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We have seen how both rational values and the rational expressions which are quotients of polynomials in the same variable both exhibit the rich structure of a field.
However, these are certainly not the only fields one might consider -- not by a long shot!
Interestingly, we can often build a new field from an existing one through a process known as field extension, where additional elements are inserted into an existing field.
For example, we know the set of all rational values $a$ forms a field.
Consider the set of all values that take the form $a + b\sqrt{2}$, where both $a$ and $b$ are rational, which we may denote by $$\mathbb{Q}(\sqrt{2}) = \{a + b\sqrt{2} \ | \ a,b \in \mathbb{Q} \}$$
If you are curious about where this definition came from -- suppose you are wondering if a field $F$ exists that includes all the rationals, but also includes $\sqrt{2}$. Quick consideration of closure alone makes us realize that $\sqrt{2}$ can't be the only added element, however. Consider any rational $b$ in $F$, closure under multiplication will require $b\sqrt{2} \in F$. Taking things further, suppose another rational value $a$ is also in $F$. Then the sum of $a$ and the earlier $b\sqrt{2}$ must also be in $F$. As such, we minimally must have all elements of the form $a+b\sqrt{2}$ where $a,b \in \mathbb{Q}$ in such a field $F$.
Okay, so it would necessary for a field including all rationals and $\sqrt{2}$ to include all the elements of $\mathbb{Q}(\sqrt{2})$, but is it sufficient? That is to say, is $\mathbb{Q}(\sqrt{2})$ itself a field? Let's find out:
Are addition and multiplication in $\mathbb{Q}(\sqrt{2})$ both closed?
Yes! Note that $(a + b\sqrt{2}) + (c + d\sqrt{2}) = (a+c) + (b+d)\sqrt{2}$ and both $(a+c)$ and $(b+d)$ are rational when $a$, $b$, $c$, and $d$ are rational. Thus addition is closed here.
Further, $(a + b\sqrt{2})(c + d\sqrt{2}) = (ac + 2bd) + (ad + bc)\sqrt{2}$ and both $(ac + 2bd)$ and $(ad+bc)$ are rational under the same assumption about $a$, $b$, $c$, and $d$ being rational. So multiplication is also closed here.
Are addition and multiplication in $\mathbb{Q}(\sqrt{2})$ both commutative?
Yes! Recall that sums and products of real values are commutative, and both $(a+b\sqrt{2})$ and $(c+d\sqrt{2})$ are real values.
Are addition and multiplication in $\mathbb{Q}(\sqrt{2})$ both associative?
Yes! Again, since we know sums and products of real values are associative, and $(a+b\sqrt{2})$ and $(c+d\sqrt{2})$ are real values.
Do additive and multiplicative identities exist in $\mathbb{Q}(\sqrt{2})$?
Yes! $0$ and $1$ serve as additive and multiplicative identities for real numbers, so they function in that same capacity here. Importantly, both $0$ and $1$ are also in $\mathbb{Q}(\sqrt{2})$ as $0 = 0 + 0\sqrt{2}$ and $1 = 1 + 0\sqrt{2}$.
Do additive and multiplicative inverses always exist in $\mathbb{Q}(\sqrt{2})$ for non-zero values?
Yes! Finding the additive inverses of such an expressions is straight-forward. Notice that $$(a + b\sqrt{2}) + (-a - b\sqrt{2}) = 0$$ Where things get more interesting is when we try to find the multiplicative inverse. We know the multiplicative inverse of any non-zero real value $x$ is $1/x$, so we must show that we can find some rational $c$ and $d$ so that for any rational $a$ and $b$ the following holds: $$\cfrac{1}{a + b\sqrt{2}} = c + d\sqrt{2}$$ As such, we want to manipulate the left side so that we no longer have a square root in the denominator-- which can indeed be done! Consider the following: $$\begin{array}{rcl} \cfrac{1}{a + b\sqrt{2}} &=& \cfrac{1}{a + b\sqrt{2}} \cdot \cfrac{a - b\sqrt{2}}{a - b\sqrt{2}}\\ &=& \cfrac{a - b\sqrt{2}}{a^2 - 2b^2}\\ &=& \cfrac{a}{a^2-2b^2} + \cfrac{-b}{a^2 - 2b^2}\sqrt{2} \end{array}$$ Then note we require $c = \cfrac{a}{a^2-2b^2}$ and $d = \cfrac{-b}{a^2 - 2b^2}$ to both be rational when $a$ and $b$ are rational and not both zero.
Note, when both $a = b = 0$ then $a^2-2b^2$ is also zero, leaving the fractions above undefined. Fortunately, we don't have to worry about this as $a = b = 0$ also means the very thing for which we are attempting to find a multiplicative inverse ($a + b\sqrt{2}$) is also zero! Recall, we don't require a multiplicative inverse for zero in a field, and can thus ignore this case.
Really, the only thing that could go wrong here is if $a^2-2b^2 = 0$ when $a$ and $b$ are not both zero, as otherwise the closure of rationals under addition, subtraction, multiplication, and non-zero division ensures the rationality of $c$ and $d$ above.
However, consider the following indirect argument. First, we don't have to worry about just $b$ being zero, as if $a^2 - 2b^2$ is zero and $b=0$, then $a=0$ too. (see the note above concerning $a=b=0$)
As such, suppose $a^2 - 2b^2 = 0$ where $b \neq 0$. This tells us $2b^2 = a^2$ and consequently $$2 = \cfrac{a^2}{b^2}$$ Taking as square root of both sides as shown below, then suggests $\sqrt{2}$ is a rational value -- which we know to be absurd. $$\sqrt{2} = \cfrac{a}{b}$$
As such, we never need to worry about this case. It never happens. Whew!
Finally, does multiplication distribute over addition in $\mathbb{Q}(\sqrt{2})$?
Yes! Here again, we know that this happens for real values, and every value in $\mathbb{Q}(\sqrt{2})$ is real.
As such, $\mathbb{Q}(\sqrt{2})$ satisfies all the requisite properties -- it too is a field!
Was there anything special about $\sqrt{2}$ here? What would happen if we considered $\mathbb{Q}(\sqrt{3}) = \{a + b\sqrt{3} | a,b \in \mathbb{Q}\}$? ..or $\mathbb{Q}(\sqrt{7})$ defined similarly? Are these also fields?
To find the multiplicative inverse of $a+b\sqrt{2}$ in the previous discussion, we had to eliminate the radical in the denominator, and did so by finding a way to multiply the denominator $(a+b\sqrt{2})$ by $(a-b\sqrt{2})$. This particular trick is one of many that we can exploit to eliminate radicals from one side of a fraction (typically the bottom), collectively known as methods to "rationalize" the denominator (or sometimes the numerator) .
Recall, square roots that were factors of a denominator could be eliminated by multiplying by a well-chosen value in the form of that square root over itself, as the example below illustrates: $$\begin{array}{rcl} \displaystyle{\frac{9x}{(x+7)\sqrt{3y}}} &=& \displaystyle{\frac{9x}{(x+7)\sqrt{3y}} \cdot \frac{\sqrt{3y}}{\sqrt{3y}}}\\\\ &=& \displaystyle{\frac{9x\sqrt{3y}}{(x+7) \cdot 3y}}\\\\ &=& \displaystyle{\frac{3x\sqrt{3y}}{y(x+7)}} \end{array}$$ Similarly, $n^{th}$ roots that were factors of a denominator could also be eliminated by multiplying by a well-chosen value of one. Only this time, the form of that "well-chosen value of one" was selected to add to the factors already present in the denominator to a point where there were only $n^{th}$ powers inside the $n^{th}$ root -- or equivalently, to a point where the denominator could be expressed as factors that were all integer powers. The following calculation provides an example of this trick: $$\begin{array}{rcl} \displaystyle{\frac{x^2+4x+3}{2\sqrt[5]{(x+1)^2 y^3}}} &=& \displaystyle{\frac{x^2 + 4x + 3}{2\sqrt[5]{(x+1)^2 y^3}} \cdot \frac{\sqrt[5]{(x+1)^3 y^2}}{\sqrt[5]{(x+1)^2 y^2}}}\\\\ &=& \displaystyle{\frac{(x^2 + 4x + 3)\sqrt[5]{(x+1)^3 y^2}}{2y(x+1)}}\\\\ &=& \displaystyle{\frac{(x+1)(x+3)\sqrt[5]{(x+1)^3 y^2}}{2y(x+1)}}\\\\ &=& \displaystyle{\frac{(x+3)\sqrt[5]{(x+1)^3 y^2}}{2y}; \quad \quad \textrm{provided $x \neq -1$}} \end{array}$$ Here's the same calculation done in terms of rational exponents: $$\begin{array}{rcl} \displaystyle{\frac{x^2 + 4x + 3}{2 (x+1)^{2/5} y^{3/5}}} &=& \displaystyle{\frac{x^2+4x+3}{2(x+1)^{2/5} y^{3/5}} \cdot \frac{(x+1)^{3/5} y^{2/5}}{(x+1)^{3/5} y^{2/5}}}\\\\ &=& \displaystyle{\frac{(x^2 + 4x + 3)(x+1)^{3/5} y^{2/5}}{2y(x+1)}}\\\\ &=& \displaystyle{\frac{(x+4)(x+1)(x+1)^{3/5} y^{2/5}}{2y(x+1)}}\\\\ &=& \displaystyle{\frac{(x+4)(x+1)^{3/5} y^{2/5}}{2y}; \quad \quad \textrm{provided $x \neq -1$}} \end{array}$$ The two tricks above can be easily modified to deal with rationalizing the numerator. However, importantly these two techniques can only eliminate roots that appear as factors of the denominator (or numerator) -- i.e., either the entire denominator (or numerator) can be expressed as a product, with the root in question being one of the expressions being multiplied together in that product.
We must appeal to a different set of tricks when the root is instead a term of either the top or bottom of the fraction in question -- i.e., the side on which the root(s) appear can be expressed as a sum, with the root in question being one of the expressions being added together in that sum.
When the denominator is a binomial with either one or both terms being square roots, we can use the trick seen in the first section. Representing this binomial as $(a+b)$, we multiply by $(a-b)$. Recalling the factorization of a "difference of squares", where $(a+b)(a-b) = a^2 - b^2$, any square roots present in $a$ and/or $b$ will be "squared away" in the product. Consider this strategy as it is employed below: $$\begin{array}{rcl} \cfrac{x^4 - y^4}{\sqrt{x} - \sqrt{y}} &=& \cfrac{x^4 - y^4}{\sqrt{x} - \sqrt{y}} \cdot \cfrac{\sqrt{x} + \sqrt{y}}{\sqrt{x} + \sqrt{y}}\\\\ &=& \cfrac{(x^4 - y^4)(\sqrt{x} + \sqrt{y})}{x-y}\\\\ &=& \cfrac{(x^2 + y^2)(x^2 - y^2)(\sqrt{x} + \sqrt{y})}{x-y}\\\\ &=& \cfrac{(x^2 + y^2)(x+y)(x-y)(\sqrt{x} + \sqrt{y})}{x-y}\\\\ &=& (x^2 + y^2)(x+y)(\sqrt{x} + \sqrt{y}); \quad \quad \textrm{provided } x \neq y\\\\ \end{array}$$
Just as the factorization of "difference of squares" inspired the trick to eliminate square roots seen in the terms of a binomial, we can eliminate one or more cube roots seen in the terms of a binomial by appealing to the factorization of a "sum/difference of cubes" -- i.e., $(x \pm y)(x^2 \mp xy + y^2) = x^3 \pm y^3$, as the following demonstrates: $$\begin{array}{rcl} \cfrac{c^2-64}{2+\sqrt[3]{c}} &=& \cfrac{c^2-64}{2+\sqrt[3]{c}} \cdot \cfrac{4 + 2\sqrt[3]{c} + \sqrt[3]{c^2}}{4 + 2\sqrt[3]{c} + \sqrt[3]{c^2}}\\\\ &=& \cfrac{(c^2-64)(4 + 2\sqrt[3]{c} + \sqrt[3]{c^2})}{(8 + c)}\\\\ &=& \cfrac{(c+8)(c-8)(4 + 2\sqrt[3]{c} + \sqrt[3]{c^2})}{(8 + c)}\\\\ &=& (c-8)(4 + 2\sqrt[3]{c} + \sqrt[3]{c^2}); \quad \quad \textrm{provided } c \neq -8\\\\ \end{array}$$ Similar factorization rules can be used to rationalize the denominator (or numerator) of a fraction. For example $(x+y) \cdot [(x-y)(x^2+y^2)] = (x^2-y^2)(x^2+y^2) = x^4 - y^4$ could be used to eliminate fourth roots in the terms of a binomial denominator (or numerator), while $(x \pm y)(x^4 \mp x^3 y + x^2 y^2 \mp xy^3 + y^4) = x^5 \pm y^5)$ can be used to eliminate fifth roots in the same.
Sometimes we can use these rules to fairly quickly rationalize denominators or numerators with even more terms, as the below calculation suggests: $$\begin{array}{rcl} \cfrac{1}{1 + \sqrt{2} + \sqrt{10}} &=& \cfrac{1}{1 + \sqrt{2} + \sqrt{10}} \cdot \cfrac{1 + \sqrt{2} - \sqrt{10}}{1 + \sqrt{2} - \sqrt{10}}\\\\ &=& \cfrac{1 + \sqrt{2} - \sqrt{10}}{-7 + 2\sqrt{2}}\\\\ &=& \cfrac{1 + \sqrt{2} - \sqrt{10}}{-7 + 2\sqrt{2}} \cdot \cfrac{-7 - 2\sqrt{2}}{-7 - 2\sqrt{2}}\\\\ &=& \cfrac{(1 + \sqrt{2} - \sqrt{10})(-7 - 2\sqrt{2})}{41} \end{array}$$ That said, the algebra involved can quickly get overwhelming when the index of the related roots to be eliminated is high.
One might also wonder -- noting that we saw relevant factorizations for rationalizing binomials involving sums and differences of odd-indexed roots, but for binomials involving even-indexed roots we only saw a relevant factorization for differences of such terms -- what can we do with the sums of such terms?We will answer that question soon! (in the coming sections)
For now however, let us turn our attention to a different variation on this idea...
What if we wanted to rationalize the denominator for something like the below! $$\frac{1}{1-\sqrt{1+\sqrt[3]{2}}}$$ Of course, wanting to rationalize the denominator of the fraction above suggests the denominator is irrational to begin with -- do we know that for sure? Certainly the first term of $1$ in the denominator is rational -- so we would really just need to show $\sqrt{1+\sqrt[3]{2}}$ is irrational to argue it is. But how do we do that?
Let $r = \sqrt{1+\sqrt[3]{2}}$, and let us try to somehow get rid of the "ugly" stuff (i.e., the various roots symbols). Certainly, if $a=b$, then $a^2 = b^2$, so $$r^2 = 1+\sqrt[3]{2}$$ We'd like to get rid of the cube root in a similar way, and if $a=b$ then certainly $a^3=b^3$ -- but if we cube both sides of this equation now, the binomial theorem tells us there will still be cube roots on the right side (think about the middle terms of $(a+b)^3$.
Noting that if $a=b$ then $a-c=b-c$, let us subtract $1$ from both sides to get around this difficulty. $$r^2 - 1 = \sqrt[3]{2}$$ Now we can cube both sides to discover $$(r^2-1)^3 = 2$$ Of course, expanding the left side with the binomial theorem, we have $$r^6-3r^4+3r^2-1 = 2$$ or equivalently, if we subtract $2$ from both sides, $$r^6-3r^4+3r^2-3 = 0$$ Interesting! This means that the value of $r$ with which we started solves $f(x)=0$ when $f(x)$ is the polynomial function given by $f(x) = x^6-3x^4+3x^2-3$, which is (nicely) a polynomial with rational coefficients and a rational constant term.
Consequently, by the factor theorem, $(x-r)$ must be a factor of $x^6-3x^4+3x^2-3$!
Recall we wanted to know if $r$ really was irrational. Suppose it wasn't -- and it was rational. Wouldn't the rational roots theorem then kick in and tell us that $r = \pm1$ or $r = \pm 3$? Of course, none of these can possibly be true (as it is easy to determine $1 \lt r \lt 2$) -- so it must be the case that $r$ is indeed irrational!
Getting back to our original task, this tells us that rationalizing the denominator of $$\frac{1}{1-\sqrt{1+\sqrt[3]{2}}}$$ is a reasonable course of action since we have now proven the denominator really is irrational. But how do we rationalize it?
As it turns out, the polynomial we found, $f(x) = x^6-3x^4+3x^2-3$ helps us immensely! Note that if we let $x=1$ we can write the fraction above as $$\frac{1}{x-r}$$ But recall, we earlier argued that $(x-r)$ must be a factor of $f(x)$, so $f(x) = q(x)(x-r)$ for some polynomial $q(x)$. It's worth noting though -- that the coefficients and/or constants of $q(x)$ might be irrational.
As a quick example, consider a different $r$ value: $r=\sqrt{2}$. Following the same process as above, we see $r^2 = 2$, so $r^2-2 = 0$. But then if $f(x)=x^2-2$, it must be that $f(r)=0$. This leads to $(x-\sqrt{2})$ being a factor of $x^2-2$ by the factor theorem. Of course if we do the long division, or simply appeal to $x^2-y^2 = (x+y)(x-y)$, we see that when $x^2-2 = (x-\sqrt{2})q(x)$, it must be that $q(x) = x+\sqrt{2}$ -- whose constant term is irrational.
Now, consider the following well-chosen multiplication by $1$: $$\frac{1}{x-r} \cdot \frac{q(x)}{q(x)} = \frac{q(x)}{(x-r)q(x)} = \frac{q(x)}{f(x)}$$ Note that $x=1$ was a rational value. (Incidently, this means $x-r \in \mathbb{Q}(r)$!) But then, given that $f(x)$ had all rational coefficients and a rational constant term, $f(x)$ must be rational by closure! The numerator might still be irrational, given the possibility it has irrational coefficients or an irrational constant term (see the italicized note above), but that is fine -- we only sought to rationalize the denominator. Indeed, rationalizing the denominator is quire straight-forward now! We need only find $q(x) = f(x)/(x-r)$, which we can do with long division (although doing so is admittedly quite bit tediuous) and then plug $x=1$ into $q(x)/f(x)$.
For the curious, the $q(x)$ resulting from the long division here is:
$$x^5 + \sqrt{1+\sqrt[3]{2}}x^4 + (\sqrt[3]{2} - 2) x^3 + \left(\sqrt[3]{2}\sqrt{1+\sqrt[3]{2}}-2\sqrt{1+\sqrt[3]{2}}\right)x^2 + (1-\sqrt[3]{2}+\sqrt[3]{4})x + \sqrt[3]{4}\sqrt{1+\sqrt[3]{2}}-\sqrt[3]{2}\sqrt{1+\sqrt[3]{2}}+\sqrt{1+\sqrt[3]{2}}$$
Evaluating $q(x)/f(x)$ when $x=1$ then gives the rationalized-denominator form $$\frac{1}{1-\sqrt{1+\sqrt[3]{2}}} = \frac{\sqrt[3]{4}\left(1+\sqrt{1+\sqrt[3]{2}}\right)}{-2}$$
Yes, finding $q(x)$ for this particular example by long division was indeed tedius and the final rationalized-denominator form was a bit gross-looking -- but the real payoff here is the beautiful (and previously hidden) connection between irrational expressions $r$ involving various roots (possibly combinations of square roots, cube roots, etc -- including some nested inside others) and polynomials $f(x)$ where $f(r)=0$. Given this relationship, we call $r$ a root of $f(x)$.
This leads directly to other questions of course! Given a polynomial, what kinds of roots can it have? How do we find these roots? We've certainly seen they can be much more exotic than a simple rational value or a single square or cube root. What about other functions, like those defined by rational expressions -- do they have roots too? What about functions involving logarithms, or exponential functions -- or both? Do these and other functions we might define have roots as well? How do we find all these different kinds of roots when they exist?
Putting all of these thoughts together, we ask for a function $f(x)$ in general, "How do we solve $f(x)=0$?" This is a big question given the variety of functions $f(x)$ might represent, and tackling it starts us on a path that will eventually take us to directly to the big results at the end of this course! The next section attempts to answer some of these questions, at least for polynomial and rational functions, $f(x)$.