# Show that if $\lfloor x+a \rfloor$ = $\lfloor x+b \rfloor, \forall x \in \Bbb R$ then $a=b$; is showing that $x+a=x+b$ enough?

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?

-
It isn't enough, does not address at all the floor function part. Hint: Pick $a=3.14$ and $b=3.2$, to see what's going on. – André Nicolas Dec 18 '12 at 16:43
@labbhattacharjee I think you might have missed the significance of the $\forall x \in \Bbb R$? – Old John Dec 18 '12 at 16:57

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?

-
the equation is true even if $a$ is not equal $b$....for $x=13.2$, $a=2.3$ and $b=2.5$ as an example. How can i prove it holds when it does not... – phi Dec 18 '12 at 17:32
Have you tried following the idea in my hint? If so, where do you get stuck? – Old John Dec 18 '12 at 17:36
@phi: Try the idea mentioned by Old John. In the example you give, pick $x=0.5$, or $0.6$. – André Nicolas Dec 18 '12 at 18:04

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.

-
Not sure if this is correct... – phi Aug 12 '13 at 18:01