Example Solution: q = 1+sqrt(2) Explanation: See solution.
Practice makes perfect
We need to find a nonzero number q such that the expression shown below is a rational number.
q(1-2^(12))
Recall that we can rewrite a rational exponent as a radical.
q(1-2^(12))
a^(12)=sqrt(a)
q(1-sqrt(2))
Therefore, we need q(1-sqrt(2))to be a rational number. The expression inside the parentheses suggests using the conjugate pair. Recall that conjugate pairs are of the form a+ sqrt(b) and a - sqrt(b). When a and b are rational numbers, the product of this conjugate pair is a rational number as well.
(a+ sqrt(b))(a - sqrt(b)) &= a^2 - (sqrt(b))^2 [0.8em]
(a+ sqrt(b))(a - sqrt(b)) &= a^2 - b
Therefore, if we choose q to be the conjugate of 1-sqrt(2) — this is q=1+sqrt(2) — the product will be a rational number.
(1+ sqrt(2))(1 - sqrt(2)) &= 1^2 - (sqrt(2))^2 [0.8em]
(1+ sqrt(2))(1 - sqrt(2)) &= 1 - 2 [0.8em]
(1+ sqrt(2))(1 - sqrt(2)) &= - 1
We can see that their product is -1, which is a rational number, as expected. However, the same would be true if q = a(1+sqrt(2)), where a is any nonzero rational number. This is because the product would then be a(-1)=- a, which would be as well a rational number, as required. Therefore, there are infinitely many solutions.
