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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

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

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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.

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

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