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MCIII Number Theory 3.8

 
 
JinTina的头像
MCIII Number Theory 3.8
JinTina - 2024年01月5日 Friday 14:24
 

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

 
WangDr. Kevin的头像
Re: MCIII Number Theory 3.8
WangDr. Kevin - 2024年01月6日 Saturday 00:24
 

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 $x$.

JinTina的头像
Re: MCIII Number Theory 3.8
JinTina - 2024年01月14日 Sunday 10:46
 

I understand now! x=(1+-sqrt(5))/2 but only the + one gives floor(x)=1, so only (1+sqrt(5))/2 works.