Welcome to our community

Be a part of something great, join today!

fractional part Eq.

jacks

Well-known member
Apr 5, 2012
226
Calculate $x$ in $\{x^2\}+\{x\} = 1$

where $\{x\} = $ fractional part of $x$
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Calculate $x$ in $\{x^2\}+\{x\} = 1$

where $\{x\} = $ fractional part of $x$
Hi jacks, :)

According to Wolfram, both solutions of the quadratic equation, \(x^2+x=1\) are solutions of \(\{x^2\}+\{x\} = 1\). That is,

\[x=-\frac{\sqrt{5}+1}{2}\mbox{ and }x=\frac{\sqrt{5}-1}{2}\]

are solutions of \(\{x^2\}+\{x\} = 1\).

Kind Regards,
Sudharaka.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Hi jacks, :)

According to Wolfram, both solutions of the quadratic equation, \(x^2+x=1\) are solutions of \(\{x^2\}+\{x\} = 1\). That is,

\[x=-\frac{\sqrt{5}+1}{2}\mbox{ and }x=\frac{\sqrt{5}-1}{2}\]

are solutions of \(\{x^2\}+\{x\} = 1\).

Kind Regards,
Sudharaka.
The first solution does not satisfy the original equation. I'm not exactly sure what WolframAlpha is doing there, but I don't think it's solving the original equation. The second solution does satisfy the original equation.

[EDIT]: See posts below for more clarification.
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Forgetting for the moment any problems about negative values of $x$, notice that for any positive integer $n$, the positive solution of the equation $x^2+x = n$ will satisfy $\{x^2\} + \{x\} = 1$, unless that solution is an integer.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
The first solution does not satisfy the original equation. I'm not exactly sure what WolframAlpha is doing there, but I don't think it's solving the original equation. The second solution does satisfy the original equation.

[EDIT]: See posts below for more clarification.
Yes, as chisigma had mentioned there is an ambiguity in the definition, however taking \(\{x\}=x- \lfloor x \rfloor\,\forall x\in\Re\), both solutions satisfy the original equation.

Forgetting for the moment any problems about negative values of $x$, notice that for any positive integer $n$, the positive solution of the equation $x^2+x = n$ will satisfy $\{x^2\} + \{x\} = 1$, unless that solution is an integer.
Indeed, but just for the curiosity, can you explain how you thought about this. :)

If we take, \(\{x\}=x- \lfloor x \rfloor\,\forall x\in\Re\) as the definition of \(\{x\}\), I think that all the roots(except integers) of the equations \(x^2+x = n\) are solutions of the original equation. That is,

\[x=\frac{-1\pm\sqrt{4\,n+1}}{2}\mbox{ where }n\in\mathbb{Z}\, \wedge \,x\notin\mathbb{Z}\]

are solutions of \(\{x^2\} + \{x\} = 1\). Furthermore by looking at this, I feel that these seem to be the only solutions of the original equation.

Kind Regards,
Sudharaka.
 

chisigma

Well-known member
Feb 13, 2012
1,704