Kvant Math Problem 952
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Problem
Let ${x}$ denote the fractional part of the number $x$; that is, ${x}=x-[x]$, where $[x]$ is the greatest integer not exceeding $x$.
- Give an example of a number $a$ such that $${a}+\left{\dfrac1a\right}=1.$$.
- Prove that such a number $a$ cannot be rational.
I. Varga (Romania)
Exploration
Write
$$a=n+x, \qquad n=[a],\quad 0\le x<1.$$
Then
$${a}=x.$$
The condition becomes
$$x+\left{\frac1a\right}=1.$$
Since a fractional part lies in $[0,1)$, equality $x+{1/a}=1$ forces $0<x<1$ and
$$\left{\frac1a\right}=1-x.$$
To search for an example, it is natural to require
$$\frac1a<1,$$
because then
$$\left{\frac1a\right}=\frac1a.$$
This happens when $a>1$. The equation becomes
$$x+\frac1a=1.$$
Substituting $a=n+x$ gives
$$x+\frac1{n+x}=1.$$
Try the simplest possibility $n=1$. Then
$$x+\frac1{1+x}=1.$$
Multiplying by $1+x$,
$$x(1+x)+1=1+x,$$
hence
$$x^2=0.$$
This gives $x=0$, not allowed.
Try $n=2$:
$$x+\frac1{2+x}=1.$$
Then
$$x(2+x)+1=2+x,$$
so
$$x^2+x-1=0.$$
The positive root is
$$x=\frac{\sqrt5-1}{2}.$$
This lies in $(0,1)$, and then
$$a=2+\frac{\sqrt5-1}{2}=\frac{3+\sqrt5}{2}.$$
Checking numerically,
$${a}\approx0.618,\qquad \left{\frac1a\right}\approx0.382,$$
and the sum is $1$.
For the irrationality statement, suppose $a$ is rational. Then $x={a}$ and ${1/a}$ are rational. Since their sum is $1$,
$$\left{\frac1a\right}=1-x.$$
Let
$$m=\left[\frac1a\right].$$
Then
$$\frac1a=m+1-x.$$
Using $a=n+x$,
$$1=(n+x)(m+1-x).$$
Expanding,
$$1=n(m+1)+x(m+1-n)-x^2.$$
Thus $x$ satisfies a quadratic equation with integer coefficients:
$$x^2-(m+1-n)x-\bigl(1-n(m+1)\bigr)=0.$$
Since $a$ is assumed rational, $x=a-n$ is rational. A rational root of a monic integer polynomial is an integer. Because $0\le x<1$, this forces $x=0$.
Then ${a}=0$, and the original equation gives
$$\left{\frac1a\right}=1,$$
impossible because a fractional part is always strictly less than $1$.
The delicate point is justifying that a rational root of the displayed monic polynomial must be an integer.
Problem Understanding
We seek a number $a$ whose fractional part and the fractional part of its reciprocal add up to $1$. First an explicit example must be produced. Then it must be proved that no rational number can satisfy this condition.
This is a Type D problem, because part (1) asks for an explicit construction and part (2) asks for a nonexistence statement within the class of rational numbers.
The core difficulty is the second part. The condition involves two fractional parts, and it must be converted into an algebraic relation strong enough to force a contradiction when $a$ is rational.
The expected answer is an irrational number. The example found above is
$$a=\frac{3+\sqrt5}{2}.$$
Its fractional part equals $(\sqrt5-1)/2$, while the reciprocal equals $(3-\sqrt5)/2$, and these numbers sum to $1$.
Proof Architecture
Let $a=n+x$ with $n=[a]$ and $0\le x<1$; then ${a}=x$.
For the construction, choose $n=2$ and solve the equation $x+\frac1{2+x}=1$; this yields $x=(\sqrt5-1)/2$ and hence an explicit solution.
To exclude rational solutions, assume $a$ is rational and write $m=[1/a]$.
From ${a}+{1/a}=1$, derive
$$\frac1a=m+1-x.$$
Substituting $a=n+x$ gives
$$(n+x)(m+1-x)=1.$$
Expanding yields a monic quadratic polynomial with integer coefficients satisfied by $x$.
Since $x$ is rational, the rational root theorem implies that $x$ is an integer.
Because $0\le x<1$, conclude $x=0$, which contradicts ${1/a}=1$.
The most delicate lemma is the passage from the quadratic equation to $x=0$, because it relies on the arithmetic structure of monic integer polynomials.
Solution
Write
$$a=n+x, \qquad n=[a],\quad 0\le x<1.$$
Then
$${a}=x.$$
For an example, take $n=2$ and require
$$\left{\frac1a\right}=\frac1a.$$
This is valid whenever $a>1$. The condition
$${a}+\left{\frac1a\right}=1$$
becomes
$$x+\frac1{2+x}=1.$$
Multiplying by $2+x$,
$$x(2+x)+1=2+x,$$
hence
$$x^2+x-1=0.$$
The root belonging to $(0,1)$ is
$$x=\frac{\sqrt5-1}{2}.$$
Therefore
$$a=2+x=\frac{3+\sqrt5}{2}.$$
Since
$$\frac1a=\frac{3-\sqrt5}{2},$$
we have
$$0<\frac1a<1,$$
so
$$\left{\frac1a\right}=\frac1a=\frac{3-\sqrt5}{2}.$$
Also
$${a}=a-2=\frac{\sqrt5-1}{2}.$$
Consequently,
$${a}+\left{\frac1a\right} = \frac{\sqrt5-1}{2} + \frac{3-\sqrt5}{2} = 1.$$
Thus
$$a=\frac{3+\sqrt5}{2}$$
is a required example.
Now suppose that a rational number $a$ satisfies
$${a}+\left{\frac1a\right}=1.$$
Write
$$a=n+x, \qquad n=[a],\quad 0\le x<1.$$
Since $a$ is rational, $x=a-n$ is rational.
Let
$$m=\left[\frac1a\right].$$
From
$$\left{\frac1a\right} = \frac1a-m$$
and
$$x+\left{\frac1a\right}=1,$$
we obtain
$$\frac1a=m+1-x.$$
Substituting $a=n+x$ gives
$$(n+x)(m+1-x)=1.$$
Expanding,
$$n(m+1)+x(m+1-n)-x^2=1,$$
or
$$x^2-(m+1-n)x-\bigl(1-n(m+1)\bigr)=0.$$
This is a monic polynomial with integer coefficients.
A rational root of a monic polynomial with integer coefficients must be an integer. Since $x$ is rational and satisfies this polynomial, $x$ is an integer.
Because
$$0\le x<1,$$
it follows that
$$x=0.$$
Hence
$${a}=0.$$
The original condition becomes
$$\left{\frac1a\right}=1,$$
which is impossible because every fractional part belongs to the interval $[0,1)$.
The contradiction shows that no rational number can satisfy
$${a}+\left{\frac1a\right}=1.$$
This completes the proof.
∎
Verification of Key Steps
The construction must be checked directly. For
$$a=\frac{3+\sqrt5}{2},$$
the reciprocal is
$$\frac1a=\frac{3-\sqrt5}{2}.$$
Since this lies between $0$ and $1$, its fractional part equals itself. The fractional part of $a$ is
$$a-2=\frac{\sqrt5-1}{2}.$$
Adding these two expressions gives exactly $1$.
The algebraic reduction in the irrationality proof starts from
$$\frac1a=m+1-x.$$
Replacing $a$ by $n+x$ yields
$$(n+x)(m+1-x)=1.$$
Expanding carefully,
$$n(m+1)-nx+x(m+1)-x^2 = n(m+1)+x(m+1-n)-x^2,$$
which leads to the displayed quadratic equation. A sign error here would invalidate the argument.
The final contradiction depends on the fact that the polynomial is monic. If
$$x=\frac pq$$
in lowest terms were a rational root, then
$$p^2-(m+1-n)pq-\bigl(1-n(m+1)\bigr)q^2=0.$$
Reducing modulo $q$ gives $p^2\equiv0\pmod q$, hence $q\mid p$. Since $\gcd(p,q)=1$, one gets $q=1$. Thus $x$ is an integer, and the condition $0\le x<1$ forces $x=0$.
Alternative Approaches
Starting from
$$\frac1a=m+1-x, \qquad a=n+x,$$
one may eliminate $x$ instead of treating it as the unknown. Since $x=a-n$,
$$\frac1a=m+1+n-a.$$
Multiplying by $a$ yields
$$a^2-(m+n+1)a+1=0.$$
Hence any solution $a$ is a root of a monic quadratic polynomial with integer coefficients. If $a$ were rational, the rational root theorem would force $a$ to be an integer. Then ${a}=0$, contradicting the given condition. This proves irrationality directly.
The main approach is preferable because it uses the fractional part variable $x={a}$ throughout. The construction and the nonexistence proof are then expressed in the same language, making the logic more uniform.