Is $x(x+1)$ is a whole number if and only if $x$ is whole (x can be irrational!).

Question

I wonder if the following statement is true:

If $x$ is a real number, then $x(x+1)$ and $\frac{x(x+1)}{2}$ are whole numbers if and only if x is a whole number.

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This is my desperate attempt to prove.

1. If x is a whole number, then $x(x+1)$ and $\frac{x(x+1)}{2}$ are obviously whole numbers.

2. Let $x=\frac{p}{q}$, where $p$ and $q$ are co-prime numbers and $q>1$. Then $p+q$ and $p$ are also co-prime, so $x(x+1)=\frac{p(p+q)}{q^2}$ is irreducible, therefore $x(x+1)$ is fractional, and so is $\frac{x(x+1)}{2}$.

3. Now let x be irrational. Then $x=\lim_{n\to\infty}{r_n}$, where $r_n$-s are rational. So $x(x+1)=\lim_{n\to\infty}{r_n(r_n+1)}$... Stop!!!

How to prove that that limit is not a whole number (presumably irrational)? I guess, it has to do with the value of denominator of $r_n$-s, if it can be shown that it grows infinitely high, but I am no so strong in the theory of irrational numbers to prove that.

Answer 1

Suppose $x(x+1)$ is a non-negative integer (whole number). Let's say $z = x(x+1)$. Then $z = x^2 + x$, so we want to rearrange this in terms of $x$. We get

$\begin{align} x^2 + x - z &= 0\\ x^2 + 2\left(\frac{x}{2}\right) + \frac{1}{4} - \frac{1}{4} - z &= 0\\ x^2 + 2\left(\frac{1}{2}x\right) + \frac{1}{4} &= \frac{1}{4} + z\\ \left(x + \frac{1}{2}\right)^2 &= \frac{1}{4} + z\\ \end{align}$

Note that $\frac{1}{4} + z$ is greater than zero because $z ≥ 0$, so we can take the square root.

$\begin{align} x + \frac{1}{2} &= \sqrt{\frac{1}{4} + z}\\ x &= \sqrt{\frac{1}{4} + z} - \frac{1}{2} \end{align}$

Now if you substitute $z = 1$ for example, you get $x = \sqrt{\frac{5}{4}}-\frac{1}{2} = something\ gross$

In the same way, your claim isn't true for the second formula, but I'll leave that for you to solve :)

Answer 2

If p is a non constant polynomial with integer coefficients and leading coefficient one and x is a real number not and integer and p(x)=0 then x is irrational .