It's not beyond ALL doubt, it's beyond a reasonable doubt. Saying "I know it really seems like I killed my wife, but dude what if we're all just brains in a vat?" isn't a reasonable argument.
Using "proof" by induction
Andrew Lopez
Aiden Watson
shut the fuck up
Samuel Gray
That is not the reason induction works on the naturals. The naturals have an axiom called the well ordered property. Every nonempty subset of the naturals has a least element, 'a' such that a=1, If P(k) is true then P(k+1) is true.
Then P(n) is true for every natural n>=1.
Proof: Let S be the set of naturals such that P is false. Assume there is an element in S. Then S has a least element 'a' because S is a nonempty subset of the naturals. By assertion a=/=1 so a>1. Since a is least, a-1 is not in S, then P(a-1) is true. But by property 2, P(a) must be true. This is a contradiction and thus our assumption that S is nonempty is false. S is empty and P(n) is true for every natural n>=1.
Josiah Perry
>relying on classical maths in 2017
when does summer end?
Brandon Brown
[math]
\underline{induction}
\\
\underline{1^{\circ}} \\
f(n)= 1+2+3+...+n = \frac{n(n+1)}{2} \\
\underline{2^{\circ}} \\
n\rightarrow (n+1): \\
\left \{
\begin{align*}
f(n+1)&=1+2+3+...+n+(n+1) \\
f(n+1)&= \frac{n(n+1)}{2} + (n+1) \\
\end{align*}
\right.
\\f(n+1) = \frac{n(n+1)}{2} + (n+1) \\
= \frac{n^2+n+2n+2}{2} = \frac{n^2+3n+2}{2} \\
= \frac{(n+1)(n+2)}{2} = \frac{(n+1)((n+1)+1)}{2} \\ \\
=> \text{if f(n) is ok, so is f(n+1)} \\
\underline{3^{\circ}} \\
f(1) = \frac{1(1+1)}{2} = 1
[/math]
Josiah Bennett
>tfw bayesian
its fun watching you plebs squabble over induction
Joseph Scott
Not a math guy, but isn't mathematical "induction" deductive reasoning?
Joshua Carter
Yes. The name induction is a very poor choice. They should really be called recursive proofs or something like that.
Aaron Foster
>Q.E.D. OP is a retard