MHB Trouble proving multiplication cancellation law fails

[skate_nerd]

The problem given says to consider the addition and multiplication \(\oplus\) and \(\odot\) on \(\mathbb{Z}_6\) defined by
$$[m]\oplus[n]=[m+n]$$ and $$[m]\odot[n]=[mn].$$I need to verify, among other things, that the multiplication cancellation law fails (if \([m]\odot[j]=[m]\odot[k]\) and \([m]\neq{0}\), then \([j]=[k]\) ).
With a bit of inspection, I began to believe that it in fact does not fail, and I thought I constructed a valid proof showing why, until my professor told me otherwise.

Suppose that \([m]\odot[j]=[m]\odot[k]\). This can be written as \([mk]=[mj]\). This holds if \([mj]=[mk+6p]\) for any integer \(p\).
So we have that
$$mj=mk+6p$$ $$6p=mj-mk$$ $$6p=m(j-k)$$ $$6\frac{p}{m}=j-k$$ \(j=k+6\frac{p}{m}\) where \(\frac{p}{m}\) is an integer. This is the same thing as writing \([j]=[k]\), because \([j]=[k+6\frac{p}{m}]\).
I am pretty sure that this proves the multiplication cancellation law holds but apparently that isn't the case, making me feel like I really don't understand this topic well...

If anybody could help show me where my logic goes south, that would be great. Thanks guys

 

[Evgeny.Makarov]

Re: trouble proving multiplication cancellation law fails

$$6p=m(j-k)$$ $$6\frac{p}{m}=j-k$$ \(j=k+6\frac{p}{m}\) where \(\frac{p}{m}\) is an integer.

The fact that $j-k$ is an integer does not imply that $p/m$ is, although it does imply that $6p/m$ is an integer.

I am pretty sure that this proves the multiplication cancellation law holds but apparently that isn't the case, making me feel like I really don't understand this topic well...

To grasp what's going on, go through the proof that cancellation law holds when there are no zero divisors. See where the absence of zero divisors is used. Then find zero divisors in $\mathbb{Z}_5$ and construct a counterexample to your statement. Go through your proof and see exactly which step is wrong.

 

[Deveno]

Re: trouble proving multiplication cancellation law fails

[2]*[3] = [0]

[0]*[3] = [0]

[2] does not equal [0].

 

[skate_nerd]

Re: trouble proving multiplication cancellation law fails

The fact that $j-k$ is an integer does not imply that $p/m$ is, although it does imply that $6p/m$ is an integer.

To grasp what's going on, go through the proof that cancellation law holds when there are no zero divisors. See where the absence of zero divisors is used. Then find zero divisors in $\mathbb{Z}_5$ and construct a counterexample to your statement. Go through your proof and see exactly which step is wrong.

A couple things:
Did you mean to say find zero divisors in $\mathbb{Z}_5$? Because the problem is considering $\mathbb{Z}_6$.
Also, you say that $j-k$ being an integer DOES imply that $6\frac{p}{m}$ is also an integer. So if this is true, then my conclusion would still prove the multiplication cancellation law does not fail, right? I concluded that $[j]=[k+6\frac{p}{m}]$, meaning \(j\) equals \(k\) plus some integer.
Would this not be sufficient proof because $6\frac{p}{m}$ has some values where it would not make $[j]=[k]$?

 

[skate_nerd]

Re: trouble proving multiplication cancellation law fails

By the way, I found a counter-example that does prove that the law fails. It fails when $[j]=[0]$, $[k]=[2]$, and $[m]=[3]$. I am just still a bit perplexed as to why the proof I constructed initially is invalid.

 

[Evgeny.Makarov]

Re: trouble proving multiplication cancellation law fails

Did you mean to say find zero divisors in $\mathbb{Z}_5$?

In $\mathbb{Z}_6$.

Also, you say that $j-k$ being an integer DOES imply that $6\frac{p}{m}$ is also an integer. So if this is true, then my conclusion would still prove the multiplication cancellation law does not fail, right? I concluded that $[j]=[k+6\frac{p}{m}]$, meaning \(j\) equals \(k\) plus some integer.

To conclude that $[j]=[k]$, you need to show that $j=k+6q$ where $q$ is an integer, not that $j=k+q$ where $q$ is an integer. You showed that $j=k+6p/m$, but $p/m$ is not necessarily an integer.