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

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

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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