Is this proof correct? Establishing that $|x+y| geq | |x| - |y| |$
$begingroup$
Using that $|a|+|b|geq|a+b|$
begin{align}
|-x|+|x+y| geq |-x+x+y| = |y|\
|-y|+|x+y| geq |-y+x+y| = |x|
end{align}
Substracting $|-x|$ from the first inequality and $|-y|$ from the second:
begin{align}
|x+y| geq |y|-|-x| \
|x+y| geq |x| - |-y|
end{align}
Using the fact that if $a geq b$ and $a geq -b$ then $a geq |b|$.
begin{align}
|x+y| geq ||x| - |-y|| = ||x| - |y||
end{align}
Is the above correct?
real-analysis elementary-number-theory inequality
asked Dec 18 '18 at 23:26
PintecoPinteco
731313
$endgroup$
add a comment |
$begingroup$
Using that $|a|+|b|geq|a+b|$
begin{align}
|-x|+|x+y| geq |-x+x+y| = |y|\
|-y|+|x+y| geq |-y+x+y| = |x|
end{align}
Substracting $|-x|$ from the first inequality and $|-y|$ from the second:
begin{align}
|x+y| geq |y|-|-x| \
|x+y| geq |x| - |-y|
end{align}
Using the fact that if $a geq b$ and $a geq -b$ then $a geq |b|$.
begin{align}
|x+y| geq ||x| - |-y|| = ||x| - |y||
end{align}
Is the above correct?
real-analysis elementary-number-theory inequality
asked Dec 18 '18 at 23:26
PintecoPinteco
731313
$endgroup$
3
$begingroup$
The proof is good!
$endgroup$
– Theo Bendit
Dec 18 '18 at 23:34
1
$begingroup$
Seems good enough!
$endgroup$
– Rebellos
Dec 18 '18 at 23:40
1
$begingroup$
Nice work. See also Help checking proof of $|x| - |y| leq |x+y|$
$endgroup$
– amWhy
Dec 18 '18 at 23:57
$begingroup$
Perfect. But I would have typed $|(|x|-|y|)|$ or $|; |x|-|y|;|$(etc.) to make it easier to read, and to avoid confusion with the functional-analysis symbol $||z||$ (also written $|z|,$ coded as |z|). You can use ; and , to add a little space between key-strokes.
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:16
add a comment |
$begingroup$
Using that $|a|+|b|geq|a+b|$
begin{align}
|-x|+|x+y| geq |-x+x+y| = |y|\
|-y|+|x+y| geq |-y+x+y| = |x|
end{align}
Substracting $|-x|$ from the first inequality and $|-y|$ from the second:
begin{align}
|x+y| geq |y|-|-x| \
|x+y| geq |x| - |-y|
end{align}
Using the fact that if $a geq b$ and $a geq -b$ then $a geq |b|$.
begin{align}
|x+y| geq ||x| - |-y|| = ||x| - |y||
end{align}
Is the above correct?
real-analysis elementary-number-theory inequality
asked Dec 18 '18 at 23:26
PintecoPinteco
731313
$endgroup$
Using that $|a|+|b|geq|a+b|$
begin{align}
|-x|+|x+y| geq |-x+x+y| = |y|\
|-y|+|x+y| geq |-y+x+y| = |x|
end{align}
Substracting $|-x|$ from the first inequality and $|-y|$ from the second:
begin{align}
|x+y| geq |y|-|-x| \
|x+y| geq |x| - |-y|
end{align}
Using the fact that if $a geq b$ and $a geq -b$ then $a geq |b|$.
begin{align}
|x+y| geq ||x| - |-y|| = ||x| - |y||
end{align}
Is the above correct?
real-analysis elementary-number-theory inequality
real-analysis elementary-number-theory inequality
asked Dec 18 '18 at 23:26
PintecoPinteco
731313
asked Dec 18 '18 at 23:26
PintecoPinteco
731313
asked Dec 18 '18 at 23:26
PintecoPinteco
731313
asked Dec 18 '18 at 23:26
PintecoPinteco
731313
asked Dec 18 '18 at 23:26
PintecoPinteco
731313
731313
3
$begingroup$
The proof is good!
$endgroup$
– Theo Bendit
Dec 18 '18 at 23:34
1
$begingroup$
Seems good enough!
$endgroup$
– Rebellos
Dec 18 '18 at 23:40
1
$begingroup$
Nice work. See also Help checking proof of $|x| - |y| leq |x+y|$
$endgroup$
– amWhy
Dec 18 '18 at 23:57
$begingroup$
Perfect. But I would have typed $|(|x|-|y|)|$ or $|; |x|-|y|;|$(etc.) to make it easier to read, and to avoid confusion with the functional-analysis symbol $||z||$ (also written $|z|,$ coded as |z|). You can use ; and , to add a little space between key-strokes.
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:16
add a comment |
3
$begingroup$
The proof is good!
$endgroup$
– Theo Bendit
Dec 18 '18 at 23:34
1
$begingroup$
Seems good enough!
$endgroup$
– Rebellos
Dec 18 '18 at 23:40
1
$begingroup$
Nice work. See also Help checking proof of $|x| - |y| leq |x+y|$
$endgroup$
– amWhy
Dec 18 '18 at 23:57
$begingroup$
Perfect. But I would have typed $|(|x|-|y|)|$ or $|; |x|-|y|;|$(etc.) to make it easier to read, and to avoid confusion with the functional-analysis symbol $||z||$ (also written $|z|,$ coded as |z|). You can use ; and , to add a little space between key-strokes.
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:16
3
3
$begingroup$
The proof is good!
$endgroup$
– Theo Bendit
Dec 18 '18 at 23:34
$begingroup$
The proof is good!
$endgroup$
– Theo Bendit
Dec 18 '18 at 23:34
1
1
$begingroup$
Seems good enough!
$endgroup$
– Rebellos
Dec 18 '18 at 23:40
$begingroup$
Seems good enough!
$endgroup$
– Rebellos
Dec 18 '18 at 23:40
1
1
$begingroup$
Nice work. See also Help checking proof of $|x| - |y| leq |x+y|$
$endgroup$
– amWhy
Dec 18 '18 at 23:57
$begingroup$
Nice work. See also Help checking proof of $|x| - |y| leq |x+y|$
$endgroup$
– amWhy
Dec 18 '18 at 23:57
$begingroup$
Perfect. But I would have typed $|(|x|-|y|)|$ or $|; |x|-|y|;|$(etc.) to make it easier to read, and to avoid confusion with the functional-analysis symbol $||z||$ (also written $|z|,$ coded as |z|). You can use ; and , to add a little space between key-strokes.
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:16
$begingroup$
Perfect. But I would have typed $|(|x|-|y|)|$ or $|; |x|-|y|;|$(etc.) to make it easier to read, and to avoid confusion with the functional-analysis symbol $||z||$ (also written $|z|,$ coded as |z|). You can use ; and , to add a little space between key-strokes.
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:16
add a comment |
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3
$begingroup$
The proof is good!
$endgroup$
– Theo Bendit
Dec 18 '18 at 23:34
1
$begingroup$
Seems good enough!
$endgroup$
– Rebellos
Dec 18 '18 at 23:40
1
$begingroup$
Nice work. See also Help checking proof of $|x| - |y| leq |x+y|$
$endgroup$
– amWhy
Dec 18 '18 at 23:57
$begingroup$
Perfect. But I would have typed $|(|x|-|y|)|$ or $|; |x|-|y|;|$(etc.) to make it easier to read, and to avoid confusion with the functional-analysis symbol $||z||$ (also written $|z|,$ coded as |z|). You can use ; and , to add a little space between key-strokes.
$endgroup$
– DanielWainfleet
Dec 19 '18 at 3:16