# *Proof* Sum of Rational and Irrational Numbers

• prelic
In summary, in a proof by contradiction, we assume the original hypothesis is true and the negated original conclusion is true. We then use logical reasoning to arrive at a contradiction, which leads us to conclude that the original hypothesis is true and the original conclusion is also true.
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## Homework Statement

Prove by contradiction: If a and b are rational numbers and b != 0, and r is an irrational number, then a+br is irrational.

In addition, I am to use only properties of integers, the definitions of rational and irrational numbers, and algebra.

You guys should also know that I am new to proofs, so if I'm breaking convention in any blatant way, please let me know.

## The Attempt at a Solution

First, I figure I need to find the negation of this statement:

Negation: There exists rational numbers a and b, b!=0, and irrational number r, such that a+br is rational.

I'm pretty sure this negation is correct, but I've been wrong before

Anyway, my proof would start like this:

Proof: To prove that for any rational numbers a,b,b!=0, and irrational number r, a+br is irrational, let's suppose not. Suppose there exists rational numbers a,b,b!=0 and irrational number r such that a+br is rational.
Let c,d,e,f be integers, by definition of rational, a+br can be rewritten as:

$$\frac{c}{d}$$ + $$\frac{e}{f}$$*r is rational.

$$\frac{e}{f}$$*r = -$$\frac{c}{d}$$

r = -$$\frac{cf}{de}$$

$$\frac{cf}{de}$$ can be written as a quotient of integers, so it is rational, therefore, our negation is false, therefore, our theorem is true?

I think I'm on the right track, but I know its messy and I might have made some logic mistakes...Also, my professor said that if we didn't have to explicitly say why b!=0, we probably weren't doing it right, and I never really did, so I have no idea...any help would be greatly appreciated.

edit-I guess if b was 0, that would make the quantity a+(br) 0, which leaves you with a, which is just a rational number

In a proof by contradiction, you assume that the original hypothesis is true and that the conclusion is false.

Your original hypothesis is "a and b are rational numbers and b != 0, and r is an irrational number". Your original conclusion is "a + br is irrational".

The negative of your original conclusion is that a + br is not irrational, or in other words that a + br is rational.

Now start by assuming original hypothesis is true and that a + br is rational (the negated original conclusion). You should arrive at a contradiction, from which you can conclude that when the original hypothesis is true, the original conclusion is true, as well.

## What is a rational number?

A rational number is a number that can be written as a ratio of two integers, where the denominator is not equal to zero. Examples of rational numbers include 1/2, 3/4, and -5/9.

## What is an irrational number?

An irrational number is a number that cannot be written as a ratio of two integers. These numbers have decimal representations that neither terminate nor repeat. Examples of irrational numbers include pi (3.14159...), e (2.71828...), and the square root of 2 (1.41421...).

## What is the sum of a rational and an irrational number?

The sum of a rational and an irrational number is an irrational number. This is because the decimal representation of an irrational number is non-terminating and non-repeating, and when added to a rational number, it cannot be simplified to a terminating or repeating decimal.

## Can the sum of two irrational numbers be rational?

No, the sum of two irrational numbers can never be rational. This is because the decimal representations of both irrational numbers are non-terminating and non-repeating, and when added, they cannot be simplified to a terminating or repeating decimal.

## How can we prove that the sum of a rational and an irrational number is an irrational number?

We can prove this by contradiction. Assume that the sum of a rational and an irrational number is a rational number. This would mean that the decimal representation of the sum would either terminate or repeat. However, this is not possible since an irrational number added to a rational number will always result in a non-terminating and non-repeating decimal. Therefore, our assumption is false and the sum of a rational and an irrational number is always an irrational number.

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