# Show that if $n$ is a positive integer and $n$ is not a perfect cube, then $\sqrt{n}$ is an irrational number.

How can I use the method of infinite descent to prove that if $$n$$ is a positive integer and $$n$$ is not a perfect cube, then $$\sqrt{n}$$ is an irrational number. This question originates from a problem in an elementary number theory book. The problem is: show that if $$n$$ is a positive integer and $$n$$ is not a perfect square, then $$\sqrt{n}$$ is an irrational number. The proof in the book is given like this:
Proof. We present the proof by contradiction. Suppose $$\sqrt{n}=\frac{p}{q}$$, where $$p$$ and $$q$$ are positive integers. Thus we get that $$p^2=nq^2$$. Due to $$n$$ is not a perfect square, there exists a positive integer $$m$$, such that $$m<\frac{p}{q}, namely $$0. We next subtract $$mpq$$ from both sides of the equation $$p^2=nq^2$$ to obtain $$p^2-mpq=nq^2-mpq$$. The equation is equivalent to the following equation: $$\frac{p}{q}=\frac{nq-mp}{p-mq}$$. Let $$p_1=nq-mp$$, $$q_1=p-mq$$. Hence $$q_1$$ is a positive integer and $$q_1. So, we get that $$p_1$$ is also a positive integer and $$p_1. As a result, we get that $$\frac{p}{q}=\frac{p_1}{p_1}$$ with $$p_1 and $$q_1. By the well-ordering property, we know that among above positive fractions whose numerators and denominators are positive integers, there is a fraction with the smallest value of the numerator. However, we have shown that from this fraction we can find another fraction with a smaller value of the numerator, leading to a contradiction. This completes the proof by the method of infinite descent. I want to prove the question I raised in this way. Suppose $$\sqrt{n}=\frac{p}{q}$$, where $$p$$ and $$q$$ are positive integers. Thus we get that $$p^3=nq^3$$. Due to $$n$$ is not a perfect cube, there exists a positive integer $$m$$, such that $$m<\frac{p}{q}, namely $$0. But in next step, I can't figure out what polynomial should I subtract. How could I continue…

• Generalize it with prime factors. If $n$ is not a perfect cube there is a prime factor $p$ whose highest power dividing $n$ is not a multiple of three. So if $p^{3k+j}|n$ then $p^j|\frac n{p^{3k})$ so .... Jun 4 '20 at 3:44
• I think the direct generalization of this method might not be appliable for irratioality of $\sqrt{n}$. Jun 4 '20 at 4:45
• Do you need to use infinite descent in your proof? If you believe the Fundamental Theorem of Arithmetic, then a proof can be given in fewer than a dozen lines. Jun 4 '20 at 6:04
• Yes, I know the Fundamental Theorem of Arithmetic will make the proof much simpler, but I need to use infinite descent in my proof.
– Bob
Jun 4 '20 at 6:46

This is basically the same way that you would prove $$\sqrt{2}$$ is rational. Let us assume $$n^{1/3} = a/b$$ for some integers $$a$$ and $$b$$, and further assume that $$a$$ and $$b$$ are as small as possible. Then $$n = a^3 / b^3 \iff n b^3 = a^3$$ Suppose the unique prime factorization of $$n$$ is $$p_1^{k_1} \cdot p_2^{k_2} \cdots p_m^{k_m}$$. Then it must be the case that $$p_1^{q_1} \cdot p_2^{q_2} \cdots p_m^{q_m}$$ divides $$a$$, where $$q_i = \lceil k_i / 3\rceil$$; otherwise $$a^3$$ could not be a multiple of $$n$$ as suggested. Since by assumption $$n$$ is not a perfect cube, at least one of the $$k_i$$ must not be a multiple of $$3$$. WLOG, let $$k_1$$ be this number that is not a multiple of $$3$$. Let us count factors of $$p_1$$ on both sides of the equation $$nb^3 = a^3$$. On the left-hand side, we have $$k_1$$ factors from $$n$$, and some factors from $$b^3$$ (we'll come back to this later). On the right-hand side, we have $$3 q_1 = 3 \lceil k_1 / 3\rceil$$, which is either $$3 k_1 + 1$$ or $$3 k_1 + 2$$ factors, depending upon whether $$k_1$$ was $$1$$ or $$2$$ modulo $$3$$. Of course, for the equation $$nb^3 = a^3$$ to hold, the number of factors of $$p_1$$ must be the same on each side. Therefore, it must be the case that $$p_1$$ divides $$b^3$$ (otherwise we simply do not have enough factors of $$p_1$$ on the left-hand side to make up for the number of $$p_1$$'s we have on the right-hand side). However, if $$p_1$$ divides $$b^3$$, then $$p_1$$ must divide $$b$$ since $$p_1$$ is a prime. But $$p_1$$ divides $$q$$ too, and as $$p_i$$ is a prime (and hence at least two), it follows that $$n = \frac{(a / p_1)^3}{(b / p_1)^3} = \frac{(a')^3}{(b')^3} \implies n^{1/3} = \frac{a'}{b'}$$ with $$a'$$ and $$b'$$ integers. Note also that $$a' < a$$ and $$b' < b$$, so we have constructed a fraction equal to $$n^{1/3}$$ with a strictly smaller numerator and denominator. Hence, $$n^{1/3}$$ must be irrational. $$\square$$