0

I wrote a proof and I'm really just here to make sure my argument is making sense.

For all x>=0, (x^2)-x is even.

Base case: x=0. (0^2)-0 is even.
 Inductive Step: x>= 0. Suppose (x^2)-x is even. Then there is an integer y so that (x^2)-x = 2y

    Now (x+1)^2 - (x+1) = x^2 +2x + 1 - x - 1.
                       = x^2 + x
                       = (x^2 - x) + 2x
                       = 2y + 2x
                       = 2(y+x)
    Thus it's even.

Does that argument make sense? It checks out to me..

  • 1
    Yup! Definitely makes sense. Just as an interesting note, observe that $x^2 - x = x(x - 1)$. So if $x \ge 0$ is even, then $x(x - 1)$ is an even time an odd so that it's even. If $x \ge 0$ is odd, then $x - 1$ is even so that $x(x - 1)$ is also even. – AlkaKadri Nov 20 '17 at 17:59
  • @AlkaKadri I'm not sure how the question got deleted.. but I am actually confused on something, I kinda did this based off an example I had. How do you go from x^2 + x to (x^2 - x) + 2x? –  Nov 20 '17 at 19:33
  • That's just an algebraic manipulation. Observe that $x = 2x - x$. So you just rewrite $x^2 + x = x^2 + (2x - x) = x^2 -x + 2x = (x^2 - x) + 2x$. Don't let the bracketing confuse you. – AlkaKadri Nov 20 '17 at 21:41

1 Answers1

0

Classic induction. Looks fine to me.

Fabian Schn.
  • 908
  • Okay I did this based on an example but now I am slightly confused, how do you go from x^2 + x to (x^2 - x) + 2x? I don't understand how my example did it I just kinda changed the numbers in the example and it still worked out –  Nov 20 '17 at 19:34
  • you added $x-x$, which is $0$. So $$ x^2+x = x^2 + x + 0 = x^2 + x + x- x + x^2 - x + 2x $$ – Fabian Schn. Nov 20 '17 at 21:51