Prove: |a-b|≤|a|+|b| using Definition of Absolute Value

[eme_girl]

Prove that for any vectors a and b, |a-b| is less than or equal to |a| + |b|

I'm kind of lost, b/c i can't see a case where |a-b| would actually result in a value being less than |a| + |b|.

I've tried doing a proof that is similar, and when I was taught, the definition of absolute value was used to prove it.

 

[vincentchan]

|x|>|y| if and only if |x|^2 > |y|^2

square both side and you will see the relationship

 

[wais]

Let a and b be any vectors. By the definition of absolute value, |x| is equal to x if x is greater than or equal to 0, and -x if x is less than 0. Therefore, |a| is equal to a or -a, and |b| is equal to b or -b.

Case 1: a and b are both positive or both negative

In this case, |a-b| is equal to a-b or -(a-b), and |a| + |b| is equal to a+b or -(a+b). Since a-b and a+b are both positive, it follows that |a-b| ≤ |a| + |b|.

Case 2: a and b have different signs

Without loss of generality, assume that a is positive and b is negative. In this case, |a-b| is equal to a-b, and |a| + |b| is equal to a-b. Again, it follows that |a-b| ≤ |a| + |b|.

Therefore, for any vectors a and b, |a-b| is less than or equal to |a| + |b|, which proves the statement.