Prove abs(x)-abs(y) is less than or equal to abs(x-y)

[r0bHadz]

Homework Statement

Prove |x|-|y| ≤ |x-y|

Homework Equations

The Attempt at a Solution

So you have 2 cases with 2 subcases in each
(1)
|x|-|y| ≤ x-y if x-y≥0

and
(2)
|x|-|y| ≤ -x+y if x-y≤0

(1.1) if x≥0 and y≥0, the result |x|-|y| = x-y is an obvious one
(1.2) if x≥0 and y≤0, |x|-|y| ≤ x-y because if y is not zero but less than zero, x-y will hold a greater value than |x|-|y|

(2.1) If x ≤0 y≥0, |x|-|y| = -x - y
(2.2) If x≤0 y≤0, |x|-|y| = -x - |-y| which ≤ -x+y

Are these all of the cases?

Not only that, is this proof valid? I feel like the method is pretty trivial. I don't see how it requires a "proof" seeing as you you have to know, for example, say a≤0, then |a| = -a, that's literally the only thing you need to know for this problem

 

[andrewkirk]

It is not all of the cases. Since there are three absolute value operators, there are $2^3=8$ cases. But there may be ways of avoiding having to consider all cases.

An easier way is to use the triangle inequality $|a+b|\leq |a|+|b|$ for a metric space, with $a=y,\ b=x-y$. Whether you are allowed to do that will depend on the context of your question.

 

[r0bHadz]

It is not all of the cases. Since there are three absolute value operators, there are $2^3=8$ cases. But there may be ways of avoiding having to consider all cases.

An easier way is to use the triangle inequality $|a+b|\leq |a|+|b|$ for a metric space, with $a=y,\ b=x-y$. Whether you are allowed to do that will depend on the context of your question.

Hmm but it seems to me the only cases should be (a≥0 b≥0) (a≥0 b≤0) (a≤0 b≥0) (a≤0 b≤0)

For example, my book states that there are only 4 cases for |a+b| ≤ |a| + |b| as well, even though there should be 2^3 cases

I guess I'll have to think this one through a little bit more.

 

[andrewkirk]

In the context of the OP, the four cases not considered are:

1.3: $y\le 0;\ x \le 0$ where $x-y> 0$
1.4: $x\le 0;\ y > 0$ where $x-y> 0$
2.3: $y\le 0;\ x > 0$ where $x-y\le 0$
2.4: $x\le 0; y \le 0$ where $x-y\le 0$

The cases 1.4 and 2.4 can be dismissed as self-contradictory. But 1.3 ($y\le x \le 0$) and 2.3 ($x< y\le 0$) need to be considered.

 

[RPinPA]

An alternate approach: Square both sides.
For $a, b \geq 0, a \leq b \iff a^2 \leq b^2$. That's pretty easy to prove. Rearrange it as $b^2 - a^2 \geq 0$ and factor.

You can assume $|x| - |y| \geq 0$ because if not, the inequality obviously holds.
So if you can establish $(|x| - |y|)^2 \leq (x - y)^2$, then the result immediately follows.

$(|x| - |y|)^2 \leq (x - y)^2 \iff x^2 + y^2 - 2|x||y| \leq x^2 + y^2 - 2xy$ and that's easy to prove in a couple of lines. Recall $a \leq |a|$ for all real numbers.

 

[r0bHadz]

In the context of the OP, the four cases not considered are:

1.3: $y\le 0;\ x \le 0$ where $x-y> 0$
1.4: $x\le 0;\ y > 0$ where $x-y> 0$
2.3: $y\le 0;\ x > 0$ where $x-y\le 0$
2.4: $x\le 0; y \le 0$ where $x-y\le 0$

The cases 1.4 and 2.4 can be dismissed as self-contradictory. But 1.3 ($y\le x \le 0$) and 2.3 ($x< y\le 0$) need to be considered.

Hmm I think I see now. for |a+b| ≤ |a| + |b| there are only 4 cases we have to worry about, but for this problem the minus sign gives us extra cases

 

[r0bHadz]

Prove: |x|-|y|≤|x-y|
(1)
If x≥0, y≥0, x-y≥0
|x|-|y| = x-y ≤ x-y

(2)
If x≥0, y≥0, x-y≤0
|x|-|y| = x-y ≤ -x+y => -y≤y is true for y≥0

(3)
If x≥0, y≤0,x-y≥0
|x|-|y| = x-(-y) => x+y ≤x-y is true for y≤0

(4)
If x≥0, y≤0, x-y≤0
|x| - |y|= x+y ≤ -x+y can only be true if x=0 because "x≥0, y≤0, x-y≤0" doesn't work for any other value

(5)
If x≤0, y≥0, x-y≥0
|x|-|y| = -x - y ≤ x-y only if x=0 because "If x≤0, y≥0, x-y≥0" doesn't work for any other value

(6)
If x≤0, y≤0,x-y≥0
|x|-|y| = -x+y ≤ x-y since y≤0

(7)
If x≤0, y≥0, x-y≤0
|x|-|y| = -x -y ≤ -x+y => -y≤y which is true

(8)
If x≤0, y≤0, x-y≤0
|x|-|y| = -x+y ≤ -x+y is true Does this seem right? Obviously this isn't the most effective way to tackle this problem but I feel like it offers a lot of insight. It should have been obvious to me that there are 2^3 cases but I guess you have to learn things the hard way..

 

[andrewkirk]

Does this seem right?

It's broadly correct, but it suffers from a common flaw known as affirming the consequent. That's where we want to prove $B$, and we prove $B\to A$ and $A$ is known to be true. It does not follow that $B$ is true. The arrow points the wrong way!

eg in both (1) and (2) a true statement is derived, but that doesn't mean that what we started with was true.

What is needed is for the endpoint of the deductions to be the inequality we are trying to prove.

For example:

For (1) we write
$$|x|-|y| = x-y = |x-y|
\to |x|-|y| \le |x-y|$$

For (2) we write
$$|x|-|y|=x-y \le 0 \le |x-y|$$

For (3) we write
$$|x|-|y|=x+ y = x-y +2y \le x-y = |x-y|$$

For (4) we start by working with the conditions of the case:
$$x\ge 0\wedge y\le 0\wedge x-y\le 0
\to
x\ge 0\wedge y\le 0\wedge x-y\le 0\wedge x-y\ge 0
\to
x=y=0
\to
|x|-|y|= 0 = |x-y|
\to |x|-|y|\le |x-y|
$$

and so on.

 

[r0bHadz]

Very well written response, and very well explained. Thank you.