# Proving this inequality

## Homework Statement

|x + y| ≥ |x| - |y| [Hint: write out x = x + y - y, and apply Theorem 3, together with the fact that |-y| = |y|]

## Homework Equations

Theorem 3: |a + b| ≤ |a| + |b|

x = x + y - y

|-y| = |y|

## The Attempt at a Solution

|x + y| ≥ |x| - |y|

x = x + y - y (don't know where to use it)

xy ≤ |xy| = |x| |y| ( I really don't know why I am taking these steps, I am pretty much following the proof of theorem 3 in this book)

2xy ≤ 2|x||y| ( does the |-y| = |y| come into play here to flip the inequality? )

(x+y)^2 = x^2 + 2xy + y^2 ≤ x^2 - 2|x||y| + y^2 = (|x| - |y|)^2

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pasmith
Homework Helper

## Homework Statement

|x + y| ≥ |x| - |y| [Hint: write out x = x + y - y, and apply Theorem 3, together with the fact that |-y| = |y|]

## Homework Equations

Theorem 3: |a + b| ≤ |a| + |b|

x = x + y - y

|-y| = |y|

## The Attempt at a Solution

|x + y| ≥ |x| - |y|

x = x + y - y (not sure why this is true, and don't know where to use it)

Is it not obvious that x + y - y = x + (y - y) = x + 0 = x?

tiny-tim
Homework Helper
hi jimmyly! x = x + y - y (not sure why this is true …

x = x + (y - y)

now can you see why it's true? … and don't know where to use it)

add some brackets in a different place tiny-tim
Homework Helper
hi jimmyly! … I just don't know what that has to do with proving it.

y - y = 0

x + (y - y) = x + 0

x + 0 = x ? @mark-44: How was it false? I mean if you could clarify that would be great. Wasn't trying to spread false information or anything.

You have to remember that |x|= x so for |-x| it becomes -(-x)= + x.

So When you change || to how you normally write it out you get: x + y>= x + y Which can then be written as:

I don't understand why I would write x + y ≥ x + y
how does the subtraction turn into addition? does this mean that x + y = |x| - |y|? I see that -(-x) = + x
but its |x| - |y| not |x| + |-y|
or is my thinking wrong?

hi jimmyly! y - y = 0

x + (y - y) = x + 0

x + 0 = x ? Thank you, I understand why x = x + y - y it is very obvious! I am confused because I have no idea what it has to do with the proof and where to apply it when proving that |x+y| >= |x| - |y|

Mark44
Mentor
You have to remember that |x|= x so for |-x| it becomes -(-x)= + x.

So When you change || to how you normally write it out you get: x + y>= x + y

Tornado Dragon's post was full of errors, so I deleted it. Right at the start he tells you to remember that |x|= x. This is not true in general, particularly if x happens to be negative.

I don't understand why I would write x + y ≥ x + y
how does the subtraction turn into addition? does this mean that x + y = |x| - |y|? I see that -(-x) = + x
but its |x| - |y| not |x| + |-y|
or is my thinking wrong?
No, it's not. You can't just ignore the absolute value signs, which Tornado Dragon seems to have done.

tiny-tim
Homework Helper
Thank you, I understand why x = x + y - y it is very obvious! I am confused because I have no idea what it has to do with the proof and where to apply it when proving that |x+y| >= |x| - |y|

apply Theorem 3 to x = x + y - y (with brackets in a suitable place) @jimmyly ignore my post I have made a few mistakes in it. I will get back to you on it once I have gone it over it again.

edit: @mark44&jimmlyl: I did make en eroneous error but I did not ignore the absolute signs like that I will redo it and post it at a later tim.e.

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okay so here is what i'm doing right now

|x+y| >= |x| - |y|
with x = x + y - y
I got
|x + y| >= |x| + |y| - |y| - |y|
cancelling the |y|
|x + y| >= |x| + |y|
am I on the right track? :)

Tornado Dragon's post was full of errors, so I deleted it. Right at the start he tells you to remember that |x|= x. This is not true in general, particularly if x happens to be negative.

No, it's not. You can't just ignore the absolute value signs, which Tornado Dragon seems to have done.

understandable

@jimmyly ignore my post I have made a few mistakes in it. I will get back to you on it once I have gone it over it again.

edit: @mark44&jimmlyl: I did make en eroneous error but I did not ignore the absolute signs like that I will redo it and post it at a later tim.e.

no worries we all make mistakes!

Dick
Homework Helper
Thank you, I understand why x = x + y - y it is very obvious! I am confused because I have no idea what it has to do with the proof and where to apply it when proving that |x+y| >= |x| - |y|

Take your theorem 3, |a+b|<=|a|+|b|. Put a=x+y and b=(-y). What happens?

tiny-tim
Homework Helper
|x + y| >= |x| + |y| - |y| - |y|

am I on the right track? :)

no!!

Theorem 3 applies to 3 things

use brackets to reduce the number of things to 3 !!

then you can apply Theorem 3 Take your theorem 3, |a+b|<=|a|+|b|. Put a=x+y and b=(-y). What happens?

so from this I got

|a + b| <= |a| + |b|
|(x+y) + (-y)| <= |x+y| + |-y|
|x + y - y| <= |x+y| + |-y|
|x| - |y| <= |x + y|

oh that just did it didn't it?!

Dick
Homework Helper
oh that just did it didn't it?!

Sure did!

Would this be classified as a direct proof? I'm trying to learn proofs on my own so this is a little bit confusing to me. Thanks everyone for helping me out!

Sure did!

Wow that's amazing. Thanks everyone! You are all wonderful

tiny-tim
Homework Helper
Would this be classified as a direct proof? I'm trying to learn proofs on my own …

jimmyly, you seem to be worrying that there's something special about proofs

there isn't

if it starts with the question and finishes with the answer, it's a proof! Dick
Homework Helper
Would this be classified as a direct proof? I'm trying to learn proofs on my own so this is a little bit confusing to me. Thanks everyone for helping me out!

Yes, it's a direct proof. Indirect proofs work by assuming what you are trying to prove is false. Then you show that leads to a logical contradiction. So what you are trying to prove must be true. Also called "proof by contradiction".

jimmyly, you seem to be worrying that there's something special about proofs

there isn't

if it starts with the question and finishes with the answer, it's a proof! Yes, it's a direct proof. Indirect proofs work by assuming what you are trying to prove is false. Then you show that leads to a logical contradiction. So what you are trying to prove must be true. Also called "proof by contradiction".

Thank you! :)