Online Course Discussion Forum
MCIII Number Theory 3.8
Hello,
Here is my solution:
floor(x)^2=xfractional(x) (given)
floor(x)^2=x(x-floor(x))
floor(x)^2+xfloor(x)-x^2=0
by quadratic formula, x=(floor(x)±sqrt(5floor(x)^2))/2
plug into the equation floor(x)≤x<floor(x)+1, (let t=floor(x))
t≤t(1+sqrt(5))/2<t+1
2t≤t+tsqrt(5)<2t+2
after solving, you get that t is actually 0, so there are no solutions.
However, the answer to this question is not no solutions, but I don't understand what I did wrong.
Thank you,
Tina Jin
The final inequality you have: $$2t \leq t + t\sqrt{5} < 2t+2,$$ this leads to $$t \leq t\sqrt{5} < t + 2.$$ The question requires that $t \geq 1$, so the first part of the inequality is always true. Now $t\sqrt{5} < t + 2$ gives $t < \dfrac{2}{\sqrt{5}-1} = \dfrac{\sqrt{5}+1}{2}$, which means $t=1$. This does provide a solution for $x$.
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