Show that the reciprocal of 3+2√2 is an irrational number
the reciprocal of 3+2√2 is 1/(3+2√2)
1/(3+2√2)
= 1/(3+2√2) * (3-2√2)/(3-2√2)
= (3-2√2)/(9 - 8)
= 3-2√2 , which is irrational
To show that the reciprocal of 3+2√2 is an irrational number, we can assume the opposite and show that it leads to a contradiction.
Let's denote the reciprocal as 1/(3+2√2) and assume it is rational. This means it can be expressed as a fraction of two integers, p and q, where q is not zero:
1/(3+2√2) = p/q
Now, we can multiply both sides of the equation by (3-2√2), which is the conjugate of (3+2√2). This will help us get rid of the square root term in the denominator:
1/(3+2√2) * (3-2√2) = p/q * (3-2√2)
Using the distributive property, we can expand both sides of the equation:
(1 * 3) + (1 * -2√2) + (2√2 * 3) + (2√2 * -2√2) = p/q * 3 + p/q * -2√2
Simplifying further:
3 - 2√2 + 6√2 - 4 * 2 = 3p/q - 2√2p/q
Combine like terms:
(3 - 8) + (4√2 - 2√2) = 3p/q - 2√2p/q
-5 + 2√2 = 3p/q - 2√2p/q
Now, let's consider the left side of the equation (-5 + 2√2) as a single term and the right side as another single term:
-5 + 2√2 = (3p/q - 2√2p/q)
Notice that the left side (-5 + 2√2) is irrational, as it involves a combination of an integer (-5) and a square root (√2).
Now, on the right side, if 3p/q and 2√2p/q were both rational, their difference (3p/q - 2√2p/q) would also be rational.
This leads to a contradiction, as we have shown that the left side (an irrational number) is equal to the right side (a rational number).
Therefore, our initial assumption that the reciprocal of 3+2√2 is rational is false. Consequently, the reciprocal of 3+2√2 must be an irrational number.