- #1
evagelos
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In proving :|x|=0[tex]\Longrightarrow[/tex] x=0,the following indirect proof is offered:
Let |x|= 0 and suppose [tex]x\neq 0[/tex] then x>0 or x<0.
For x>0 |x| =x and since |x|=0 ,x=0
For x<0 |x| = -x and since |x|=0 ,x=0
So in either case x=0 ,which is a contradiction since we assumed [tex]x\neq 0[/tex]
Hence x=0.
Is there a direct shorter proof??
Let |x|= 0 and suppose [tex]x\neq 0[/tex] then x>0 or x<0.
For x>0 |x| =x and since |x|=0 ,x=0
For x<0 |x| = -x and since |x|=0 ,x=0
So in either case x=0 ,which is a contradiction since we assumed [tex]x\neq 0[/tex]
Hence x=0.
Is there a direct shorter proof??