Proving that ($x^2-xyz$) is rational when all three $x, y$ and $z$ are rational

Prove that $x^2-xyz\,$ is rational when $x,y,z$ are all rational.

Should I prove this question by proving that multiplication or addition of rational numbers are rationals or there are better ways to this? Thanks in advance.

2 Answers

It depends upon the level and already known information. If you haven't proved yet that rational numbers are closed under multiplication and addition (so you cannot use it), you must begin from the definition of rational numbers:

Let $x=\dfrac ab$, $y=\dfrac cd$ and $z=\dfrac uv$, where $a$, $b$, $c$, $d$, $u$ and $v$ are all integers, and $b$, $d$ and $v$ are not zero.

Proceed by evaluating the fraction and then show that the numerator and denominator are also going to be integers, and the denominator will be non-zero.

If you are allowed to use the closure of rational numbers under addition and multiplication, you can directly say that $x^2\equiv x\cdot x$ is a rational number, and then show that $xyz$ is a rational number and continue similarly.

Hope this helps. Ask anything if not clear :)

Just use the fact that all rational numbers can be represented as $\frac{p}{q}$ for rational $p,q$. If $x=\frac{x_1}{x_2}, y=\frac{y_1}{y_2}, z=\frac{z_1}{z_2}$ for integers $x_1,x_2,y_1,y_2,z_1,z_2$, then $$x^2-xyz=\frac{x_1^2}{x_2^2}-\frac{x_1y_1z_1}{x_2y_2z_2}=\frac{x_1^2y_2z_2-x_2x_1y_1z_1}{x_2^2y_2z_2}$$ Which is an integer divided by an integer. Since an integer is also a rational number ($\mathbb{Z}\subseteq\mathbb{Q}$), this is also a rational number.