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Does the floor function hold for this property:

$$\left \lfloor \frac{a}{b} \right \rfloor + \left \lfloor \frac{c}{d} \right \rfloor =\left \lfloor \frac{c+a}{b*d} \right \rfloor$$

Is this a valid expression using the floor function in mathematics ?

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    $\begingroup$ You mean: $\lfloor \frac ab\rfloor + \lfloor \frac cd\rfloor= \lfloor \frac {a+c}{bd}\rfloor$? Try $\frac ab=\frac 54=\frac cd$. $\endgroup$ – lulu Oct 14 '16 at 20:12
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    $\begingroup$ Try a = b = c = 1, d = 3. $\endgroup$ – user144221 Oct 14 '16 at 20:12
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What you've written makes no sense. Well, it make sense but there's no reasoning behind it. Let $a=b=c=d $ then you have $[a/b]+[c/d]=1+1=2$ and $[\frac {a+c}{bd}]=[2/b] =2,1,0$ if $b=1,2,3^+$. The two sides have nothing to do with each other so it's a silly question.

I think you meant does $[a/b] + [c/d] =[a/b + c/d] =[\frac {ad+bc}{bd}] $. This is false but it's reasonable to ask and isn't "silly".

Let $a=nb+j $, $c=md+k $

$[a/b]+[c/d]= n+m $

$[a/b + c/d]=[n+j/b+m+k/d]=n+m+[j/b+k/d]$. $0\le j/b <1$ and $0\le k/d <1$ so $0\le j/b+k/d <2$ so $[j/b+k/d]=0,1$.

So $[a/b+c/d]= $ either $[a/b]+[c/d] $ or $[a/b]+[c/d] +1$.

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