Any proof?

D

damortis

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Nov 9, 2020
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So i 've been trying to prove [([x]/n)]=[x/n] but nothing came out .Can you help me?Thank you!
 
"nothing came out" - Please provide slightly more information about what is going in:

What is "n"?
What is "x"
What does "[]" mean?
 
By any chance are you trying to show that

[MATH]n \in \mathbb Z^+, x \ge 0 \implies \left \lfloor \dfrac{ \lfloor \ x \ \rfloor}{n} \right \rfloor = \left \lfloor \dfrac{ x }{n} \right \rfloor.[/MATH]
What have you tried?
 
Sure we can help you! But first you need to tell us what you need help with AND provide the whole entire problem as given to you. Showing what you tried will really get helpers to respond to your post.
 
damortis said:
So i 've been trying to prove [([x]/n)]=[x/n] but nothing came out .Can you help me?Thank you!
Click to expand...

It was a lot of fun checking this out in Desmos: https://www.desmos.com/calculator/oja8cfmd97

It looks like it doesn't require x to be non-negative, but n definitely has to be a positive integer. You need to clearly state the problem, with its conditions, before trying to prove it.

But we'll need to see an attempt at a proof in order to help with that part. What is it that "nothing came out" of?
 
tkhunny said:
"nothing came out" - Please provide slightly more information about what is going in:

What is "n"?
What is "x"
What does "[]" mean?
Click to expand...
So:n is any natural number
:x can be any number
:[ ] stands for integer part for example [2.7] is 2
[5.3] is 5 and so on
 
JeffM said:
By any chance are you trying to show that

[MATH]n \in \mathbb Z^+, x \ge 0 \implies \left \lfloor \dfrac{ \lfloor \ x \ \rfloor}{n} \right \rfloor = \left \lfloor \dfrac{ x }{n} \right \rfloor.[/MATH]
What have you tried?
Click to expand...
Basically that's the same thing but x can be any given number eg. 1,-5,3.5...
 
Jomo said:
Sure we can help you! But first you need to tell us what you need help with AND provide the whole entire problem as given to you. Showing what you tried will really get helpers to respond to your post.
Click to expand...
So the probelm is to prove that: [[x]/n] = [x/n] for every natural number (n) except for 0 and for every real number (x)
[ ] stands for integer part for example [2.7] is 2
[5.3] is 5 and so on
 
damortis said:
So the probelm is to prove that: [[x]/n] = [x/n] for every natural number (n) except for 0 and for every real number (x)
[ ] stands for integer part for example [2.7] is 2
[5.3] is 5 and so on
Click to expand...
You still haven't given us the main thing we've asked for: the work you said you did that didn't lead anywhere.

We need to see what kind of proof you are attempting, and how close you came to a valid proof. Questions about proof require context; we can't help without knowing what definitions or theorems you have available, and what proof techniques you will understand. And the best help we can give is to make small adjustments to things you already know.
 
Dr.Peterson said:
You still haven't given us the main thing we've asked for: the work you said you did that didn't lead anywhere.

We need to see what kind of proof you are attempting, and how close you came to a valid proof. Questions about proof require context; we can't help without knowing what definitions or theorems you have available, and what proof techniques you will understand. And the best help we can give is to make small adjustments to things you already know.
Click to expand...
Im going to let you know tomorrow cause right now im about to hit the sack.
 
Here is a hint

[MATH]n \text { an integer} > 1 \text { and } x \text { a real number}.[/MATH]
[MATH]\therefore \exists \text { integer } s \text { such that } s \le \dfrac{x}{n} < s + 1[/MATH].
 
JeffM said:
Here is a hint

[MATH]n \text { an integer} > 1 \text { and } x \text { a real number}.[/MATH]
[MATH]\therefore \exists \text { integer } s \text { such that } s \le \dfrac{x}{n} < s + 1[/MATH].
Click to expand...
This is what i did but i'm stuck hereIMG_20201110_090644.jpg
 
Dr.Peterson said:
You still haven't given us the main thing we've asked for: the work you said you did that didn't lead anywhere.

We need to see what kind of proof you are attempting, and how close you came to a valid proof. Questions about proof require context; we can't help without knowing what definitions or theorems you have available, and what proof techniques you will understand. And the best help we can give is to make small adjustments to things you already know.
Click to expand...
This is where i am:IMG_20201110_090644.jpg
 
Your first line isn't really what you mean, as that can't be true for any real number m; the LHS defines one particular number! What you mean is something like this:

Let n be a positive integer and x a real number. Then let [MATH]m = \left \lfloor \dfrac{ \lfloor \ x \ \rfloor}{n} \right \rfloor[/MATH]. ...​

(Observe that m will in fact be an integer, not "any real number"!)

Your second line would be a valid inference if you replaced [MATH]x[/MATH] with [MATH]\lfloor x \rfloor[/MATH]. Is that what you intended? If so, then you can now use the fact that [MATH]\lfloor x \rfloor[/MATH] is an integer to take a next step.

But your third line does not follow yet! There are several integers between mn and mn + n, and [MATH]\lfloor x \rfloor[/MATH] might be any of them.
 
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