Real Numbers — Exercise 1.2

Method: Proof by contradiction. A prime dividing an integer square also divides that integer.

1. Suppose \(\sqrt5=a/b\), with integers a,b, b ≠ 0 and gcd(a,b)=1.
2. Squaring gives \(a^2=5b^2\). Thus 5 divides a², so 5 divides a. Write a=5k.
3. Substituting gives \(25k^2=5b^2\), hence \(b^2=5k^2\). Therefore 5 also divides b.
4. Now a and b share the factor 5, contradicting the assumption that they are coprime.

Assume the whole expression equals a rational number r. Rearrange to isolate the square root:

\[3+2\sqrt5=r\]

\[\sqrt5=\frac{r-3}{2}\]

Subtracting an integer from a rational number and dividing by the nonzero integer 2 produces a rational number. This would make √5 rational, contradicting the result proved in Question 1.

Use that √2 and √5 are irrational, together with closure of rational numbers under subtraction and division by a nonzero rational number.

(i) \(1/\sqrt2\)

Suppose \(1/\sqrt2=r\) is rational. It is nonzero, so its reciprocal 1/r is rational. But \(1/r=\sqrt2\), a contradiction.

(ii) \(7\sqrt5\)

Suppose \(7\sqrt5=r\) is rational. Dividing by 7 gives \(\sqrt5=r/7\), which would be rational. This contradicts the irrationality of √5.

(iii) \(6+\sqrt2\)

Suppose \(6+\sqrt2=r\) is rational. Subtracting 6 gives \(\sqrt2=r-6\), which would be rational. This is a contradiction.

Why √2 is irrational: If √2=a/b in lowest terms, then a²=2b² forces a=2k. Substituting gives b²=2k², so b is even too. The common factor 2 contradicts lowest terms.