# Is this a valid proof? Discrete Mathematics

I feel like there are many solutions for some proofs, and I want to make sure I'm still getting everything correct. This is the problem:

$$m^2 = n^2 (P)\quad\text{if and only if}\quad m = n (Q)\quad\text{or}\quad m = -n (Z)$$

So I need to prove $P \Rightarrow (Q \lor Z)$ and $(Q \lor Z) \Rightarrow P$ because this is a bi-conditional statement, yes? This is my proof.

Proving $P \Rightarrow (Q \lor Z)$:

Let's assume $m$ doesn't equal $n$ (proving by contra-position). It is said that $m^2 = n^2$. Take the square root of both sides and you are left with $m = n$. However, we are assuming $m$ doesn't equal $n$ so there is a contradiction.

And I pretty much did the reverse for $(Q \lor Z) \Rightarrow P$ by squaring both sides to reach the contradiction.

Is this a valid proof or is something going over my head?

• It appears like you are using circular reasoning. – GoodDeeds Mar 8 '16 at 20:08
• You may be forgetting about negative numbers. Note that for example $(-3)^2=3^2$. – André Nicolas Mar 8 '16 at 20:09
• @AndréNicolas so how would I go about solving this? – Laugh7 Mar 8 '16 at 20:11
• The easier part: If $m=-n$, then $m^2=(-n)^2=n^2$. For the other direction, maybe use $m^2-n^2=(m+n)(m-n)$. – André Nicolas Mar 8 '16 at 20:13

So I need to prove $P \to (Q \vee Z)$ and $(Q \vee Z) \to P$ because this is a bi-conditional statement, yes?

Yes.

Let's assume $m$ doesn't equal $n$ (proving by contra-position). It is said that $m^2 = n^2$. Take the square root of both sides and you are left with $m = n$. However, we are assuming $m$ doesn't equal $n$ so there is a contradiction.

There are a couple of flaws here. First, the outline of what you did is "We want to prove $R$. So assume $\neg R$. Then, blah blah blah, $R$. But we assumed $\neg R$. This is a contradiction." But you only contradicted your original assumption $\neg R$. In the middle, you proved $R$. So in proofs like that you don't need the proof-by-contradiction framework. You just did a direct proof.

Or did you? You said $m^2 = n^2 \implies m = n$ “by taking the square root of both sides.” But this isn't valid. For $2^2 = (-2)^2$ while $2 \neq -2$. So something went wrong, and it was that you lost a solution by assuming $m$ and $n$ were both positive.

With nonlinear algebraic equations, it's often safer to set the equation equal to zero and factor. In this case: $$m^2 = n^2 \implies m^2 - n^2 = 0 \implies (m-n)(m+n) = 0$$ Two numbers have a product of zero if and only if one of them is zero. So either $m-n = 0$ (in which case $m=n$) or $m+n=0$ (in which case $m=-n$).

Others have pointed out the flaws in your proof.

Hint for a correct proof: Show that $$(n-m)(n+m)=n^2-m^2$$

So if $n^2=m^2$ then what?

Bigger hint: You need to know (or prove) the rule that $ab=0$ if and only if $a=0$ or $b=0$.