I have been struggling with this proof for a while, and every explanation I've found about it just didn't click for me. You put this together in a way that no other UA-cam lecturer could. Thank you for saving my sanity brother!
*Everyone,* here is a modified approach: The base case is for n = 5. 2^5 vs. 5^2 32 > 25 So, the base case is satisfied. The inductive step. Assume it is true for n = k, for k >= 5: That is, assume 2^k > k^2. Then, show it is true for n = k + 1. That is, show 2^(k + 1) > (k + 1)^2. Take the inequality in what we are assuming and multiply each side by 2 so that it resembles closer to the inequality that we want to show: Assume: 2^k > k^2 2*2^k >2*k^2 2^(k + 1) > 2k^2 We ultimately need to show 2^(k + 1) is greater than (k + 1)^2 for k >= 5. If we can show that 2k^2 is greater than (k + 1)^2, then we can use the transitive property to finish this. 2k^2 vs. (k + 1)^2 2k^2 vs. k^2 + 2k + 1 2k^2 - k^2 - 2k vs. 1 k^2 - 2k vs. 1 k^2 - 2k + 1 vs. 1 + 1 (k - 1)^2 vs. 2 By inspection, you can see that for k >= 3, the left-hand side is greater than 2. So, 2^(k + 1) > 2k^2 > (k + 1)^2. By transitivity, 2^(k + 1) > (k + 1)^2. Thus, by the Principle of Mathematical Induction, I have shown that 2^n > n^2 for all integers n greater than or equal to 5.
this one is straight out of an advanced calculus book haha! it is a problem in chapter one of Kenneth A. Ross: Analysis the Theory of Calculus -- love it!
Thank you for this very informative video!Im freshman in college and we had induction topic 3-4 weeks ago and i remember doing this exercise or similar to this one in class. My question is instead of step by step making it smaller in 6:51 can we just say 2k+1 is less than k² because k is starting from 5? Does that work too?
You can actually do that what i did : >k^2 + k^2 | Take out k^2 = k x k > k^2 + k *k | k*k = 5k because K>=5 > k^2 + 5k | split into 2k + 3k >k^2 + 2k + 3k | 3k>1 > k^2 + 2k + 1
I had also some idea. 2^n-n^2>0, (2^n/2-n)(2^n/2+n)>0, So this Proof Is ekvivalent with the Proof 2^n/2>n by induction... sqrt2*2^n/2>n+1 for n+1, sqrt2=(1+ something Positive). Sometimes I like combinations Proofs.
Great content man, thank you so much for the explanation. Now would the proof change if instead of strictly greater than we had a greater or equl than? Like 2^n >= n^2
U need to ask yourself what you don’t get: is it the principle of induction? The principle of calculating the inequality or maybe something simpler like the calculation rules ( I know you probably didn’t need this but once u break down what u undertaken and what u don’t it becomes much easier for you to help itself understand)
Well basically that's the point. In mathematical induction for inequalities, as long it doesn't disrupt the thing which in this case, the greater than sign, you can basically make up ur own variables as long it makes the statement true. But not every variables, it only works when u try to proof it. By which in this example the 4k, because we know that k>4 which is essentially k=5, so sub it into the k^2, we get 25 right, then why it's 4k, because if we sub it into the 4(20), it's correct right 25>20 so there u go
U could say that it's 3k, just that remember to fact it out that it will become 2k + k for the next step, I don't think it works for 2k, it might work but there's gonna some explanation involved as well so just follow the variables from the (k+1)^2 expanded version
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I have been struggling with this proof for a while, and every explanation I've found about it just didn't click for me. You put this together in a way that no other UA-cam lecturer could. Thank you for saving my sanity brother!
probably the best explanation on youtube. great stuff! thanks!
seriously bro , i agree
I love your videos man! From Italy 🇮🇹, never stop learning
Thanks! Will do!
I'm here coz i went to the math sorcerer and he completely confused me😂😭
thank you very much you definitly need more views
*Everyone,* here is a modified approach:
The base case is for n = 5.
2^5 vs. 5^2
32 > 25
So, the base case is satisfied.
The inductive step. Assume it is true for n = k, for k >= 5:
That is, assume 2^k > k^2.
Then, show it is true for n = k + 1.
That is, show 2^(k + 1) > (k + 1)^2.
Take the inequality in what we are assuming and multiply each side by 2
so that it resembles closer to the inequality that we want to show:
Assume: 2^k > k^2
2*2^k >2*k^2
2^(k + 1) > 2k^2
We ultimately need to show 2^(k + 1) is greater than (k + 1)^2 for k >= 5.
If we can show that 2k^2 is greater than (k + 1)^2, then we can use the
transitive property to finish this.
2k^2 vs. (k + 1)^2
2k^2 vs. k^2 + 2k + 1
2k^2 - k^2 - 2k vs. 1
k^2 - 2k vs. 1
k^2 - 2k + 1 vs. 1 + 1
(k - 1)^2 vs. 2
By inspection, you can see that for k >= 3, the left-hand side is greater than 2.
So, 2^(k + 1) > 2k^2 > (k + 1)^2.
By transitivity, 2^(k + 1) > (k + 1)^2.
Thus, by the Principle of Mathematical Induction, I have shown that
2^n > n^2 for all integers n greater than or equal to 5.
thank you for making this video. It was really useful for me.
absolute legend saving me from cs
😂 love the Einstein insert!!
Perfect
this one is straight out of an advanced calculus book haha! it is a problem in chapter one of Kenneth A. Ross: Analysis the Theory of Calculus -- love it!
took me a bit to not be confused but i get it, amazing
Great ❤️
Love from india
Mujhe bhi batana please last step smjh nahin aaya 😢
love from india nice explanation
Thanks sir.Waiting for maths induction for questions in Matrix form
You should share an example. It helps. That would be Linear Algebra
@@PrimeNewtons ok sir
That little tmr song/ intro was so sweet
That was a neat solution. Thank you!
You're welcome!
Thank you for this very informative video!Im freshman in college and we had induction topic 3-4 weeks ago and i remember doing this exercise or similar to this one in class. My question is instead of step by step making it smaller in 6:51 can we just say 2k+1 is less than k² because k is starting from 5? Does that work too?
You can actually do that what i did :
>k^2 + k^2 | Take out k^2 = k x k
> k^2 + k *k | k*k = 5k because K>=5
> k^2 + 5k | split into 2k + 3k
>k^2 + 2k + 3k | 3k>1
> k^2 + 2k + 1
so technically you're not wrong actually
00:59 👌
thank you
Very nice approach end explanation.
I'll present another approach, for the proof that:
[∃ m ∈ N , m > 4 , 2^m > m^2] ⟹ [n = m + 1 ⟹ 2^n > n^2]
Proof:
2^m > m^2 ⟹ 2^m - m^2 > 0
Therefore, for n = m + 1 we obtain:
2^n - n^2 = 2^(m+1) - (m+1)^2
= 2⋅2^m - m^2 - 2m - 1
= 2(2^m - m^2) + (m-1)^2 - 2
> (m-1)^2 - 2 > (4-1)^2 - 2
= 7
> 0
Therefore: 2^n > n^2 ∎
Many thanks sir
Thank you, this was a great explanation I finally got it.
where'd you get your hat? i might need one...
I had also some idea. 2^n-n^2>0, (2^n/2-n)(2^n/2+n)>0, So this Proof Is ekvivalent with the Proof 2^n/2>n by induction...
sqrt2*2^n/2>n+1 for n+1, sqrt2=(1+ something Positive). Sometimes I like combinations Proofs.
Great content man, thank you so much for the explanation. Now would the proof change if instead of strictly greater than we had a greater or equl than? Like 2^n >= n^2
does it also work if i turn it into 2^k + 2^(k+1)>(k+1)^2?
🎉🎉 thank you!!
k^2> 2k+1
What I don't get is how can put least value of k=4 in above eq?
I still don't understand 😭
Sir please improve on your board I can't see
U need to ask yourself what you don’t get: is it the principle of induction? The principle of calculating the inequality or maybe something simpler like the calculation rules ( I know you probably didn’t need this but once u break down what u undertaken and what u don’t it becomes much easier for you to help itself understand)
@@No.School.dk_Colur-- This is not partial texting. Spell out all of your words.
May I ask that if it doesn't work for k =5,Can I work out to prove k=6and if k=6works out does that mean P(n)is true for k greater than 4.
Please answer me.I feel so frustrating with inequalities induction.
There are so many ways and I am dizzy now.
Why do u choose 4k which is not 5k?
Why do u want to reduce?
Explain me please.
Is it normal to reduce and can I reduce almost every inequalities like that, sir?
Sir, will itbe wrong if I try to calculate
=k square plus 5k
=k square plus 2k plus 3k
=ksquare plus 2k plus 1
Discrete math is a different beast 😅
Good work. But not clear to me.
Hmm why not use
n > 2log2(n)
WAITT I GET IT NOW LOL TYSM
i will never understand induction with inequalities cus i can wrap my brain around the fact, that i can js change the value like that 😭
Well basically that's the point. In mathematical induction for inequalities, as long it doesn't disrupt the thing which in this case, the greater than sign, you can basically make up ur own variables as long it makes the statement true. But not every variables, it only works when u try to proof it. By which in this example the 4k, because we know that k>4 which is essentially k=5, so sub it into the k^2, we get 25 right, then why it's 4k, because if we sub it into the 4(20), it's correct right 25>20 so there u go
U could say that it's 3k, just that remember to fact it out that it will become 2k + k for the next step, I don't think it works for 2k, it might work but there's gonna some explanation involved as well so just follow the variables from the (k+1)^2 expanded version
My prob is how do you know k²
Make a video specifically for me because I still don't get it 😢
still do not understand...good work though🙂
Toan gi bo ich
this reminds me of curry (not racially motivated)
I don't understand your teaching