Online Course Discussion Forum
MCIII Number Theory 3.12 and 3.13
Hello,
Hello! I did 3.12 problem using two ways, both are very tedious:
1. Using the quadratic formula to get x in terms of floor(x), then using floor(x)≤x<floor(x)+1
2. Finding floor(x) in terms of x and using x-1<floor(x)≤x
Both of them involve radicals and one of them requires plugging in multiple numbers back into the equation.
Is there a less tedious way to do this?
For 3.13, I added floor(x)+floor(1/x) to both sides to get:
floor(x)+floor(1/x)+1=x+1/x
then by using the fact that [a]+[b]≤[a+b] (floor(x)=[x])
(floor(x)+floor(1/x))+1(=x+1/x)≤floor(x+1/x)+1
so x+1/x≤floor(x+1/x)+1 which is trivial since (let y=x+1/x) y-1<floor(y)≤y, add one to both sides y<floor(y)+1
When I get an inequality that is trivial, does that mean the solution is all real numbers or does it just mean that I used the wrong inequality, an inequality that doesn't solve the problem?
Thanks,
Tina Jin
For 3.12:
It is sometimes easier if you let $n=\lfloor x\rfloor$. Then the equation becomes $$x^2 - 4x + n = 0.$$ For the equation to have real solutions, the discriminant has to be nonnegative. This gives a small range for $n$ (which is the integer part of $x$), and you can try each possible value of $n$ to solve for $x$.
For 3.13:
You already got a very important result: $x + 1/x = \lfloor{x}\rfloor + \lfloor 1/x \rfloor + 1$, which is an integer. So what kind of integers can equal $x + 1/x$ for real number $x$? (In fact it can be any integer, positive or negative, except for $-1, 0, 1$. You can prove this fact easily.) Assume $x + 1/x = k$, where $k$ is an integer, and solve for $x$. Then it will turn out that $k=\pm 2$ don't work either, but everything else works. That solves part (1). Use the result in Part (1) to prove part (2).
Hello,
I still need some help on this question. I tested some values of k like 3, 4, 5, and -3, -4, -5, and they do work, but how do I prove it?
Thanks,
Tina Jin
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