In discrete structures course a few years ago as a freshman

hairygrape
hairygrape

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?"

Attached: 1511014789907.jpg (5 KB, 225x225)

Skullbone
Skullbone

MATHEMATICIANS ABSOLUTELY AND ETERNALLY BTFO
How can these pseudologicians ever recover?

Burnblaze
Burnblaze

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

TalkBomber
TalkBomber

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

BlogWobbles
BlogWobbles

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

Techpill
Techpill

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

MPmaster
MPmaster

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

happy_sad
happy_sad

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

Poker_Star
Poker_Star

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

TalkBomber
TalkBomber

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

Deadlyinx
Deadlyinx

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

BlogWobbles
BlogWobbles

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

Skullbone
Skullbone

Nice LARP

Emberfire
Emberfire

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

Sir_Gallonhead
Sir_Gallonhead

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

w8t4u
w8t4u

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.

askme
askme

see

Skullbone
Skullbone

no just assuming ever k in series to infinity

DeathDog
DeathDog

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.

Spazyfool
Spazyfool

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.

Nude_Bikergirl
Nude_Bikergirl

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

StrangeWizard
StrangeWizard

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

SniperWish
SniperWish

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.

PurpleCharger
PurpleCharger

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

5mileys
5mileys

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

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