Proving |x + y| ≥ |x| - |y| using Theorem 3 and the fact that |-y| = |y|

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Homework Help Overview

The discussion revolves around proving the inequality |x + y| ≥ |x| - |y|, utilizing Theorem 3, which states that |a + b| ≤ |a| + |b|, and the property that |-y| = |y|. Participants are exploring how to apply these concepts to the given inequality.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to manipulate the expression x = x + y - y and question its relevance to the proof. There are discussions about the implications of rewriting expressions and how to apply Theorem 3 correctly. Some participants express confusion about the relationship between absolute values and the steps taken in the proof.

Discussion Status

The discussion is active with participants sharing their thoughts and clarifying each other's points. Some have offered guidance on how to apply Theorem 3, while others are questioning the validity of certain steps and assumptions. There is a mix of interpretations being explored, but no explicit consensus has been reached yet.

Contextual Notes

Participants are grappling with the application of absolute value properties and the conditions under which Theorem 3 can be applied. There are mentions of potential errors in reasoning and the need for careful handling of absolute values in the context of the proof.

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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
 
Last edited:
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jimmyly said:

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?
 
hi jimmyly! :smile:
jimmyly said:
x = x + y - y (not sure why this is true …

add some brackets …

x = x + (y - y)

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

add some brackets in a different place :wink:
 
hi jimmyly! :smile:
jimmyly said:
… I just don't know what that has to do with proving it.

y - y = 0

x + (y - y) = x + 0

x + 0 = x ? o:)
 
@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.
 
Tornado Dragon said:
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?
 
tiny-tim said:
hi jimmyly! :smile:


y - y = 0

x + (y - y) = x + 0

x + 0 = x ? o:)

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|
 
Tornado Dragon said:
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.



jimmyly said:
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.
 
jimmyly said:
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) :wink:
 
  • #10
@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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  • #11
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? :)
 
  • #12
Mark44 said:
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

Tornado Dragon said:
@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!
 
  • #13
jimmyly said:
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?
 
  • #14
jimmyly said:
|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 :smile:​
 
  • #15
Dick said:
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|
 
  • #16
oh that just did it didn't it?!
 
  • #17
jimmyly said:
oh that just did it didn't it?!

Sure did!
 
  • #18
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!
 
  • #19
Dick said:
Sure did!

Wow that's amazing. Thanks everyone! You are all wonderful
 
  • #20
jimmyly said:
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! :smile:
 
  • #21
jimmyly said:
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".
 
  • #22
tiny-tim said:
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! :smile:

Dick said:
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! :)
 

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