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i want to show that the multiplicative semigroup of the natural numbers without zero are cancellative, that is: For all $h,x,y$, $hx=hy \implies x=y$

I tried proving this by induction over $h$ and letting $x,y$ fixed.

For $h=1$ it is trivial.

But now I’m stuck in the induction step, I gotta show that $(h+1)x=(h+1)y \implies x=y$ But I don’t know how to do that, I tried to write it like this $hx+x=hy+y$ but that doesn’t help :(

Can anyone help me?

J.-E. Pin
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Blue2001
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  • The way to do this with induction is to induct on $x$. In the induction step, you have $hx + h = hy$. From this it follows that $y$ can't be $1$, so say $y = y' + 1$. Then $hx = hy'$ so now you can apply the inductive hypothesis and deduce $x + 1 = y$. – Izaak van Dongen Jun 14 '22 at 20:35
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    You're having trouble because you're inducting on $h$ rather than $x$ and $y$!

    After all, if we aren't using any special properties of $x$ and $y$, then your same proof would show that $hx = hy \implies x=y$ for $x$ and $y$ coming from any additive semigroup (after all, in any additive semigroup we can define an action $hx = \underbrace{x + x + \cdots + x}_{h \text{ times}}$, and since your proof is inducting on $h$, it would work in this setting).

    Of course, this theorem is false in some semigroups (I'll leave this as an easy exercise) so induction on $h$ can't possibly work.

     – HallaSurvivor Jun 14 '22 at 20:36

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Performing a proof by contraposition.

Without loss of generality, suppose that we have $x \gt y$. It means that it exists $a \in \mathbb N$ such that $x=y+a$. But then $hx = h y + h a$ and therefore $hx \neq hy$. And we're done.

mathcounterexamples.net
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