>in discrete structures course a few years ago as a freshman

>learning proof by induction

>"alright, so you just assume it's true at k and show that it works at k+1"

>guy raises his hand

>"professor... why can't we just assume it's true at k+1?"

# In discrete structures course a few years ago as a freshman

MATHEMATICIANS ABSOLUTELY AND ETERNALLY BTFO

How can these pseudologicians ever recover?

But you can, and then you gotta prove it's true at k and k+1

you'd have to prove it at k+2 if you assume k+1 is true

I'm assuming they prove that if it's true for k+1 it's true for k+2

I still don’t for my own life understand proof by induction

What? We learned induction in the 2nd year of high school, that was fun stuff.

Or maybe even 1st, damn can't remember now.

Most CS degrees require a Discrete Math course, where they are taught proof by induction, which breaks their brainlet brains.

You don't have to assume it's true at k, you have to prove it eventually.

yes but this dumb nigger thought he could assume everything he wanted to prove and the proof would be done

It's just a way of describing the natural numbers.

Nice LARP

How do you prove that 2^n > n^2 for all n > 4?

That's not how induction works retard.

>prove it works at some arbitrary number (usually one or zero)

>prove that IF it works at k, then it works for k+1

proof by induction is pretty much "prove its true because it is" so you just replace all of your n with k, k + 1, etc to where ever your base cases are and the math always works out. If it doesnt then youre either a brainlet or the proof dont work.

see

>no just assuming ever k in series to infinity

Those are the people that clearly don't understand what's going on.

Maybe the prof was a bit unclear but students need to at least try.

C1: 2^n > n^2 when n = 5

2^5 > 5^2

64 > 25

C2: n^2 > n+1 when n > 4

n^2 > n+1

n^2-n-1 > 0

n > ϕ

n > 4

C3: 2^(n+1) > (n+1)^2 when n > 4

2^n > n^2

2*2^n > 2*n^2 = n^2+n^2 > n^2+n+1 when C2 holds

2*2^n > n^2+n+1 when C2 holds

2^(n+1) > (n+1)^2 when C2 holds

C1 is the base case and C3 is the induction step, proving that 2^n > n^2 when n > 4

No assumptions necessary.

C3 should read 2^(n+1) > (n+1)^2 when 2^n > n^2 and n > 4

>goes to lectures

it's you who's the idiot. why waste hours staring at a wall when you could be staring your computer screen at home? don't make no fucking sense

How? Literally how? You show it's true for at least one value, then show that each previous case forces the next one to be true also.

It's like the most intuitive shit in the world. Are you, by chance, a middle school student? Not joking, genuine question.

I was never able to do the last part of proof by induction, ergo the n=(n+1) part because I never managed to do that algebraic simplification of the sum

Not unexpected if it's the first time you're doing mathematical induction. Stop being rude to people who are trying to learn.