If $x$ a positive integer so that $\forall y\in\mathbb{Z}$, $x\mid y$, then $x=1$. Why does taking $y=1$ complete the proof?

Question

The statement is this:

If $x$ is a positive integer so that $\forall y\in\Bbb Z$, $x|y$, then $x=1$.

So the solution given says take $y=1$. Then $x|1$, and then ended up saying then $xk=1$ for some integer $k$, and since $x$ is positive, $k$ is also positive. So that means that $xk\geq x$ which implies $x=1$ since x is positive.

So what if we take $y=2$?? Well, I mean, sure it works if $x=1$, but what about $y=3$? $y=15$??? We can list an infinite amount of $y$s, so why does just stating one possible $y$ complete this proof?

Answer 1

The assumption is that the condition holds for ALL $y$ in $\mathbb{Z}$, so it certainly holds if $y$ happens to be $1$.

Answer 2

$x|y$ for any integer $y$

Now if you choose to study $y=2$ for example you'll get that $x=1$ or $x=2$ (since $x|y$)

Which means that you won't have a specific answer if you choose the case $y=2$

Also for example if you take $y=15$ you'll get that $x=1,3,5,$ or $15$.

So again you don't have a specific value for $x$

But if you took $y=1$ you'll get that $x=1$ since $x|y$

In other words, (The statement is true for all integers) $\implies$ (it is true for $y=1$) $\implies$ ($x=1$)

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Another way to look at it is, the statement is true for all integers so it is true for $y=1$ and $y=2$ and $y=3$ and ...

Which gives you that ($x=1$) and ($x=1,$ or $2$) and ($x=1,$ or $3$) and ....

So to get the value of $x$ you must take the intersection of all these sets in paranthesis which is $1$

So $x=1$