Decimals, Percents, and Real Numbers
7.5 Real Numbers
Real Numbers
As we further extend our definition of what numbers can be, we have to look at the spaces between the numbers we already know. We have populated the natural numbers, whole numbers, integers, and even rational numbers. Yet there are still holes in the number line left by these numbers. One such question led to the downfall of the Pythagoreans.
Fact: Historical note
The discovery of irrational numbers is usually attributed to Pythagoras, more specifically to the Pythagorean Hippasus of Metapontum, who produced a proof of the irrationality of $\sqrt{2}$. The story goes that Hippasus discovered irrational numbers when trying to represent $\sqrt{2}$ as a fraction. Pythagoras believed in the absoluteness of numbers and could not accept the existence of irrational numbers.
Imagine that you are shown a square and told that the area contained within is $9$ square units. What must the side lengths be? Now suppose that another square is shown that has area $2$ square units. What must the side lengths be?
Here is a classic algebraic proof that there is no rational number that answers the second question.
Proof: Why $\sqrt{2}$ is irrational
Assume for contradiction that such a number exists. This means, if $\sqrt{2}$ is rational, then it must be able to be written as a fully reduced fraction.
$$\sqrt{2}=\frac{a}{b}$$
where $a$ and $b$ are integers, $b\neq0$, and $\gcd(a,b)=1$.
Then, based on what we know about rational numbers, it must be the case that,
$$2=\left(\frac{a}{b}\right)^2$$ $$2=\frac{a^2}{b^2}$$
and multiplying by $b^2$ gives
$$2b^2=a^2.$$
This shows that $a^2$ is even, so $a$ must be even. Ask yourself: why? And if $a$ is even, then by definition $a=2k$ for some integer $k$. Substituting gives
$$2b^2=(2k)^2=4k^2.$$
Dividing by $2$ gives
$$b^2=2k^2.$$
So $b^2$ is even, which means $b$ is even. But then both $a$ and $b$ are even, contradicting the assumption that $\frac{a}{b}$ was fully reduced. At the very least $\gcd(a,b) \geq 2$ which contradicts our assumption that $\sqrt{2}$ is rational.
Therefore, it is impossible for $2=\left(\frac{a}{b}\right)^2$, so $\sqrt{2}$ cannot be written as a rational number.
Definition: Irrational Number
An irrational number is a real number that cannot be expressed as a fraction of two integers. The decimal representation of an irrational number is non-terminating and non-repeating.
Definition: Real Number
A real number is any number that can be represented by a decimal.
MyOpenMath: Identifying types of numbers
To find some more irrational numbers, we need to talk about squareroots. Irrational numbers first start appearing to students when they begin working with area, like the example above. In truth, there are two numbers that could be squared to give $9$. The number $3$ was our original solution, but $-3$ is also a solution. In terms of area, however, the negative solution does not make much sense. So we define not just the square root, but the principal square root.
Definition: Principal Square Root
The principal square root of a number is the nonnegative number whose square is the original number. For real numbers $a$ and $x$, the statement $\sqrt{x}=a$ means $a^2=x$ and $a\ge0$.
Example: Square roots and equations
Find the following.
- The square roots of $144$
- $\sqrt{144}$
- $\sqrt{\frac{4}{9}}$
- The solutions of $x^2=5$
Show Solution
- The square roots of $144$ are $-12$ and $12$.
- $\sqrt{144}=12$ because the principal square root is nonnegative.
- $\sqrt{\frac{4}{9}}=\frac{2}{3}$.
- The solutions of $x^2=5$ are $x=\pm\sqrt{5}$.
Warning: Square-root caution
The square root of a number is always nonnegative. The square root of a negative number is not a real number. In particular,
$$\sqrt{x^2}=|x|.$$
MyOpenMath: Irrational and square roots
Other Roots
In general, the $n$th root of a number is the number that, when raised to the $n$th power, gives the original number. The principal $n$th root of a number is the nonnegative number whose $n$th power is the original number. For real numbers $a$ and $x$, the statement $\sqrt[n]{x}=a$ means $a^n=x$ and $a\ge0$.
Example: Other roots
Find the following.
- The cube root of $64$
- $\sqrt[3]{64}$
- $\sqrt[3]{\frac{8}{27}}$
- The solutions of $x^5=32$
Show Solution
- The cube root of $64$ is $4$.
- $\sqrt[3]{64}=4$.
- $\sqrt[3]{\frac{8}{27}}=\frac{2}{3}$.
- The real solution of $x^5=32$ is $x=2$.
MyOpenMath: Higher-order roots
Estimating Square Roots
As we saw earlier, the square root of $2$ is not a rational number. It is, however, a real number. That means that it can be represented by a decimal, and we can estimate its value. Initially guessing, we might try $1.5^2=2.25$ and $1.4^2=1.96$. The number $1.4$ is too small, and $1.5$ is too large. So we know that $\sqrt{2}$ is between $1.4$ and $1.5$. Instead of repeatedly guessing, we can use a method to approximate square roots.
Newton's Method for Calculating Square Roots
Fact: Newton's method steps
- Make an initial guess for the square root. The closer the better.
- Divide the original number by the guess.
- Average the result of the division and the guess to get a new guess.
- Repeat steps 2 and 3 until the desired accuracy is achieved.
Example: Approximate $\sqrt{2}$
Use Newton's method to approximate the square root of $2$ to $5$ decimal places.
Show Solution
Start with the guess $1.4$.
| Guess | Quotient | New Guess | New Guess Squared |
|---|---|---|---|
| $$1.4$$ | $$\frac{2}{1.4}=1.42857$$ | $$\frac{1.4+1.42857}{2}=1.414285$$ | $${(1.414285)}^2\approx2.000202$$ |
| $$1.414285$$ | $$\frac{2}{1.414285}\approx1.41421356$$ | $$\frac{1.414285+1.41421356}{2}\approx1.41424928$$ | $${(1.41424928)}^2\approx2.0000000001$$ |
So
$$\sqrt{2}\approx1.41421.$$
As you can see, this method is extremely efficient. With only two iterations of the algorithm, we have found $\sqrt{2}$ to several decimal places. This is an impressive result, and it shows the power of Newton's method. We can now calculate an approximation of irrational square roots to any desired level of accuracy.
Example: Newton's Method
Approximate $\sqrt{3}$ to at least $5$ decimal places using Newton's method.
MyOpenMath: Newton method for square roots
Real Numbers, Operations, and Their Properties
So far we have described natural numbers $\mathbb{N}$, which are a subset of whole numbers $\mathbb{W}$, which are a subset of integers $\mathbb{Z}$, which are a subset of rational numbers $\mathbb{Q}$, which are a subset of real numbers $\mathbb{R}$. The real numbers are the most general set of numbers we discuss in this course. They include all the numbers we have studied so far, but they are still not all of the numbers.
Theorem: Properties of Real Numbers
For all real numbers $a$, $b$, and $c$, the following properties hold.
- Closure: $a+b$, $a-b$, and $ab$ are real numbers, and $\frac{a}{b}$ is a real number when $b\ne0$.
- Commutative Property: $a+b=b+a$ and $ab=ba$.
- Associative Property: $(a+b)+c=a+(b+c)$ and $(ab)c=a(bc)$.
- Distributive Property: $a(b+c)=ab+ac$.
- Identity Property: $a+0=a$ and $a\cdot1=a$.
- Inverse Property: $a+(-a)=0$ and $a\cdot\frac{1}{a}=1$ when $a\ne0$.
- Multiplication Property of Zero: $a\cdot0=0$.
- Density Property: For any two real numbers $a$ and $b$ with $a<b$, there is a real number $c$ such that $a<c<b$.
Example: Number between two reals
Find a repeating decimal between $\sqrt{2}$ and $1\frac{4}{9}$.
Show Solution
First approximate the endpoints:
$$\sqrt{2}\approx1.4142,\qquad 1\frac{4}{9}=1.4444\ldots$$
Now choose a repeating decimal between them, for example
$$1.\overline{42}=1.424242\ldots$$
This number lies strictly between $1.4142$ and $1.4444\ldots$.