## Number Systems: Problem Sheet 6

In Example 3.9 we used unique factorization to show that while $\sqrt{3}$ and $\frac{2148105}{1240209}$ are both equal to $1.73205080756$ to twelve decimal places, they are not the same number.

Proof of Claim 3.10. More generally, suppose for a contradiction that $\sqrt{3} = m/n$ where $m \in \mathbb{N}$ and $n \in \mathbb{N}$. Multiply through by $n$ and square to get $3n^2 = m^2.$

Now consider the prime factorizations of each side. In particular, we will look at the exponent of the prime 3. Let $e$ be the exponent of 3 in the prime factorization of $n$ and let $B = n/3^e$. Similarly, let $d$ be the exponent of 3 in the prime factorization of $m$, and let $A = m/3^d$. Since $m = 3^d A$ and $n = 3^e B$ we have

$3\times 3^{2e} \times B^2 = 3^{2d} \times A^2.$

On the left-hand side the exponent of 3 is $2e+1$ and on the right-hand side the exponent of 3 is $2d$. This contradicts unique factorization.

Connection with Example 3.9. In Example 3.9 we had $m = 2148105$ and $n=1240209$ and the equation $3n^2 = m^2$ became

$3 \times 3^4 \times 41^2 \times 3361^2 = 3^2 \times 5^2 \times 71^2 \times 2017^2.$

On the right-hand side the exponent of 3 is 2 and on the left-hand side the exponent of 3 is 5. As expected, this contradicts unique factorization. (Of course you could also multiply out both sides, and show that the left-hand side is $4614355091043$, whereas the right-hand side is $4614355091025$, but looking at the powers of 3 uses much less calculation!)

Question 4. Now try to adapt the proof above to show that $\sqrt[3]{5}$ is irrational.

Background. The fraction $\frac{2148105}{1240209}$ is an unusually good approximation to $\sqrt{3}$. Such good approximations can be found using a version of Euclid’s Algorithm, to be seen later in the course. You could also try searching for convergents or continued fractions on the web.