x and y are real numbers
asked Oct 19, 2016 in Algebra 1 Answers by anonymous

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Let x=(m1/m2)+p and y=(n1/n2)+q, where m1, m2, n1, n2 are integers and m2,n2≠0. The fractions are rational by definition of rational numbers; p and q are irrational so cannot be expressed as the quotient of two integers.

x+y=(m1/m2)+(n1/n2)+p+q=(m1n2+m2n1/m2n2)+p+q=M/N+p+q where M is an integer = m1n2+m2n1 and N is also an integer = m2n2. Therefore M/N is rational and p+q is irrational. So x+y is irrational. However x and y are individually irrational; neither is rational.

Let p=0 so x is rational. In this case x+y=M/N+q which is irrational because of q. The sum is irrational because y is irrational. Similarly, if q=0, x+y is irrational because x is irrational.

If p=q=0 then x+y is rational.

All cases of rationality have been covered. If x+y is irrational, therefore, either x or y are irrational, or both are irrational, so it is not true that at least one is rational. Rather, it is true that at least one is irrational.


answered Oct 19, 2016 by Rod Top Rated User (416,500 points)
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