# Need help proving a couple theorems

• Jan 20th 2009, 10:27 PM
noles2188
Need help proving a couple theorems
Prove the following theorems.

1) If x is any number, then 0*x = 0.

2) If x is any number, there do not exist two numbers y and y' such that x*y = 1 and x*y' = 1.

Use any of the following axioms to help prove the above theorems:
1) If each of x and y is a number, then x*y is a number.
2) If each of x and y is a number, then x*y = y*x.
3) If each of x, y, and z is a number, then (x*y)*z = x*(y*z).
4) There is a number U such that if x is any number, then U*x = x.
• Jan 20th 2009, 11:09 PM
Jhevon
Quote:

Originally Posted by noles2188
Prove the following theorems.

1) If x is any number, then 0*x = 0.

this is actually an interesting one. you usually see this as an axiom. not always, but a lot of the time.

$\displaystyle 0 \cdot x = 0 \cdot (U \cdot x)$ ....axiom (4)

$\displaystyle = 0 \cdot \underbrace{(U + U + \cdots + U)}_{x \text{ times}}$

$\displaystyle = 0 \cdot U + 0 \cdot U + \cdots + 0 \cdot U$.....distributive property

$\displaystyle = 0 + 0 + \cdots + 0$ .......axiom (4)

$\displaystyle = 0$

Quote:

2) If x is any number, there do not exist two numbers y and y' such that x*y = 1 and x*y' = 1.
i suppose you mean distinct numbers y and y', of course.

assume to the contrary that there does exist distinct numbers y and y' such that

x*y = 1.........(a) and,
x*y' = 1 ........(b)

subtracting (b) from (a) we get

x*y - x*y' = 0

=> x(y - y') = 0

thus, x = 0 or y = y'

in either case, we have a contradiction. (do you see why? in particular, why does x = 0 yield a contradiction?)