Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
@labbhattacharjee I think you might have missed the significance of the $\forall x \in \Bbb R$? –  Old John Dec 18 '12 at 16:57

2 Answers 2

up vote 3 down vote accepted

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?

share|improve this answer
    
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.

share|improve this answer
    
Not sure if this is correct... –  phi Aug 12 '13 at 18:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.