Recall that a real number r is called rational if there are integers n and d so

Question

Recall that a real number r is called rational if there are integers n and d so that r = n/d. Prove the following statement using either the contrapositive or contradiction. If a is a non-zero rational number and b is irrational, then ab is irrational. Suppose a is a real number with a notequalto 0 and a notequalto 1. Use mathematical induction to prove that for all n elemetnsof N we have sigma^n_i = 0 a^I = 1 - a^n + 1/1 - a Let b_0 = 1 and, for n > 0, let b_n = 5b_n - 1 - 2. What are the first five terms of the sequence b_0, b_1, b_2, b_3, b_4, ...? Prove that for n elementsof N, b_n = 5^n + 1/2

Explanation / Answer

3) we will prove by contradiction

a is rational = j/k ,where j and k are integer

assume that ab is rational ,which means it can be written in the form of m/n where m and n are integers

ab = m/n

now divide both side by a

so b = m/(na) = m/(n*j/k) = mk/nj

observe that mk as well as nj are integers

since b is rational ,it can not be written as ratio of two integers .

hence there is contradiction

hence our assumption that ab is rational is wrong

so ab is irrational QED

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