Absolute value of two numbers

Question

I want to verify and to check with you this inequality |a| > |b| implies |a-b| > 0 Is it correct?

Answer 1

Direct proof:

$|a|>|b|\implies$

$(a>b)\vee(a<b)\implies$

$a \neq b\implies$

$a-b\neq0\implies$

$|a-b|\neq0\implies$

$|a-b|>0$

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Proof by contradiction:

$|a-b|\not>0$

$|a-b|=0$

$a-b=0$

$a=b$

$|a|=|b|$

$|a|\not>|b|$

Answer 2

$|a| > |b| \Rightarrow a \not= \pm b \Rightarrow a - b \not= 0 \Rightarrow |a - b| > 0$

Answer 3

In fact $a\ne b$ implies $|a-b|>0$.

Just notice that:

• $a\ne b$ $\Rightarrow$ $a-b\ne0$
• $x\ne 0$ $\Rightarrow$ $|x|>0$
• Use $x=a-b$.