# Uniqueness proof

1. Oct 31, 2012

### nicnicman

To get the following proof I followed another similar example, but I'm not sure if it's correct. Does this proof properly show existence and uniqueness?

Show that if x is a nonzero rational number, then there is a unique rational number y such that xy = 2

Solution:
Existence: The nonzero rational number y = 2/x is a solution of xy = 2 because x(2/x) = 2 = x(2/x) - 2 = x - x = 0.

Uniqueness: Suppose s is a nonzero rational number such that xs = 2. Then, xy =2 = xy - 2 = 0 and xs = 2 = xs - 2 = 0. Then:

xy - 2 = xs - 2
xy = xs
y = s

This would be a complete proof wouldn't it?

2. Oct 31, 2012

### Klungo

It could be cleaner.

First of all, use -> instead of = (for implication) since
xy = 2 = xy - 2=0 is a contradiction. Whereas xy = 2 -> xy-2=0 works.

Second, for the first part. It is sufficient to say xy=2. That is, let y=2/x, then xy=(x)(2/x)=2. You must make the restriction x =/= 0.

Third, Try to prove the uniqueness part of the proof more cleanly. Think about what you've done versus xs=2 and what xy is.

3. Nov 1, 2012

### Bacle2

If you're allowed to use the field properties of the rationals, you can then also show that

multiplicative inverses are unique , so that :

xy=2 -> x-1xy=x-12 , so that y=2x-1 (by

commutativity).

Last edited: Nov 1, 2012