Prove that log_2 3 is irrational. Please use the following predicates: (i) Q(x):

Question

Prove that log_2 3 is irrational. Please use the following predicates: (i) Q(x): x is rational" defined as follows exist p, q elementof Z, q notequalto 0 and x = p/q (ii) I(x): " x is irrational" defined as follows Q(x). You may also use (without proof) the following facts (and nothing else, save for rules of elementary algebra): (iii) Definition of logarithm: For any positive real number x, logarithm with base 2 of is called the real number y such that x = 2^y. (v) Forall x elementof R^+, x > 2 log_2 x > 0. (v) The product of two even integers is even: also the product of two odd integers is odd.

Explanation / Answer

Let us assume that log2 3 is rational or Q(log2 3)

=> log2 3 = p/q

=> 2p/q = 3

Raising to the power q

=> 2p = 3q

2p = 2*2*2.... (p times)

As the product of two even integers is even, 2*2 is even, (2*2)*2 is even and so on

=> 2p is even.

3q = 3*3*3.... (q times)

As the product of two odd integers is odd, 3*3 is odd, (3*3)*3 is odd and so on

=> 3q is odd.

But we have 2p = 3q.

Thus we arrive at a contradiction

=> ~Q(log2 3) or I(log2 3)

=> log2 3 is irrational.

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