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.

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