Can anyone help me show that:
$||x|-|y||≤|x-y|$
I am new to proofs and I am not sure how I can show something as trivial as this!
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3Hint: $|x+z| \leq |x| + |z|$. Now take $z =$ something useful – Prahlad Vaidyanathan Nov 25 '13 at 08:57
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http://www.cut-the-knot.org/arithmetic/algebra/ComplexNumberInequalities.shtml – lab bhattacharjee Nov 25 '13 at 09:25
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1Assuming that the OP intends $;x,y \in \mathbb R;$, this is a duplicate of http://math.stackexchange.com/questions/193938/proving-absolute-value-inequalities. – MarnixKlooster ReinstateMonica Nov 25 '13 at 13:18
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You should prove the fact that for all $a,b \in \mathbb{R}$, $|a-b| = |b-a|$. Then:
By the triangle inequality, for all $a,b \in \mathbb{R}$, we have $|a+b| \leq |a| + |b|$.
Let $a=y$ and $b=x-y$. Those are in $\mathbb{R}$ because addition is closed. Then we have $|x| = |y+x-y|$ and $|y+x-y| \leq |y| + |x-y|$, i.e. $|x| \leq |y| + |x-y|$.
So $|x| - |y| \leq |x-y|$.
Now let $a=x$ and $b=y-x$. Then we have $|y| = |x+y-x|$ and $|x+y-x| \leq |x| + |y-x|$, i.e. $|y| \leq |x| + |x-y|$, i.e. $|y| - |x| \leq |x-y|$.
It follows that $-|x-y| \leq |x|-|y|$. Therefore, $-|x-y| \leq |x|-|y| \leq |x-y|$, which is equivalent to $||x|-|y|| \leq |x-y|$.
Newb
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Warning: I am incredibly tired and may have made typoes or other such mistakes. The form of the proof is perfectly correct, but I may have switched around $x$s and $y$s. Please read carefully and make sure everything makes sense. – Newb Nov 25 '13 at 09:49
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Hint: consider four cases, (1) $x > 0$ and $y > 0$, (2) $x < 0$ and $y > 0$, (3) $x > 0$ and $y < 0$, (4) $x < 0$ and $y < 0$.
Math Student 020
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Those $>$ signs should be $\ge$ signs, to cover the cases $x=0$ and/or $y=0$. – TonyK Nov 25 '13 at 08:58