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I've decided to learn the basics of proofs and here is my first attempt. Could I improve or simplify my proof in any way? Is my formal language correct? Thanks!

Let $n$ be any integer. $n(n+2) = (n+1)^2 - 1$ I will prove this identify by induction.

First, check with $n=1$;

$1 \times 3 = 2^2 + 1 \equiv 3 = 3$

Inductive step: Assume that the identity is true for n = k;

$k(k+2) = (k+1)^2 - 1$

When $n = k + 1$;

$(k+1)((k+1)+2) = ((k+1)+1)^2 - 1$

$\equiv k+1(k+3) = (k+2)^2 - 1$

Let $n = k + 1$;

$n(n+2) = (n+1)^2 - 1$

  • 0
    It's like using induction to prove $n \times n = n^2$.2014-06-20

3 Answers 3

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Your proof is correct (except for parentheses missing around $k+1$ somewhere in the induction step), but since you're looking for advice: you could be clearer halfway through the induction step, like: "the l.h.s. yields ..., while the r.h.s. yields ...", making it easier to read. Finally Alo Mestanogullari's comment is spot on; if you'd present this proof more formally, you could use terms like "because of the distributive law..." etc.

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Let $n$ be an integer. $n(n+2) = n^2 + 2n = n^2 + 2n + 1 - 1 = (n+1)^2 - 1$

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Using $a^2 - b^2 = (a-b)(a+b)$ we have $(n+1)^2 - 1 = (n+1)^2 - 1^2 = (n+1-1)(n+1+1) = n(n+2).$