# Thread: Distance - absolute value proofs

1. ## Distance - absolute value proofs

Hey all, how would one go about proving the following?

abs|x - z| <= abs|x - y| + abs|y - z|

abs|x - y| >= abs| abs|x| - abs|y| |

where we know the following,

N (naturals), considered as a subset of R (reals), is not bounded above.
Z (integers) is not bounded above.
Z (integers) is not bounded below.
For each epsilon > 0, there exists n in N (naturals) such that 1/n < epsilon

Absolute value is defined to be x if x >= 0 and to be -x if x < 0.

Also, x<y if and only if x^2 < y^2
abs|x| < abs|y| if and only if x^2 < y^2

For all x,y in R (reals):

abs|x| = 0 if and only if x = 0
abs|xy| = abs|x| * abs|y|
- abs|x| <= x <= abs|x|
abs|x+y| <= abs|x| + abs|y|
If -y < x < y then |x| < |y|

Also, if you let x be real such that 0 <= x <= 1, and let m,n be natural such that m >= n, then x^m <= x^n

Thank you!

2. First I would prove the triangle inequality for two elements

$|a+b| \le |a|+|b|$

By the absolute value property we have the

$-|a|\le a \le |a|$ and $-|b|< b < |b|$
add these two equations together to get

$-(|a|+|b|) \le a+b \le (|a|+|b|)$
Now by definition of the absolute value again you get

$|a+b| < |a|+|b|$

Now here is a hint for the 2nd part

$|x-z|=|x-y+y+(-z)|$
Can you finish from here?