# 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.
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Usamah Prosser

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 $x+y=c/d$

$x+a/b=c/d$
$x=c/d-a/b$
$x=\left(bc-ad\right)/bd$ 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 .