2 question I don't understand after consider. How plus 2 power k
I clicked the video without looking at the channel but immediately recognized Eddie Woo the moment he said the word "here."
Majorly overcomplicated.
When you had 2^(k+1)-(k+1)^2 you could simply rearrange to 2(2^k-k^2) + (k-1)^2 - 2 which is certainly positive for k>=4
Shouldn't it be "n = { 4,5,6,... }," not "{ n = 4,5,6,... }"?
Ok anyone watching this video: before you go and comment about not understanding something, take the time to carefully watch the ENTIRE thing. It will all makes sense :D
But it's complicated for no reason...
Can you do a video for 4^n > n^4 ??
I am from India my ? How can you return as n square is equal to 2n+1 .but why
This is the question Anita ! Can anyone answer ?
@@danieldorsz1047 yes dear
@@anitaojwani so I was hoping you have found the answer because your comment was posted 6 months ago haha is anyone else interested ? Give some answers people !
why is it 2k+3
Because when he plugged in k + 1 into 2k + 1, he got 2(k + 1) + 1 which becomes 2k + 2 + 1 which is 2k + 3!
Cheers bro
thank you so much for this
it because when you simplfy 2 (k+1)+1 it will give you 2k+3
Sir why did you replace 2^k by K^2 ?
+Lopang Wayne Moalosi because 2^k is greater than k^2 by assumption, you replace 2^k by k^2 in the next row and replace sing = with greater than :D 13:27 check again :)
Thanks a lot...i see!
It now makes sense.
+Lopang Wayne Moalosi np :D
thanks for this video! now i am even more confused haahah
why is 2^k>2k+1
I don't get it
7:54 - Where did the 2k - 1 come from? I thought it was 2k + 1.
+Angela Fawn Leach continued from the previous steps... 2(2k+1)-2k-3 ..multiply 2 to get rid of parens making 4k+2-2k-3 then combine like terms... 4k-2k+2-3 making 2k-1
Is induction the only way to solve this kind of problem...
This is genuinely not an easy thing to grasp.
You could use some calculus I suppose...
The best explanation I watched so far
n^2 = 2n+1 ????
2^3 is smaller than 3^2
The domain is {n ≥ 4}.
What level of algebra is this?
and what grade will I learn this? These proofs are so interesting
Basic😂
Really
for any positive integer n, 6n - 1 is divisible by 5.
eh nope 6*4 -1 = 23 not divisible by 5
Dude thank you. I needed this explanation ❤️
You are my savior.
@ 8:46
7 is >= 0
why can you make 2^(k+1) - (2k+3) >=0
does it mean that it can also be =1? since it's >=0?
but it should also be >=7 right?
I can somehow understand the whole video but I want to fully grasp the concepts. I think there's something wrong somewhere in my understanding. Enlighten me please.
How can it affect, or does the EXACT number at the LHS really affect the whole equation? or just the category it belongs to matters (negative, positive,etc)?
Demostró que 2^n > 2n+1.
Pero debía demostrar 2^n > n².
😬
you are honestly the best math teacher
Send for me some video
how 2.2^k = 2^k+ 2^k ?????
is this a standard way to depict multiplication?
+Максим Марков i thought/think the standard way is *
+Remavas ...on a computer :)
Максим Марков 2x2 is the same as 2+2. I think it only works with 2.
+Joshua L thanks but now its 2016 hahahha
Thank you for this wonderful video. I learnt a lot:D
2021
thank you very much .that was helpful
Let me get another beer ...
yes - nice.
微分して増減調べればいいのでは？
More videos plz it's interesting
I love mathematics!
i got it......Thanks a lot Sir.......
Thanks
thank you from palistine
Thank you!!
Thank you :D
That problem is too easy
Prove this one
(2n)! > n^n
How can n^2 be replaced by 2n+1? You didn't explain the first step and this whole video is senseless
i have the same question
It does if you watch the whole video.
If you're attempting this proof by yourself, you would have started with what he did towards the end of the video. And then realised you had to prove the lhs was greater than or equal to 2n+1
First, plug in some numbers if you don't get the Algebra. That's the best way!
Then you will understand in no time
Wow. That was great
Thank you, I understand now :D
Good work, buddy.
thanks !!!
thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you reeaaaaaalllllyyyyyyyy muuuuuuch!!!!!!
jaja , really i was the only one who brings the algebra homework right, and my teacher give an extra point on my exam thanks to you :D
o ga o.. oponu oshi
thanks boss
problem is ur first teacher
Thwis syde wud be bigga
I love you
wait wha?
Not helpful at all man. Even made more confused