Induction Proofs Involving Inequalities.

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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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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.

montanasebastiano
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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.

zaidahsan
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Even better than my university professor <3 thanks sir!

AmateurThings
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I watched yours video many many times .you are amazing for understanding each and every step very crystal clear. You are made for mathematics.

anshulkumar
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As a math teacher candidate in Oklahoma struggling through Number Theory thank you!

davidwoznerable
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Brilliant explanation, you have helped an eighth-grader comprehend this beautiful mathematical proof example.

particleonazock
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This video has helped me to understand MI, thanks Dr Trefor!

ShanaAngliang
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It was nice that your example required an adjustment to the base case!

georgelaing
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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! :)

Kenspectacle
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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.

ericsabacan
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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?

Shannxy
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Why not just use ‘greater than or equal to’ in the last step instead of all that extra work?

MrConverse
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shouldn't it be for k ≥ 1 instead of k > 1

slientsoul
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What software does he use to do the writing?

iamrxheem
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Can u help me in this
2^(n+1)<1+(n+1)2^n for n>=1

suvayudas
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I'm the most confused individual 😂😂😂😂😅😅😅😅😅

luvochiya
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Oy bruv why didn't you just start the problem with the given domain n > 1

BsTonCeltics
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Why not just use ‘greater than or equal to’ in the last step instead of all that extra work?

MrConverse