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Prove or disprove that $\lfloor$ xy $\rfloor$ = $\lfloor$ x $\rfloor$ $\lfloor$ y $\rfloor$ for all real numbers x and y.

Any advice to get past the mental block I have would be nice. :)

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    $\begingroup$ It's good to start by trying some examples. For instance, what happens when $x=2$ and $y=\frac{1}{2}$? $\endgroup$ – carmichael561 Nov 1 '16 at 21:36
  • $\begingroup$ ... think x = [x] plus a fractional part. This says ([x] + frac(x))([y]+ frac(y)) = [xy] + frac(xy). So frac(xy) = [x]frac(y) + [y]frac(x) + frac(x)frac(y) < 1. Is there any reason to believe that. Can you see a case where it might be larger than 1? $\endgroup$ – fleablood Nov 1 '16 at 22:02
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I will disprove this claim.

Let x be a real number that equals 1.5.

Let y be a real number that equals 2.

$\lfloor$ (1.5)(2) $\rfloor$ = 3

$\lfloor$ 1.5 $\rfloor$ $\lfloor$ 2 $\rfloor$ = 2

3 does not equal 2, therefore I have found a counterexample.

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Find an easy counterexample. For instance, $x=1.5$, $y=2$.

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To be more mathematical:

$x = [x] + \{x\}$ where $ [x] \in \mathbb Z; 0 \le \{x\} < 1$. (you'll have to take my word for it that these values always exist and they are unique; that's a proof for another time.)

This also implies $[x] \le x < [x]+1$.

Note: $xy = ([x] + \{x\})([y]+\{y\}) = [x][y] + \{x\}[y] + [x]\{y\} + \{x\}\{y\}$.

So $[xy] = [x][y]$ implies

$[([x] + \{x\})([y]+\{y\})] = [[x][y] + \{x\}[y] + [x]\{y\} + \{x\}\{y\}] = [x][y]$

This implies

$[x][y] \le [x][y] + \{x\}[y] + [x]\{y\} + \{x\}\{y\} < [x][y] + 1$ which implies

$0 \le \{x\}[y] + [x]\{y\} + \{x\}\{y\} < 1$

Is there ANY reason for this to be true?

Well $\{x\}$ and $\{y\}$ must be less than $1$ but $[x]$ and $[y]$ can be any integers we like.

So let, say, $[x]=[y]=10$ and $\{x\}=\{y\}= \frac 12$

So have $\{x\}[y] + [x]\{y\} + \{x\}\{y\}= \frac 1210 + 10\frac 12 + \frac 12 \frac 12 = 10\frac 14 > 1$

And so $[10 \frac 12*10\frac 12] = [110\frac 14] = 110$

While $[10 \frac 12][10\frac 12] = 10*10 =100$.

These are not equal.

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