Help me i dont understand how to solve this one

ALIAHMAD

ALIAHMAD

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Jul 5, 2022
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[2x]=[x]
Find the values of x satisfying the above equation.
Where [.] Denots greatest integer function

Actually i am not able to get how to solve it
I tried but I got stuck
 
Last edited:
ALIAHMAD said:
[2x]=[x]
Find the values of x satisfying the above equation.
Where [.] Denots greatest integer function

Actually i am not able to get how to solve it
I tried but I got stuck
Click to expand...
Hint 1:

For positive [imath]m[/imath],
\(\displaystyle \lfloor{mx\rfloor}=\lfloor{x\rfloor}+ \left\lfloor x + \frac{1}{m} \right\rfloor + \dots + \left\lfloor x + \frac{m-1}{m} \right\rfloor \\\)
 
ALIAHMAD said:
[2x]=[x]
Find the values of x satisfying the above equation.
Where [.] Denots greatest integer function

Actually i am not able to get how to solve it
I tried but I got stuck
Click to expand...
I see you have unstuck your [CAPS] key.
 
Consider [imath]\left|x\right|<0.5[/imath]
 
ALIAHMAD said:
[2x]=[x]
Find the values of x satisfying the above equation.
Where [.] Denots greatest integer function

Actually i am not able to get how to solve it
I tried but I got stuck
Click to expand...
I'd start by trying to graph each side on the same pair of axes. That would either give me the solution (which I could confirm by algebraic methods), or at least a hint as to what to consider. At the least, it would give you a better sense of how these functions work.

Algebraically, I'd probably break it into cases.

Possibly you've done all this, In that case, we'll want to see where you got stuck!
 
ALIAHMAD said:
Yes which is 2=1 which is not possible
Click to expand...
Before you can even consider solving your problem, you need to be able to solve the one I gave you.

I can only hope that what you wrote was based on the fact that you thought x=1 and found out that was wrong.
2x=x is a very simple equation that you should be able to solve. And yes, it does have a solution.
 
I
Steven G said:
Before you can even consider solving your problem, you need to be able to solve the one I gave you.

I can only hope that what you wrote was based on the fact that you thought x=1 and found out that was wrong.
2x=x is a very simple equation that you should be able to solve. And yes, it does have a solution
Click to expand...
I got it, I made a mistake, I am sorry
I understand that dividing things by zero can give you unreal results
 
Dr.Peterson said:
I'd start by trying to graph each side on the same pair of axes. That would either give me the solution (which I could confirm by algebraic methods), or at least a hint as to what to consider. At the least, it would give you a better sense of how these functions work.

Algebraically, I'd probably break it into cases.

Possibly you've done all this, In that case, we'll want to see where you got stuck!
Click to expand...
i did it in cases , just realised what I was doing wrong, rectified it and got the answer
thanks man
 
ALIAHMAD said:
Okay okay then it will be x belongs to [-1/2,1/2)
Click to expand...
I assume you've checked over your work to make sure there aren't any other errors, and that you won't make the same little mistake next time ...

If you'd like, you can show us your work, so we can see if there are any subtle errors in the method. But you probably did fine.

And next time you have a question, of course, you'll want to show your work from the start, so that if you just made some little slip, we can save time by just pointing it out.
 
BigBeachBanana said:
Hint 1:

For positive [imath]m[/imath],
\(\displaystyle \lfloor{mx\rfloor}=\lfloor{x\rfloor}+ \left\lfloor x + \frac{1}{m} \right\rfloor + \dots + \left\lfloor x + \frac{m-1}{m} \right\rfloor \\\)
Click to expand...
Here's my solution.
[math] \lfloor2x\rfloor = \lfloor x \rfloor\\ \underbrace{\lfloor{x\rfloor}+ \left\lfloor x + \frac{1}{2} \right\rfloor}_{\text{Hermite's Identity}} = \lfloor x \rfloor\\ \left\lfloor x + \frac{1}{2} \right\rfloor=0 \iff 0 \le x + \frac{1}{2}<1\\ \therefore x\in \left[-\frac{1}{2},\frac{1}{2}\right) [/math]
 
Dr.Peterson said:
I assume you've checked over your work to make sure there aren't any other errors, and that you won't make the same little mistake next time ...

If you'd like, you can show us your work, so we can see if there are any subtle errors in the method. But you probably did fine.

And next time you have a question, of course, you'll want to show your work from the start, so that if you just made some little slip, we can save time by just pointing it out.
Click to expand...
Dr. Peterson
I am very grateful that i found you.
As i am new in this platform I didn't knew how this works
But now i have some idea of it
Thank you so much ???
 
BigBeachBanana said:
Here's my solution.
[math] \lfloor2x\rfloor = \lfloor x \rfloor\\ \underbrace{\lfloor{x\rfloor}+ \left\lfloor x + \frac{1}{2} \right\rfloor}_{\text{Hermite's Identity}} = \lfloor x \rfloor\\ \left\lfloor x + \frac{1}{2} \right\rfloor=0 \iff 0 \le x + \frac{1}{2}<1\\ \therefore x\in \left[-\frac{1}{2},\frac{1}{2}\right) [/math]
Click to expand...
When did you learn this identity ?
I mean in which grade
 
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