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I have to prove the following:

There exists a positive real number $a$ so that for all real numbers $x$ if $x$ -$\lfloor x \rfloor < a$, then $\lfloor 3x \rfloor = 3 \lfloor x \rfloor$.

I have attempted to do this in several ways. I got the furthest when I attempted to use three separate cases, where $x$ is an integer, $x$ is a positive real number that is not an integer, and $x$ is a negative real number that is not an integer. In all cases, $x$ was a particular but arbitrarily chosen value of the given type.

This approach failed when $x$ was assumed to be a negative real number that was not an integer. Intuitively, I know that $x - \lfloor x \rfloor $ is going to be greater than 0 because $\lfloor x \rfloor$ will always be less than $x$, but I have no idea where to go from there in proving the property.

I feel like my attempted approach might have been the wrong one from the start, but I don't know how else to prove this.

Thanks in advance for any help.

monevus3
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  • It’s easiest if you first figure out what $a$ ought to be and then prove that you’re right. Do you have any idea of what $a$ is? – Brian M. Scott Feb 02 '15 at 00:02

1 Answers1

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HINT: If $x-\lfloor x\rfloor<a$, then $\lfloor x\rfloor\le x<\lfloor x\rfloor+a$, so $$3\lfloor x\rfloor\le 3x<3\lfloor x\rfloor+3a\;.$$ If $\lfloor 3x\rfloor=3\lfloor x\rfloor$, this becomes

$$\lfloor 3x\rfloor\le 3x<\lfloor 3x\rfloor+3a\;.$$

On the other hand, you know that $\lfloor 3x\rfloor\le 3x<\lfloor 3x\rfloor+1$, so ...

Brian M. Scott
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