How Can You Prove the Triangle Inequality Using Case Analysis in Discrete Math?

[cameron_c83]

Discrete Math -- Proof methods

Homework Statement

Prove |x-y| ≤ |x| + |y| for all real numbers x and y (where |x| represents the
absolute value of x, which equals x if x≥0 and equals -x if x<0). prove by cases

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

? You state the problem but say absolutely nothing else! Have you not even tried?

The problem itself says "prove by cases". What are the cases here?

 

[cyby]

Well, how many cases are there? You have cases for x $\geq$0 and x < 0, as well as the same for y. There seems to be 4 cases to work with...

 

[cameron_c83]

okay u right I should have postd my notes, but I wasnt sure so here it is :

1) x positive and y positive
2) x negitive and y positive
3) x positive and y negitive
4) x negitive and y negitive


case 1) p1 → q ,,, x-y ≤ x+y is true ,, for example 6-(+3) ≤ 6 + 3
case 2) p2 → q ,,, -x-y ≤ -x+y is true,, for example -6 - +3 ≤ -6 + 3
case 3) p3 → q ,,, x - (-y) ≤ x + (-y) is false ,,, for example 6 -(-3) = 9 and 6 +(-3) = 3
case 4) p4 → q ,,, -x-(-y) ≤ -x+(-y) is false ,,, for example -6 - (-3) = -3 and -6 + -3 = -9

so here are 4 cases and all possibilities,
please let me know what is wrong .

 

[HallsofIvy]

okay u right I should have postd my notes, but I wasnt sure so here it is :

1) x positive and y positive
2) x negitive and y positive
3) x positive and y negitive
4) x negitive and y negitive


case 1) p1 → q ,,, x-y ≤ x+y is true ,, for example 6-(+3) ≤ 6 + 3

Just posting an example does not prove it is true for all positive x and y.

case 2) p2 → q ,,, -x-y ≤ -x+y is true,, for example -6 - +3 ≤ -6 + 3
case 3) p3 → q ,,, x - (-y) ≤ x + (-y) is false ,,, for example 6 -(-3) = 9 and 6 +(-3) = 3

If x= 6, y= -3, |x+ y|= |6-(-3)|= 9 |x|+ |y|= 6+ 3= 9, NOT 6+(-3). $|x-y|\le |x|+ |y|$ is true in this example.

case 4) p4 → q ,,, -x-(-y) ≤ -x+(-y) is false ,,, for example -6 - (-3) = -3 and -6 + -3 = -9

If x= -6 and y= -3, then |x- y|= |-6-(-3)|= |-3|= 3 while |x|+ |y|= 6+ 3= 9. 3< 9. No, $|x- y|\le |x|+ |y|$ is true in this example. You are consistently forgetting the absolute values on the right side.

But "prove by cases" does not mean give examples! Examples cannot prove anything.

For example, if $x\ge 0$ and $y\ge 0$, in order to look at |x- y| we still have to consider two more cases:
1) x> y. Then |x- y|= x- y which is smaller than x. But |x|+ |y| is greater than x: $|x- y|\le |x|\le |x|+ |y|$.
2) y> x. Then |x- y|= |y- x|= y- x which is smaller than y. But |x|+ |y| is greater than y: [itex]|x- y|\le y\le |x|+ |y|.

so here are 4 cases and all possibilities,
please let me know what is wrong .