# Prove that if $x^3 + 3x + 3$ is irrational then $x$ is irrational, by proving the contrapositive

I don't understand how to do this considering that the contrapositive of $$x^3$$ is irrational. For example $$2$$ to the cube root is irrational, but I am trying to prove that is is rational

• $(A\text{ is irrational})\to(B\text{ is irrational})$ is equivalent to $(B\text{ is rational})\to(A\text{ is rational})$, which in this case is trivial. – Jack D'Aurizio Oct 3 '18 at 21:30

Suffices to show the following:

Suppose $$x$$ is rational. Then $$f(x)=x^3+3x+3$$ is also rational.

We claim that $$A(x)=x^3$$ is rational if $$x$$ is rational. Indeed, the set of rationals is closed under multiplication, and $$x^3 = x \times x \times x$$.

Likewise, as 3 is rational, it follows that $$B(x) = 3x$$ is rational if $$x$$ is rational.

However, for each $$x$$ we note that $$f(x)=A(x)+B(x)+3$$. Then if $$x$$ is rational then $$f(x)=A(x)+B(x)+3$$ is the sum of 3 rational numbers $$A(x),B(x)$$ and 3. As the set of rationals is closed under addition, it follows that $$f(x)$$ is rational as well.

A more pedestrian proof (following @Mike's first step):

Suppose $$x$$ is rational, and the quotient of two integers $$x = \frac{a}{b}.$$ Then

\begin{align} f(x) &= x^3 + 3x + 3 \\[8pt] &= \left(\frac{a}{b}\right)^3 + 3\frac{a}{b} + 3\frac{1}{1} \\[8pt] &= \frac{a^3}{b^3} + 3\frac{a}{b}\frac{b^2}{b^2} + 3\frac{1}{1}\frac{b^3}{b^3} \\[8pt] &= \frac{a^3}{b^3} + 3\frac{ab^2}{b^3} + \frac{3b^3}{b^3} \\[8pt] &= \frac{a^3 + 3ab^2 + 3b^3}{b^3} \end{align} which is a quotient of two integers (namely, $$a^3 + 3ab^2 + 3b^2$$ and $$b^3$$). Hence $$f(x)$$ is a rational as well.

• Thanks, @MichaelHardy; the line spacing's much nicer now. – John Hughes Oct 3 '18 at 21:49
• I'm glad you like it. – Michael Hardy Oct 3 '18 at 21:51
• Not only do I like it, I now know how to do it myself in the future. – John Hughes Oct 3 '18 at 22:13