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If i show that $x+a=x+b$ only if $a=b$, does that prove that the above is also true?

$ x+a=x+b \iff x+a-x-b=0 \iff a-b=0 \implies b=a$ also is this any good?

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    @labbhattacharjee I think you might have missed the significance of the $\forall x \in \Bbb R$?2012-12-18

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That argument isn't going to work, since at no point are you using any properties of the floor function.

What will work is something like this (only a hint):

If $a \ne b$ then we can assume (without loss of generality) that $a < b$, and then we can take some $y$ such that $ a < y < b$. Can you then find some expression for $x$ which will make sure that $x+a$ and $x+b$ will have different integer parts?

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    @phi: Try the idea mentioned by Old John. In the example you give, pick $x=0.5$, or $0.6$.2012-12-18
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Let [x] be the whole part of x,

[x+a] = [x+b] for every real x => b-1 < [x+a]-x <= b and a-1 < [x+b]-x <= a for every real x this means that b-1 <= a-1 and a-1 <= b-1 so a = b.

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    Not sure if this is correct...2013-08-12