Lovellss

2022-06-25

Fraction and Decimal: Reciprocal of x's non-integer

The reciprocal part of $x$'s non-integer decimal part equals $x+1$, and $x>0$. What is $x$?

Solution: I tried this way-

Let's $n$= integer part of $x$

$1/x-n=x+1$

or, $1=(x-n)(x+1)$

or, $1={x}^{2}+x-nx-n$

or, ${x}^{2}+(1-n)x-(n+1)=0$

but, stucked here. Is there any other way?

The reciprocal part of $x$'s non-integer decimal part equals $x+1$, and $x>0$. What is $x$?

Solution: I tried this way-

Let's $n$= integer part of $x$

$1/x-n=x+1$

or, $1=(x-n)(x+1)$

or, $1={x}^{2}+x-nx-n$

or, ${x}^{2}+(1-n)x-(n+1)=0$

but, stucked here. Is there any other way?

Esteban Johnson

Beginner2022-06-26Added 15 answers

To complete Ashvin Swaminathan's answer:

$x=\frac{n-1\pm \sqrt{{n}^{2}+2n+5}}{2}$

Because $x>0$, we take

$x=\frac{n-1+\sqrt{{n}^{2}+2n+5}}{2}$

Since the discriminant ${n}^{2}+2n+5=(n+1{)}^{2}+4\ge 4$, then there are indeed infinitely many solutions, and these depend on $n=\lfloor x\rfloor $

$x=\frac{n-1\pm \sqrt{{n}^{2}+2n+5}}{2}$

Because $x>0$, we take

$x=\frac{n-1+\sqrt{{n}^{2}+2n+5}}{2}$

Since the discriminant ${n}^{2}+2n+5=(n+1{)}^{2}+4\ge 4$, then there are indeed infinitely many solutions, and these depend on $n=\lfloor x\rfloor $

landdenaw

Beginner2022-06-27Added 8 answers

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