Induction Proofs Involving Inequalities.
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- Опубліковано 7 лют 2025
- We work through an induction example where we are proving an inequality. We have to decide what is the right way to make an inequality of our own in the calculation. We also play around with a technicality surrounding what the basis really is.
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![2^n is greater than n^2. Strategy for Proving Inequalities. [Mathematical Induction]](http://i.ytimg.com/vi/UE009vD3PEI/mqdefault.jpg)
I appreciate you going through the extra steps proving 2k > k + 1 for k > 1. I thought it wasn't necessary at first but after rewatching I understand the importance of this step.
Even better than my university professor
I watched yours video many many times .you are amazing for understanding each and every step very crystal clear. You are made for mathematics.
As a math teacher candidate in Oklahoma struggling through Number Theory thank you!
Brilliant explanation, you have helped an eighth-grader comprehend this beautiful mathematical proof example.
Great to hear!
It was nice that your example required an adjustment to the base case!
Hello Dr. Bazett,
Thank you for these thought out videos. The idea of using the ladder analogy is really amazing.
Also since this resource is quite useful, and as some have pointed out the transitive inequality in the comments. I want to elaborate this so it becomes obvious, and helps someone who might find it a bit confusing.
In the basis case you proved that
2^0 > 0 | k = 0
Then assuming 2^k > k we want to show that 2^(k+1) > (k+1).
This can be directly shown from the transitive inequality, which I like to write in the form of a chain.
2^(k+1) = 2 . 2^k = 2^k + 2^k
We apply the assumption on the first of the two terms on the left side.
Then,
2^(k+1) > k + 2^k [ From the Assumption i.e. 2^k > k ]
2^(k+1) > k + 2^k >= k + 1 [ Since 2^k >= 1 for all k >= 0 ]
Then from the transitive property of inequalities the first term on the left side is greater than the term in the middle, which is equal to or greater than the term on the right side. Thus the term on the left is necessarily greater than the term on the right.
2^(k+1) > k + 1. Q.E.D
Assuming the assumption we have thus proved the induction.
GOAT
Classy
This video has helped me to understand MI, thanks Dr Trefor!
Glad it was helpful!
What your age bro?
Hello, I have a question, how did we arrive to k+k > k+1? I am still confused on what does that conclusion is trying to achieve? and how did we get to that conclusion? many thanks in advance! :)
Make k + 1 = to some variable, and you will see that the induction step follows the induction hypothesis by stating that 2^some variable > some variable which is exactly like 2^K > K, but for k > 1.
Why does it turn to 2k and not 2^k instead?
Is it not k is greater or equal to 1 that we need. Since we are to use 1. Why use k greater than 1
If (2^n > n) actually holds true for n = 0,1
Why does the induction steps lead to having to replace it with (k > 1) instead?
I'm confused as well >.< I feel like it should have been k>=0 because I've never seen a case where you have k>1 but also still prove k=0,1 unless it was something like the Fibonacci sequence. But you only need to have individual cases there bc those are special cases, whereas here, like you said, it holds true for k=0,1 as well.
Hi Sir. I got interested with the way you explained this lesson. I'm doing a project and I find this video very useful for students, may I have the permission to use your video. Thank you very much.
Glad it helped!
Why not just use ‘greater than or equal to’ in the last step instead of all that extra work?
shouldn't it be for k ≥ 1 instead of k > 1
What software does he use to do the writing?
@@DrTrefor Awesome thanks. We're actually learning this topic in class and I came across this video. Gonna recommend it.
Can u help me in this
2^(n+1)=1
i love u bazett
I'm the most confused individual 😂😂😂😂😅😅😅😅😅
how did 2 change to k in 4.18
Trefor Bazett
Ohh..I get it now..Thanks so much
Old comment but I'm replying if anyone else was confused. I was confused too but basically the assumption is 2^k is greater than k. His equality has 2 * 2^k and he can thus assume also that 2 * 2^k is greater than 2 * 2^k
tnx
Oy bruv why didn't you just start the problem with the given domain n > 1
Why not just use ‘greater than or equal to’ in the last step instead of all that extra work?