Use proof by Contradiction to prove that the sum of an irrational number and a rational number is irrational.

Use proof by Contradiction to prove that the sum of an irrational number and a rational number is irrational.

Question
Discrete math
asked 2020-11-09
Use proof by Contradiction to prove that the sum of an irrational number and a rational number is irrational.

Answers (1)

2020-11-10
A real number r is rational if and only if there exists two integer a and b such that \(r = a/b\) ( b is not equal to zero) Let us assume that x is an irrational number and y is a rational number . Suppose for the sake of contradiction that \(x + y\) is not irrational \(y - a/b\) since s is irrational , x cannot be written as the ratio of two integers Since \(x + y\) in not irrational, \(x + y\) is rational . By the definition of rational, there exists integers c and d not equal to zero such that \(x + y = c/d\)
\(\text{since} y = a/b\)
\(x + a/b = c/d\)
\(x = c/d -a/b\)
\(x = (bc - ad)/bd\) Since a,b,c,d are integers bc - ad and bd are also integers. Moreover bd is zero as b and d are both non zero. However , this then implies that x is a rational number (as it is written as the ratio of tow integers). which is in contradiction with the fact that with the fact that x is an irrational number. Thus our supposition that \(x + y\) is not irrational is false. Which means that \(x + y\) is irrational .
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