Online Course Discussion Forum

MCIII Number Theory 3.9

 
 
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MCIII Number Theory 3.9
by Tina Jin - Friday, 29 December 2023, 12:47 PM
 

Hello,

I am a little bit stuck on this problem. Here are my steps so far: (floor(x)=[x])

x^2-1<floor(x^2)≤x^2

x^2-1<([x]^2)≤x^2

-1<[x]^2-x^2≤0

Also, x-1<[x]≤x

(x-1)^2=x^2-2x+1<[x]^2≤x^2

-2x+1<[x]^2-x^2≤x^2

What am I supposed to do next?

Thanks,

Tina Jin

 
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Re: MCIII Number Theory 3.9
by Dr. Kevin Wang - Friday, 29 December 2023, 10:01 PM
 

For questions involving $\lfloor x \rfloor$, it is sometimes easier to fix this floor value, because as the integer part of $x$, one integer corresponds to infinitely many $x$.  Before using algebra to solve anything, try a few small numbers for $\lfloor x\rfloor$.

This problem requires that $x\ge 0$, so it means $\lfloor x\rfloor \geq 0$.  What if $\lfloor x\rfloor = 0$?  The equation becomes $$\lfloor x^2 \rfloor = 0.$$  This means $0\leq x^2 < 1$, and thus $0\leq x < 1$.

Now what if $\lfloor x \rfloor = 1$?  The equation becomes $$\lfloor x^2 \rfloor = 1,$$ and so $1 \leq x^2 < 2$, which means $1\leq x < \sqrt{2}$.

So, in general, we can let $\lfloor x\rfloor = n$, and solve it in terms of $n$.  The full solution includes all these separate intervals.

Picture of Tina Jin
Re: MCIII Number Theory 3.9
by Tina Jin - Friday, 5 January 2024, 2:30 PM
 

Hello,

I understand now! I think the solution is:

n≤x<sqrt(n^2+1) 

Thank you!

Tina Jin