Question

Let’s say x is an irrational number and y is a rational number. The rational number y can be written as y=a/b, where a and b are integers and b=0. Leave the irrational number x as x since it can’t be written as the ratio of two integers. Prove the hypothesis using proof by contradiction. In other words try to show that x+y equals a rational number instead of an irrational number. Let the sum equal m/n where m and n are integers and n=0. Plug in the values to get this equation x+a/b=m/n. Solve for the equation for x

Answer 1

A proof by contradiction is one in which we assume the opposite of what we want to prove is true, and then explain how that assumption leads to a contradiction.

Here, we want to prove that the sum of an irrational number and a rational number is irrational, so we're going to assume the sum is rational and see why that can't possibly be true.

The question sets up this equation as a starting point:

`x+(a)/(b) =(m)/(n)`

and we're told that a, b, m, and n are all integers. We can now subtract
`(a)/(b)` from both sides to get the equation in the form

`x=(m)/(n)-(a)/(b)`

And here's where we run into our contradiction:
`(a)/(b)-(m)/(n)`, being the difference of two rational numbers, gives us a rational number as a result, but we also stated at the beginning that x was an irrational number. If we accept that, our equation now makes the absurd claim that

irrational number = rational number

which is like saying that peanut butter = jelly, black = white, or up = down; it just doesn't make sense!

Since our reasoning was airtight, the only thing that could've led us down this tragic path was our assumption that the sum of an irrational and a rational number was rational, so we know now that the sum of a rational number and an irrational number must be irrational.