prove that |a-x|=|x-a| where a is any constant for all values of x?

Question

Proof:

Answer 1

If the value in an absolute value is positive, it remains the same. If it is negative, it is multiplied by -1.

Either a x. (I think this is called trichotomy?)

If a = x, then both sides equal 0.

If a > x, then a-x is positive, and x - a is negative, so...
a - x = -1(x - a)

a - x If a > x, then x-a is positive, and a - x is negative, so...
-1(a - x) = x - a
x - a = x - a

a - x = a - x

QED

Answer 2

Let x be any real number
Let a be a real number
a-x > 0 if a>x
a-x < 0 if a x-a >0 if x>a
x-a <0 if x Take a>x, then a-x >0 and x-a < 0, but |a-x| > 0 by defn and x-a>0 so that |a-x| = |x-a|
Now take the other case, and you are done