The number is irrational. Proof If were rational, then there would be integers p
ID: 2983229 • Letter: T
Question
The number is irrational. Proof If were rational, then there would be integers p and q such that = p/q. We assume that the fraction p/q is in lowest terms so that p and q have no common factors. Note that Hence p2 = 2q2. This makes p2 even. Hence p must be even, since the square of an odd number would be odd. Thus p = 2k where k is an integer. Substituting: 4k2 = 2q2 2k2 = q2. Thus, q must be even also. But this contradicts the assumption that p and q have no common factors, proving our theorem. This exercise in a continuation of Exercise 8. Prove that is irrational. Hint: Repeat the proof of Theorem 3, using class instead of even and odd.Explanation / Answer
Let's suppose ?3 is rational.
This means there are two unique integers p and q, that areprimestoward each other (i.e. they do not share any integer divisor except 1), and such as:
?3 = p/q
If we square that equality:
(?3)