### Definition 4.2.1.

A real number, \(r\text{,}\) is rational if there exist \(a, b\in \mathbb{Z}\) such that \(r=\frac{a}{b}\) and \(b\neq 0\text{.}\) Notation: \(\mathbb{Q}\) is the set of rational numbers.

In this section we introduce the formal definition of a rational number and use it practice with the technique of direct proof.

A real number, \(r\text{,}\) is rational if there exist \(a, b\in \mathbb{Z}\) such that \(r=\frac{a}{b}\) and \(b\neq 0\text{.}\) Notation: \(\mathbb{Q}\) is the set of rational numbers.

A real number, \(r\text{,}\) that is not rational is irrational. There is not a nice notation for the irrational numbers. We will use \(\mathbb{R}\setminus\mathbb{Q}\), which is the set of real numbers “minus” the set of rationals.

To determine if a given number is rational, we need to be able to find a way to write it as a fraction of integers. To prove a number is rational is really a type of existence proof--we need to show \(a, b\in \mathbb{Z}\) exist. To prove a number is not rational, we need to show there is no possible way to write it as a fraction of integers. Also, keep in mind that rational and irrational numbers first need to be real numbers. It is possible that a number that is not rational is not a real number, and thus, not irrational either.

Is O is rational?

Yes: \(0=0/1\text{.}\)

Is 2 rational?

Yes: \(2=2/1\text{.}\)

Is \(9/5\) rational?

Yes: is already has the form of an integer over an integer.

Is \(2/0\) rational?

No: the denominator cannot be 0. In fact, \(2/0\) is not a number.

You have seen some common examples of irrational numbers in previous courses: \(\sqrt{2}, \pi, e\text{.}\) It is, in fact, challenging to prove these are irrational. We will see the proof that \(\sqrt{2}\) is irrational later in this course.

In your previous experience with rational and irrational numbers, you may have seen the property that rational numbers are ones with terminating or repeating decimal expansions, while irrational numbers have non-terminating and non-repeating decimal expansions. The next couple of examples explore this property.

Show \(0.2345\) is a rational number.

We need to find \(a, b\in \mathbb{Z}\) such that \(0.2345=\frac{a}{b}\) and \(b\neq 0\text{.}\) We can use what we know about decimals: for example, \(0.1=1/10; 0.01=1/100;\) etc. Thus, \(0.2345=2345/10000\text{.}\) Letting \(a=2345, b=10000\text{,}\) we can see that \(a, b\in \mathbb{Z}, b\neq 0\text{.}\)

Show the repeating decimal \(0.\overline{123}\) is a rational number.

We need to find \(a, b\in \mathbb{Z}\) such that \(0.\overline{123}=\frac{a}{b}\) and \(b\neq 0\text{.}\) This one is trickier than the last example and requires a new technique. First, let \(x=0.\overline{123}=0.123123\ldots\text{.}\) Then multiply both sides of \(x=0.123123\ldots\) by 1000, so that \(1000x=123.123123\ldots\text{.}\) We chose 1000 in order to get one set of the repeated digits in front of the decimal point. Now we subtract the two equations from each other:

\begin{align*}
123.123123\ldots&=1000x\\
-(0.123123\ldots&=x)
\end{align*}

Resulting in \(123=999x\text{.}\) Now we just solve for \(x\text{:}\) \(x=123/999\text{.}\)

Letting \(a=123, b=999\text{,}\) we can see that \(a, b\in \mathbb{Z}, b\neq 0\text{.}\)

Prove the following numbers, \(x\text{,}\) are rational by finding integers, \(a\) and \(b\) so that \(x=\frac{a}{b}\text{.}\)

The next theorem gives us an example of how to prove more general statements with rational numbers.

The sum of two rational numbers is rational.

Let \(r, s\) be rational. Show \(r+s\) is rational.

Since \(r\) is rational, it can be written as \(\frac{a}{b}\) for some \(a, b\in \mathbb{Z}, b\neq 0\text{.}\) Similarly, since \(s\) is rational, it can be written as \(\frac{p}{q}\) for some \(p, q\in \mathbb{Z}, q\neq 0\text{.}\) (Note, we need to use different letters since \(r\) and \(s\) are not necessarily the same.) Now,

\begin{equation*}
r+s=\frac{a}{b}+\frac{p}{q}.
\end{equation*}

Finding a common denominator,

\begin{equation*}
r+s=\frac{aq+bp}{pq}.
\end{equation*}

Then \(aq+pb, pq \in \mathbb{Z}\) and, since \(p, q\neq 0, pq\neq 0\text{.}\) Therefore, \(r+s\) is rational.

Once we have proven a theorem, we can use it to prove additional statements. Note, a corollary is just a theorem that follows almost directly from a previous theorem.

The double of a rational number is rational.

Let \(r\) be a rational number. We want to show \(2r\) is rational. But \(2r=r+r\text{.}\) By Theorem 4.2.6, \(r+r\) is rational.

Let \(r, s\) be rational numbers.

Prove \(r/2\) is rational.

Prove \(5r-2s\) is rational.

Prove or disprove \(\frac{1}{r}\) is rational.

True.

False.

True or false: \(-9\) is rational.

True.

False.

True or false: \(-\frac{6}{13}\) is rational.

True.

False.

True or false: \(0.\overline{3}\) is rational.

True.

False.

True or false: \(\frac{6}{0}\) is rational.

True.

- \(\frac{6}{0}\) is not a real number.
False.

- \(\frac{6}{0}\) is not a real number.

True or false: \(\frac{6}{0}\) is irrational.

True.

False.

True or false: \(14.5467\) is rational.

True.

False.

True or false: \(49.1122\overline{45}\) is rational.

True.

False.

True or false: \(0.\overline{9}\) is rational.

True.

- Although this is false, we need more proof techniques to prove it.
False.

- Although this is false, we need more proof techniques to prove it.

True or false: \(\sqrt{3}\) is rational.

True.

- \(\sqrt{4}=2\)
False.

- \(\sqrt{4}=2\)

True or false: \(\sqrt{4}\) is rational.

Prove the following numbers are rational by writing them as a ratio of two integers.

- \(\displaystyle \frac{3}{7}+\frac{5}{9}\)
- \(\displaystyle 351.549249249\ldots\)

The zero product property says that if a product of two real numbers is 0 then one of the numbers must be zero.

- Write this property formally using quantifiers and variables.
- Write the contrapositive of your answer in (a).
- Write an informal version (without symbols and variables) of your answer to part (b).

Prove every integer is a rational number.

Find the mistakes in the following “proof.”

Theorem: The sum of any two rational numbers is rational.

“Proof”: Suppose \(r\) and \(s\) are rational numbers. If \(r+s\) is rational, then by definition of rational \(r+s=a/b\) for integers \(a\) and \(b\) with \(b\neq0\text{.}\) Since \(r\) and \(s\) are rational, \(r=i/j\) and \(s=m/n\) for integers \(i, j, m\) and \(n\) with \(j\neq 0\) and \(n\neq 0\text{.}\) Thus

\begin{equation*}
r+s=\frac{i}{j}+\frac{m}{n}=\frac{a}{b},
\end{equation*}

which is the quotient of two integers with nonzero denominator. Thus, it is rational.

Consider the statement: The square of any rational number is a rational number.

- Write the statement formally using a quantifier and a variable.
- Determine whether the statement is true or false and justify your answer.

Determine if the following statements are true or false. For *true* statements provide a proof. For *false* statements provide a counterexample *and* determine if a small change would make the statement true. If so, correct the statement and provide a proof of the new statement.

- The quotient of any two rational numbers is a rational number.
- If \(r\) and \(s\) are rational numbers then \(\frac{r+s}{2}\) is a rational number.

Suppose \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\) are integers and \(a\neq c\text{.}\) Suppose also that \(x\) is a real number that satisfies the equation

\begin{equation*}
\frac{ax+b}{cx+d}=1.
\end{equation*}

Must \(x\) be rational? If so, express \(x\) as a ratio of two integers. Is the condition \(a\neq c\) important?

Solve for \(x\text{.}\)

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