Basic Uses of Mathematical Induction - Number Theory

I forgot to say this, but if you have better alternatives or more concise solutions, please leave them in the comments below! I love seeing different ways of approaching a problem.

[@kylelohsmathchannel7369]

I forgot to say this, but if you have better alternatives or more concise solutions, please leave them in the comments below! I love seeing different ways of approaching a problem.

[@shubhmishra3338]

Cool vid

[@saptarsh]

A simple application of induction. 3^(n+1) = 3.3^n = 3^n + 3^n + 3^n > n^2 (because you assume this in induction) + n^2 (since n^2 - 2n > 0 for all n > 2, and you know that 3^n > n^2 holds for n = 2) + n^2 > n^2 + 2n + 1 = (n+1)^2. No need of anything else. Simple logic

[@kylelohsmathchannel7369]

Thank you for the comment! It looks like there are much better ways to solve this inequality; I was inefficient in my methodology. Indeed, your solution is superior to mine

[@vxllabh09]

i took 2 base cases for n = 1 and n = 2, easily verified then proceeded in the following manner:-
(n)^2 + (n-1)^2 > 2 (for all n>=2)
2n^2 - 2n + 1 > 2
2n^2 > 2n + 1
Now, 3^n > n^2 and hence 2*3^n > 2n^2
Thus, 2*3^n > 2n + 1
And, 3^(n+1) > 2*3^n + n^2 > (n+1)^2
(I feel like this is slightly shorter and easier to notice given it took me very little time to realise the same.)
Great video!

[@audreyloh4842]

Quick little inequality

[@rafinhapereira]

I think if you use calculus this promblem is almost trivial and you can prove that it holds for all real positive numbers. Just take f(x)=xlog(3)-2log(x). This function has a global minimum on 2/log(3) on wich is postive.

[@kylelohsmathchannel7369]

That is also true. This was supposed to be more of a number theory-esque video, but yes! Thank you for commenting; I love seeing alternative solutions to the posed problem!

[@empajerek4193]

Cool video, but on thumbnail you forgot to add + symbol to integer's sign

[@kylelohsmathchannel7369]

I will make that correction 😂

[@creamy798]

Assume 3^k+1k^2 by 3. This gives 3^k+1>3k^2. This implies that 2k^2

[@kylelohsmathchannel7369]

Thank you for commenting!
Your proof is mostly correct, but there are a few minor technicalities. The initial claim was 3^k > k^2 and the induction step was 3^(k+1) > (k+1)^2. Contradiction, by definition, would require 3^(k+1)

[@creamy798]

@@kylelohsmathchannel7369 Thank you for identifying a mistake I made! I overlooked that when it could be equal to.