The set of real numbers is the set you are likely familiar with from your previous math courses, particularly algebra and calculus. These are all the numbers found on the number line, such as \(0, 1, -3, 1/2, \pi, \sqrt{2}\text{,}\) etc. Recall, we use the notation \(\mathbb{R}\) for the set of real numbers.
A real number, \(r\text{,}\) is rational if there exist \(a, b\in \mathbb{Z}\) such that \(r=\frac{a}{b}\) and \(b\neq 0\text{.}\) The set of rational numbers is represented by \(\mathbb{Q}\)
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.
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 SectionΒ 3.4.
In your earlier dealings 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.
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, \(0.1=1/10; 0.01=1/100\text{.}\) Thus, \(0.2345=2345/10000\text{.}\) Letting \(a=2345, b=10000\text{,}\) we can see that \(a, b\in \mathbb{Z}, b\neq 0\text{.}\)
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:
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,
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.
We introduce the idea of divisibility for integers. It is important to understand that this concept involves only integers and is not the same thing as division. Divisibility is a property while division is an operation.
Important note: \(d\mid n\) is a relationship between two integers. It is a statement that is either true or false. It does NOT equal a number. It is NOT the same thing as a fraction. It is not an equation, but it can be translated to the equation \(n=dk\) for some integer \(k\text{.}\)
If \(d\) does not divide \(n\text{,}\) then for every \(k\in \mathbb{Z}, n\neq dk\text{.}\) This can be difficult to show in general, but if \(n\) and \(d\) are specific integers, you could show \(\frac{n}{d}\) is not an integer. This is the ONLY time a fraction can show up in proofs about divisibility.
Let \(a, b, c\in \mathbb{Z}\text{.}\) Assume \(a\mid b\) and \(b\mid c\text{.}\) Then by definition, \(b=ak\) for some \(k\in \mathbb{Z}\) and \(c=bj\) for some \(j\in \mathbb{Z}\text{.}\) Substituting the first equation into the second, \(c=(ak)j=a(kj)\text{.}\) Since \(kj\in\mathbb{Z}\text{,}\)\(a\mid c\text{.}\)
βProofβ: 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
Determine if the following statements are true or false. For true statements provide a proof. For false statements provide a counter example AND determine if a small change would make the statement true. If so, correct the statement and provide a proof of the new statement.
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
A fast food chain has a contest in which a card with numbers on it is given to each customer who makes a purchase. If some of the numbers on the card add up to 100, then the customer wins $100. A certain customer receives a card with the numbers