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

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SUMMARY

The discussion centers on proving the inequality |x + y| ≥ |x| - |y| using Theorem 3, which states that |a + b| ≤ |a| + |b|. Participants explore the application of the theorem by rewriting x as x + y - y and utilizing the property |-y| = |y|. The conversation reveals confusion regarding the manipulation of absolute values and the correct application of the theorem, ultimately leading to a consensus that the proof is direct and valid.

PREREQUISITES
  • Understanding of absolute value properties
  • Familiarity with Theorem 3: |a + b| ≤ |a| + |b|
  • Basic algebraic manipulation skills
  • Knowledge of direct proof techniques in mathematics
NEXT STEPS
  • Study the properties of absolute values in depth
  • Practice applying Theorem 3 in various mathematical proofs
  • Learn about direct and indirect proof methods
  • Explore examples of inequalities involving absolute values
USEFUL FOR

Students learning mathematical proofs, particularly those studying inequalities and absolute values, as well as educators seeking to clarify proof techniques in a classroom setting.

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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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