0

I'm having issues proving that the multiplicative identity is unique on the integers. Heres what I have so far,

EDIT:

Suppose $\exists \ \theta_{1},\theta_{2} \ such \ that \ \theta_{1} \neq \theta_{2}$

$\theta_{1} = [(x+1,x)]$

$\theta_{2} = [(y+1,y)]$

$\theta_1 \otimes \theta_2 = [(m+1,m)] \otimes [(k+1,k)]$

$= [((m+1)(k+1) + (m)(k),(m)(k+1)+(m+1)(k))]$

$= [(mk + m + k + 1 + mk, mk + m + mk + k)]$

Since,

$mk + m + k + 1 + mk + m = mk + m + mk + k + m + 1$

We can say that

$(mk + m + k + 1 + mk, mk + m + mk + k)$ ~ $(m + 1,m)$

Thus $\theta_1 = \theta_1 \otimes \theta_2$

Similar argument for $\theta_2$

How do I go about showing that $\theta_1 = \theta_2$?

Gwagh
  • 1
  • If you want to prove there is only one object with some property, you cannot take two objects with that property and assume that they are equal, otherwise you are already assuming what you've not proven. – user21820 Dec 03 '14 at 02:34
  • I see, thanks for the heads up. Any idea on how I could proceed otherwise? – Gwagh Dec 03 '14 at 02:37
  • Adhvaitha has already provided an answer. I didn't want to as it would have been good for you to try yourself. Note that Adhvaitha's answer does not prove that it is unique, but merely that there is at most one multiplicative identity. You also need to prove that there is at least one. – user21820 Dec 03 '14 at 02:44
  • I don't see why you have "[(x,x+1)]", and where did $\alpha,\gamma,\theta$ come from? Also, the cancellation property is more complicated than what you are trying to prove. – user21820 Dec 03 '14 at 02:48
  • @user21820 I've already proven that the multiplicative identity exists. Just not sure where to go in terms of showing at most one exists. – Gwagh Dec 03 '14 at 03:14
  • Then use the trick in Adhvaitha's answer, because cancellation does not work if you choose the wrong $\alpha$. – user21820 Dec 03 '14 at 03:16
  • Why will the cancellation property NOT work if I choose the wrong alpha? – Gwagh Dec 03 '14 at 03:21
  • $0 \times 1 = 0 = 0 \times 2$. Obviously you can't cancel to get $1 = 2$. – user21820 Dec 03 '14 at 03:22
  • Ok, so we may not use the cancellative property. So the entire argument is constructed via adhvaitha's hint? – Gwagh Dec 03 '14 at 03:32
  • Yes. You can use the same trick to prove the uniqueness of additive inverses. – user21820 Dec 03 '14 at 03:59

1 Answers1

3

HINT $$\theta_1 = \theta_1 \otimes \theta_2 = \theta_2$$

Adhvaitha
  • 5,441
  • I think we are along the right track, would you have any suggestions on how to show the last steps? – Gwagh Dec 03 '14 at 04:05
  • 1
    @eidial this follows 100% from definitions. So if you are still having trouble, you need to review the definition of an identity. – rschwieb Dec 03 '14 at 04:20