Show that the reciprocal of 3+2√2 is an irrational number
I too don't know. Can someone tell me the answer.
To show that the reciprocal of 3+2√2 is an irrational number, we need to assume the number is rational and derive a contradiction.
Let's assume that the reciprocal of 3+2√2 is a rational number, denoted by p/q, where p and q are integers and q is not equal to 0.
So, (3+2√2)^(-1) = p/q
Taking the reciprocal of both sides, we get:
1/(3+2√2) = q/p
Now, let's rationalize the denominator by multiplying the numerator and denominator by the conjugate of 3+2√2, which is 3-2√2:
1/(3+2√2) * (3-2√2)/(3-2√2) = q/p
Simplifying this expression, we have:
(3-2√2)/(9-8) = q/p
(3-2√2)/(1) = q/p
3-2√2 = q/p
Rearranging the equation, we have:
3 = 2√2 + q/p
Since q/p is a rational number (which we assumed), and 2√2 is an irrational number, the sum on the right side of the equation is irrational.
However, the left side of the equation is 3, which is a rational number.
This is a contradiction, as a sum of an irrational number and a rational number cannot be rational.
Therefore, our assumption that the reciprocal of 3+2√2 is rational must be false.
Hence, the reciprocal of 3+2√2 is an irrational number.