Q1. Prove the following statement: for all integers a and b, if a is even and b
ID: 3731908 • Letter: Q
Question
Q1. Prove the following statement: for all integers a and b, if a is even and b is a multiple of 3, then ab is a multiple of 6. Q2. If m and n are integers and neither m nor n is zero, is (m n)/mn a rational number? Q3. Use proof by contradiction to show that the sum of any rational number and any irrational number is irrational. Hint: You can use the following definition to assist you answer Q1 and Q2 Definition: A real number ris rational if, and only if, it can be expressed as a quotient of two integers with a nonzero denominator. A real number that is not rational is irrational. More formally, if ris a real number, then ris rational eointegers a and b such that r= and bz 0.Explanation / Answer
Q1. Given a is even i.e. we can represent a as 2k where k is some integer.
And also given that b is multiple of 3 i.e. we can represnt b as 3k' where k' is an integer.
Multiplying a and b = ab = 2k* 3k' = 6 kk' which is a multiple of 6 (proved).
Q2. We know A real number r is rational if, and only if, it can be expressed as a quotient of two integers with a nonzero denominator. As m and n are integers and neither of them zero therefore the denominator 'mn' is also non-zero and numerator is 'm+n' which is real. Therefore we can say that (m+n)/mn is rational number.
Q3. Assume that x is a irrational number and sum of x with a rational number say a/b is rational number say c/d, where a,b,c,d are intergers and b and d != 0.
We can write a/b + x = c/d. Also, x = c/d-a/b = (cb-ad)/bd and b.d is not equal to zeros and since a,b,c,d are intergers (cb-ad) is also an integer making (cb-ad)/bd a rational number which is contradiction as we have defined x to be a irrational number. Therefore a contradiction and thus making the assumption wrong. Hence sum of rational and irrational number is an irrational number.
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