Theorem 8.2.1.
Let \(a\) be a positive real number. If a real number \(k\) is a solution to \(x^2=a\text{,}\) then \(-k\) is also a solution.
Section 8.2 Simple Roots and Irrationality
Subsection 8.2.1 Principal Roots
Think about the equation \(x^2=4\text{.}\) How many solutions does this equation have? We know that \(2^2=2 \rimes 2=4\text{,}\) but also \((-2)^2= (-2) \times (-2) = 4\text{.}\) Notice that since we are multiplying two negative numbers (that happen to be the same number), the result is positive.
There is nothing special about the number \(4\) here; in fact, this holds for any real number \(a > 0\text{.}\) We will write this in a theorem.
Proof.
Assume that \(k\) is a solution to \(x^2=a\text{.}\) This means that \(k^2=a\text{.}\) We need to show that \(-k\) is also a solution to \(x^2=a\text{.}\)
Consider \((-k)^2\text{.}\) Now:
\begin{equation*}
(-k)^2=(-k)\times (-k)
\end{equation*}
\begin{equation*}
=[(-1) \times k] \times [(-1) \times k] = [(-1)\times(-1) \times k^2
\end{equation*}
\begin{equation*}
=[1] \times k^2 = 1 \times a = a
\end{equation*}
since we are assuming \(k^2=a\text{.}\) Thus we have just showed that \((-k)^2=a\) and thus \(-k\) is a solution to \(x^2=a\) as well.
Example 8.2.2.
The numbers \(7\) and \((-7)\) are square roots of \(49\) since \(7^2=49\) and \((-7)^2=49\text{.}\)
The numbers \(\frac{1}{4}\) and \(-\frac{1}{4}\) are square roots of \(\frac{1}{16}\) since \(\left(\frac{1}{4}\right)^2=\frac{1}{16}\) and \(\left(-\frac{1}{4}\right)^2=\frac{1}{16}\text{.}\)
The number \(0\) is *the* square root of \(0\) since \(0^2=0\) and also since \(-0=0\) there is no other square root.
We have just shown that if there is a solution to the equation \(x^2=a\) then there are two solutions to this equation; the other being its negative. But for any real number \(a\text{,}\) is there always a solution to \(x^2=a?\)
Checkpoint 8.2.3.
If \(a\) is a negative real number, why is it impossible for \(x^2=a\) to have any solutions? What about \(x^n=a\) if \(n\) is an even natural number?
Hint.
Checkpoint 8.2.4.
Using the previous theorem as a model, prove that if \(n\) is an even natural number, \(a\) is a positive real number, and \(k\) is a solution to \(x^n=a\) then \((-k)\) is also a solution to \(x^n=a\text{.}\)
Checkpoint 8.2.5.
How do we know that there are at most two square roots? Prove by contradiction that this is impossible. Then modify your proof to show that there are at most two \(n\)th powers for an even natural number \(n\text{.}\)
Hint 1.
Let’s explore how many roots there are for odd roots, for example cube roots. Indeed, we’ll show that there is exactly one odd root of \(a\) for all real numbers \(a\text{.}\) Let’s look at a few examples first:
Example 8.2.6.
What are all the solutions to \(x^3=-8\text{?}\) First we realize that \(x\) must be negative since we are multiplying \(x\) by itself an odd number of times and this results in an odd number. We also know that \(2^3=8\text{,}\) so we can deduce that \(x=-2\) is a solution to \(x^3=-8\text{.}\) In fact, it is the only real number that is a solution to this, and thus the only cube root of \(-8\) is \(-2\text{.}\)
Indeed, this example basically shows the only behaviour that odd roots have; there is exactly one odd root for any \(a \in \mathbb{R}\text{.}\) The next exercise will get you to show this always holds for cube roots, and then extended to all odd roots.
Checkpoint 8.2.7.
Let \(a\) be a real number. Show that there is exactly one solution to the equation \(x^3=a\text{.}\)
Hint.
Checkpoint 8.2.8.
Now that we understand the ideas of roots, we need an easy way of writing down which root we mean. The idea is that we want a function that uniquely expresses which nth root of a number \(a\) that we mean. We call this the principal root of \(a\) and write this as \(\sqrt[n]{a}\) or \(\sqrt{a}\) for the principal square root. We now go on to define what this means for both even and odd roots:
If \(n\) is even, then the principal \(n\)th root of \(a\) is the positive solution to \(x^n=a\) if \(a > 0\text{,}\) and is \(o\) if \(a=0\text{.}\) Note there are no even roots if \(a\) is negative.
If \(n\) is odd then \(\sqrt[n]{a}\) is the unique root of the equation \(x^n=a\) for any real number \(a\text{.}\)
Note that for even roots, choosing the positive root as the principal root is simply a choice that mathematicians decided on; there is nothing really special about the positive root; we simply picked one to make the principal root a function.
Example 8.2.9.
-
Even if we know there are two solutions to the equation \(x^2=4\text{,}\) we say that \(\sqrt{4}=2\text{.}\)
-
We say that the two solutions to the equation \(x^2=14\) are \(x=\sqrt{14}\) and \(x=-\sqrt{14}\) since the principal root is always positive. Note that since \(14\) isn’t a square number we cannot write this as an integer. Later, we will learn about simplifying roots.
-
We know that \(x^3=8\) has one solution since the exponent 3 is odd, so we say that \(\sqrt[3]{8}=2\text{.}\) Similarly we say that \(\sqrt[3]{-8}=-2\text{.}\)
Subsection 8.2.2 nth Powers
For this section, we will deal with the principal root of natural numbers, but these results also hold for other roots, if they exist. We will show this in the following section.
In this chapter we will work towards showing that the only \(n\)th roots that are not irrational are those that are \(n\)th powers. We’ll show that every other \(n\)th root cannot be written as a rational number by contradiction; we’ll try to, and then show that two quantities that should be equal cannot be.
In order to show this, we’ll first have to prove a property about \(n\)th powers; indeed, that they are exactly numbers where each prime in its prime factorization has an exponent with a factor of \(n\text{.}\) Let’s do some examples to illustrate this:
Example 8.2.10.
Consider \(s=144\text{.}\) We know that \(144\) has a prime factorization of \(144=2^4 \dot 3^2\text{.}\) And since the exponents \(4\) and \(2\) are both even (have a factor of 2) then we can use our exponent rules and write \(144 = 2^4 \dot 3^2 = (2^2 \dot 3)^2=12^2\text{.}\)
Also, we know that \(72=2^3 \dot 3^2\) is not a square number since there is one of the exponents that is not even. So there is no way to "factor out" a \(2\) from each exponent.
Let’s show this in general for any square and non-square number.
Theorem 8.2.11.
Let \(n\) be a natural number. The number \(m\) is a square number if and only if each exponent of its prime factorization has a factor of \(2\text{.}\)
Proof.
Assume \(m\) is a square number. Then \(m=a^2\) for some natural number \(a\text{.}\) Since \(a\) is a natural number it has a prime factorization, say \(a=p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k}.\text{.}\) Then \(m=a^2 = (p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k})^2=\) and thus each exponent of the primes in the prime factorization of \(n\) has a factor of \(n\text{.}\)
Now assume that every exponent in the prime factorization of \(m\) is even. Thus \(m=p_1^{2e_1}p_2^{2e_2}\ldots p_k^{2e_k} = (p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k})^2=\text{.}\) This means that the number \(a = p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k} \) is such that \(a^2=m\) and this means that \(m\) is a square number.
Now that we’ve shown this for square numbers, can you work out how to generalize this for any \(n\)th power?
Checkpoint 8.2.12.
Modify the proof above to prove the result "The number \(m\) is an \(n\)th power if and only if each exponent in its prime factorization has a factor of \(n\)".
Remember that if we have a conditional statement "if P then Q" it is logically equivalent to its contrapositiveL "if not Q then not P". Thus we can express a biconditional as "If P then Q *and* if not P then not Q". We can rewrite the result in the exercise above to be the following theorem that will be very useful for us:
Theorem 8.2.13.
Let \(m\) be a natural number. If \(m\) is an \(n\)th power then each exponent in its prime factorization has a factor of \(n\text{.}\) If \(m\) is not an \(n\)th power, then there is a prime number in its prime factorization with an exponent with no factor of \(n\text{.}\)
Example 8.2.14.
Consider \(m=5400\text{.}\) You can check that the prime factorization of \(m\) is \(m=2^3 \dot 3^3 \dot 5^2\text{.}\) Since the prime \(5\) has an exponent that is not divisible by 3, we know that \(5400\) is not a cube (3rd power).
However, if \(m=27000=2^3 \dot 3^3 \dot 5^3\) we know that \(m\) is indeed a cube, and \(m=(2\dot 3 \dot 5)^3 = 30^3\text{.}\)
Subsection 8.2.3 Proving Irrationality of nth Roots
Note that in this section we will assume that the set of natural numbers includes 0. We will now work on proving what we set out to prove: that all \(n\)th roots of numbers that are not \(n\)th powers are indeed irrational. A key result that we will use to help us is the Fundamental Theorem of Arithmetic, which states that every natural number has a unique prime factorization.
Let’s first explore the more general case with square roots:
Example 8.2.15.
Let \(n\) be a non-square number. This means that in its prime factorization, there is at least one prime that has a power that isn’t a multiple of \(2\text{.}\) For example, if \(n=60=2^2 \times 3 \times 5\text{.}\) Here both \(3\) and \(5\) have an exponent of one, which isn’t a multiple of \(2\text{.}\) We’ll pick the smallest of these primes with such an exponent and call that prime \(p_*\) with exponent \(2e_*+1\) for some \(e_* \in \mathbb{N}_0\) and we can write \(n=p_*^{2e_*+1}k\) where \(k\) is the "rest" of the factors of \(n\text{.}\) In our example, \(p_*=3\) and \(e_*=1\) and \(60=3^1 \times (2^2 \times 5)\) and thus \(k=2^2 \times 5\) in this example.
We will show this by contradiction; let’s pretend that we can indeed write \(\sqrt{n}\) as a rational number: \(\sqrt{n}=\frac{a}{b}\) for some appropriate integers \(a,b\text{.}\) In our example, we let \(\sqrt{60}=\frac{a}{b}\text{.}\)
Let’s arrange our equation so that we have no roots or denominators: this means multiplying both sides by \(b\) and then squaring both sides of the equation:
\begin{equation*}
\sqrt{n}=\frac{a}{b} \rightarrow \sqrt{n}b=a \rightarrow nb^2=a^2.
\end{equation*}
In our example:
\begin{equation*}
\sqrt{60}=\frac{a}{b} \rightarrow \sqrt{60}b=a \rightarrow 60b^2=a^2.
\end{equation*}
Let’s count the number of factors of \(p_*\) on both sides of the equation. On the RHS, by Theorem 8.2.13 we know that \(a^2\) has an even number of \(p_*\text{,}\) and thus the exponent \(e\) of \(p_*\) in the prime factorization of \(b^2\) is a factor of \(2\text{,}\) so say \(e=2r\) for some natural number \(r\text{.}\) So \(b^2=p_*^{2r}j\) where \(j\) are the "rest" of the factors of \(n\text{.}\)
However, on the LHS, by Theorem 8.2.13 we also know that \(b^2\) has an even exponent for \(p_*\) in it’s prime factorization, say \(2s\) for some natural number \(s\) and similarly to \(a^2\) we can say \(b^2=p_*^{2s}\ell\) for some integer \(ell\text{,}\) while \(n\) itself has an odd exponent for \(p_*\text{.}\) Thus
\begin{equation*}
nb^2 = (p_*^{2e_*+1}k)(p_*^{2s}\ell) = p_*^{2e_*+1+2s}k\ell = p_*^{2(e_*+s)+1}kl.
\end{equation*}
In our example:
\begin{equation*}
60b^2 = (3^{1} (2^2 \times 5))(3^{2s}\ell) = 3^{2s+1}[(2^2 \times 5)\ell].
\end{equation*}
Checkpoint 8.2.16.
Use the above as a guideline to show that \(\sqrt{2}\) cannot be written as a rational number.
Hint.
Look carefully at the result above. It focused on square roots, but the idea can be modified to work for any \(n\)th root.
Checkpoint 8.2.17.
Modify your argument for showing that \(\sqrt{2}\) is irrational to show that \(\sqrt[3]{2}\) is irrational.
Hint 1.
Hint 3.
Hint 4.
Checkpoint 8.2.18.
-
Use the previous exercises as a guide to show that \(\sqrt[3]{m}\) is irrational for a non-cubic number \(m\text{.}\)
-
Use the previous exercises as a guide to show that \(\sqrt[n]{m}\) is irrational for a number \(m\) that is not a \(n\)th power.
