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I'm starting to do mathematical proofs by myself, and I have found this:

$\quad\forall x,y \in \mathbb{Z}\quad$ if $\quad a \cdot x = b \cdot y \quad$ then $\quad a = b = 0$

Proof:

Lets prove it by contradiction:

If $\quad a\neq b \Rightarrow a \cdot x = b \cdot y \Rightarrow \dfrac{a}{b}=\dfrac{y}{x} \neq 1 \quad$ but if $\quad x=y\neq 0 \Leftrightarrow \dfrac{y}{x}=1$

If $\quad a\neq 0 \Rightarrow a \cdot x = b \cdot y \Rightarrow \dfrac{a}{b}=\dfrac{y}{x} \neq 0 \quad$ but if $\quad y=0, x\neq 0 \Leftrightarrow \dfrac{y}{x}=0$

Therefore: $\quad\forall x,y \in \mathbb{Z}\quad$ if $\quad a \cdot x = b \cdot y \quad$ then $\quad a = b = 0$

Question

Is the assumption and its proof well stated? In the assumption I want to say that you can take any $x$ or $y$ in $\mathbb{Z}$ but I feel like writing $\quad\forall x,y \in \mathbb{Z}\quad$ doesn't clarify this.

DaguCode
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  • Hint for a different type of proof: $$ax=by, ay=bx\implies(a-b)(x+y)=0\implies a=b\implies a(x-y)=0\forall x, y\implies a=0\implies a=b=0.$$ – insipidintegrator Dec 11 '23 at 17:44
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    It's better if you avoid division while proving. That way you also don't need to worry about dividing by $0$. First substitute $x=y=1$ to get $a=b$. Then substitute $x=1, y=2$ to get that $a=0$. – Mark Dec 11 '23 at 17:50
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    Take $x=a$ and $y=-b$ – lhf Dec 11 '23 at 17:57

2 Answers2

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Another proof: Use $x=1,y=0 $ to conclude that $a=0$. Now use $x=0, y=1$ to get $b=0$,

Jens Schwaiger
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Taking $x=a$ and $y=-b$ in $ax=by$ yields $$ a^2+b^2=0. $$ This implies $a=b=0$. This finishes the proof.

Dietrich Burde
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  • I'm struggling with proofs that involve taking particular cases to prove a general statement. My thought is: if I take $x=a$ and $y=-b$ I will get $a=b=0$, but, to me, this doesn't prove that if I take another $x$ and $y$ the statement will hold. What I'm doing wrong? – DaguCode Dec 11 '23 at 21:49
  • You are doing wrong with the for all statement. It says that we have $ax=by$ for all $x,y\in \Bbb Z$. And for all really means for all $x,y$, so in particular for special choices, like $x=0$, $y=1$, or $x=1$,$y=0$. It is enough to obtain $a=b=0$ for one particular choice. Other choices will not contradict this, but just give less, e.g., for $x=y=1$ we obtain $a=b$. Also good. – Dietrich Burde Dec 11 '23 at 22:29
  • Now I understand, my problem was with the fact that there are only a few special choices that gives you all the information. Back again with the original post, the problem with the proof is that $x$ and $y$ could be $0$? I know that it's convenient to avoid divisions by $0$, but I don't see where is this a problem in this particular proof. For example, I don't have any problem with $ \dfrac{y}{x} = \infty \neq 1$. – DaguCode Dec 12 '23 at 09:59
  • You do have a problem, since $\frac{y}{0}$ is not defined. There is no infinity in $\Bbb Z$. – Dietrich Burde Dec 12 '23 at 10:10
  • And if I also clarify that $\dfrac{y}{x} \in [-\infty, \infty]$? – DaguCode Dec 12 '23 at 11:12
  • Inlcuding infinity and rules for this problem is like "using a sledgehammer to crack a nut". – Dietrich Burde Dec 12 '23 at 13:36