exist unique rational numbers a and b with x = ar + b . Solution Since r is fixe

Question

exist unique rational numbers a and b with x = ar + b.

Explanation / Answer

Since r is fixed irrational set S consist of irrational numbers. Now you will ask why? s, t belongs to rational number, 'sr' is product of rational and an irrational number so 'sr' is irrational and t is rational so 'sr+t' is irrational because sum of rational and irrational number is irrational. Hence set S consist of irrational numbers. Let x belongs to S then x is irrational. By construction of set S, there exits rational numbers a,b such that x=ar+b. We will prove this by contradiction Let a,b is not unique. Hence there exist a1,b1 such that x=a1r+b1 Then ar+b=a1r+b1 => (a-a1)r+(b-b1)=0 Important thing here is that (a-a1)r is irrational and (b-b1) is rational, but there sum is zero. This is possible only if both are individually zero. a-a1=0 =>a=a1 and b-b1=0 =>b=b1 Hence there exist unique a,b such that for any x belong to S x=ar+b Note: I have used few obvious result, I will prove it here 1)Product of rational and irrational number is always irrational proof: Let a is rational and b is irrational. Suppose that ab is rational number. ab=p/q (q not equal to 0) b=p/qa Hence b is also rational but this contradicts the assumption that b is irrational. Hence product of rational and irrational number is always irrational. 2) Sum of rational and irrational number is always irrational. Proof:- let a be rational and b be irrational we have to find what a+b is let us assume that a+b is rational then there exist p,q such that a+b=p/q b=(p/q)-a which is rational. but this contradicts the assumption that b is irrational. Hence sum of rational and irrational number is always irrational.

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