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## Main Question or Discussion Point

Hi,

I'm trying to proof that:

[tex]\left\lfloor \frac{\left\lfloor x \right\rfloor}{y} \right\rfloor= \left\lfloor \frac{x}{y} \right\rfloor[/tex] for the specific case where y is an integer.

At the recommendation of somebody who I discussed the problem with, here's how I started:

[tex] \lfloor x \rfloor \le x < \lfloor x \rfloor + 1[/tex]

[tex] \frac{\lfloor x \rfloor}{y} \le \frac{x}{y} < \lfloor \frac{x}{y} \rfloor + \frac{1}{y} [/tex]

[tex] \left\lfloor \frac{\lfloor x \rfloor}{y} \right\rfloor \le \left\lfloor \frac{x}{y} \right\rfloor [/tex]

And from there, prove that the case of [tex] \left\lfloor \frac{\lfloor x \rfloor}{y} \right\rfloor < \left\lfloor \frac{x}{y} \right\rfloor [/tex] is impossible, leaving [tex] \left\lfloor \frac{\lfloor x \rfloor}{y} \right\rfloor = \left\lfloor \frac{x}{y} \right\rfloor [/tex]

Any ideas?

Thanks for your help,

John

I'm trying to proof that:

[tex]\left\lfloor \frac{\left\lfloor x \right\rfloor}{y} \right\rfloor= \left\lfloor \frac{x}{y} \right\rfloor[/tex] for the specific case where y is an integer.

At the recommendation of somebody who I discussed the problem with, here's how I started:

[tex] \lfloor x \rfloor \le x < \lfloor x \rfloor + 1[/tex]

[tex] \frac{\lfloor x \rfloor}{y} \le \frac{x}{y} < \lfloor \frac{x}{y} \rfloor + \frac{1}{y} [/tex]

[tex] \left\lfloor \frac{\lfloor x \rfloor}{y} \right\rfloor \le \left\lfloor \frac{x}{y} \right\rfloor [/tex]

And from there, prove that the case of [tex] \left\lfloor \frac{\lfloor x \rfloor}{y} \right\rfloor < \left\lfloor \frac{x}{y} \right\rfloor [/tex] is impossible, leaving [tex] \left\lfloor \frac{\lfloor x \rfloor}{y} \right\rfloor = \left\lfloor \frac{x}{y} \right\rfloor [/tex]

Any ideas?

Thanks for your help,

John