Linear Algebra HW #1

We wish to prove \(\sqrt{15} \notin \mathbb {Q}\)

We will do this through a proof by contradiction. Let us assume we can represent \(\sqrt{15} = \frac{a}{b} \text{, where } a,b \in \mathbb{Z}\). We can reduce the expression \(\frac{a}{b}\) completely so that it is identical in value but so that a and b are coprime, sharing no common factors except 1. By the fundamental theorem of algebra, every integer can be expressed as a product of primes, so we will represent these explicitely as \(a = a_{1}^{n_{1}} \times a_{2}^{n_{2}} \times \text{. . .} \times a_{k}^{n_{k}}\) and \(b = b_{1}^{n_{1}} \times b_{2}^{n_{2}} \times \text{. . .} \times b_{k}^{n_{k}}\). Now if \(\sqrt{15} = \frac{a}{b}\), then \[15 = (\frac{a}{b})^2 = \frac{a^2}{b^2} = \frac{a_{1}^{2n_{1}} \times a_{2}^{2n_{2}} \times \text{. . .} \times a_{k}^{2n_{k}}}{b_{1}^{2n_{1}} \times b_{2}^{2n_{2}} \times \text{. . .} \times b_{k}^{2n_{k}}}\] \[\Rightarrow 15(b_{1}^{n_{1}} \times b_{2}^{n_{2}} \times \text{. . .} \times b_{k}^{n_{k}}) = a_{1}^{n_{1}} \times a_{2}^{n_{2}} \times \text{. . .} \times a_{k}^{n_{k}}\] which implies one of the \(a_{i}\) is 15. Let us choose \(a_{1} = 15\) and reorder as necessary. Then \[15_{1}^{2n_{1}-1} \times a_{2}^{2n_{2}} \times \text{. . .} \times a_{k}^{2n_{k}} = b_{1}^{2n_{1}} \times b_{2}^{2n_{2}} \times \text{. . .} \times b_{k}^{2n_{k}}\] implying that B is divisible by 15. But then a and b share a multiple of 15, and we already reduced them so they are coprime, which is a contradiction. Therefore \(\sqrt{15} \notin \mathbb {Q}\) as desired.

We are given that \(\alpha = \sqrt[3]{2} \), and \(F = \{a_{o} + a_{1}\alpha + a_{2}\alpha^2 | a_{i} \in \mathbb {C}\}\).

Now we would like to show that if \(a, b \in F\), then \(a+b \in F\), \(ab \in F\), ie. F is closed under multiplication and addition. We will start with addition.

First we will define a as \(a = a_{o} + a_{1}\alpha + a_{2}\alpha^2\) and b similarly as \(b_{o} + b_{1}\alpha + b_{2}\alpha^2\). Now when we add them, we can see: \[a+b = a_{o} + a_{1}\alpha + a_{2}\alpha^2 + b_{o} + b_{1}\alpha + b_{2}\alpha^2=\] \[=(a_{o} + b_{o}) + (a_{1}+ b_{1})\alpha + (a_{2}+ b_{2})\alpha^2\]

Since \(\mathbb{Q}\) is a subfield of \(\mathbb {C}\), as stated in lecture, it is defined to be closed under addition, so we can redefine the coefficients as \[(a_{o} + b_{0}) \equiv c_{0}, \text { where }c_{0} \in \mathbb {Q}\] \[(a_{1} + b_{1}) \equiv c_{1}, \text { where } c_{1} \in \mathbb {Q}\] \[(a_{2} + b_{2}) \equiv c_{2}, \text { where } c_{2} \in \mathbb {Q}\]

This allows us to redefine \(a+b\) as \[c_{o} + c_{1}\alpha + c_{2}\alpha^2, c_{i} \in \mathbb {Q}\]

This is of the form for F, so \(a+b \in F\).

Now we will explore the closure of F under multiplication. Using the same definitions for \(a_{0}\) and \(b_{0}\), we see that \[(a_{o} + a_{1}\alpha + a_{2}\alpha^2 )*(b_{o} + b_{1}\alpha + b_{2}\alpha^2 ) =\] \[= a_{0}b_{0} + \alpha (a_{0}b_{1} + a_{1}b_{0}) + \alpha^2 (a_{0}b_{2} + a_{1}b_{1}) +\alpha^3(a_{1}b_{2} + a_{2}b_{1}) + a_{2}b_{2}\alpha^4\]

Consider \(\alpha^3\). \(\alpha = \sqrt[3]{2}\), so \(\alpha^3 = 2 \in \mathbb {Q}\). Therefore ab can be written as \[ab = a_{0}b_{0} + \alpha(a_{0}b_{1}+ a_{1}b_{0}) + \alpha^2(a_{0}b_{2} + a_{1}b_{1}) + 2(a_{1}b_{2}+a_{2}b_{1}) +2a_{2}b_{2} \alpha =\] \[=(a_{0}b_{0} + 2(a_{1}b_{2} + a_{2}b_{1})) + \alpha (a_{0}b_{1} + a_{1}b_{0} + 2a_{2}b_{2}) + \alpha^2 (a_{0}b_{2} + a_{1}b_{1})\]

Since \(2, a_{i}, b_{i} \in \mathbb{Q}\) and \(\mathbb {Q}\) is closed under multiplication and addition, being a subfield of \(\mathbb {C}\), we can redefine coeficcients as \[C_{0} \equiv (a_{0}b_{0} +2(a_{1}b_{2}+a_{2}b_{1}))\] \[C_{1} \equiv (a_{0}b_{1} + a_{1}b_{0} + 2a_{2}b_{2})\] \[C_{2} \equiv (a_{0}b_{2} + a_{1}b_{1})\]

So \(ab\) can be rewritten as \[c_{o} + c_{1}\alpha + c_{2}\alpha^2, c_{i} \in \mathbb {Q}\]

which is in the form of F, so \(ab \in F\)

We wish to show that if \(a_{0} + a_{1} \alpha + a_{2} \alpha^2 =0\), where \(a_{i} \in \mathbb {Q}\), then all \(a_{i} = 0\). This is the equivalent to saying this is a linearly independent combination of alphas. Since all \(a_{i} \in \mathbb{Q}\), each \(a_{i}\) can be rewritten as the quotient of two integers. Therefore, we will denote \[a_{0} = \frac{b_{0}}{c_{0}}, \text { where }b_{0},c_{0} \in \mathbb {Z}\], \[a_{1} = \frac{b_{1}}{c_{1}}, \text { where } b_{1},c_{1} \in \mathbb {Z}\], \[a_{2} = \frac{b_{2}}{c_{2}}, \text { where } b_{2},c_{2} \in \mathbb {Z}\] This is more clear, but not as manageable, so let us try to manipulate this into terms of all integers by exploiting the zero on the right side of the equation. Thus our original equation can be rewritten by multiplying through by the product of the denominators as \[c_{1}c_{2}b_{0} + c_{0}c_{2}b_{2} \alpha + c_{0}c_{1}b_{2} \alpha^2 =0\] Since all of these coefficients are integers and the set of integers are closed under multiplication, we can redefine \[c_{1}c_{2}b_{0} = d_{0}\], \[c_{0}c_{2}b_{2} = d_{1}\], \[c_{1}c_{0}b_{2} = d_{2}\] so that our original equation can be rewritten as \[d_{0} + d_{1} \alpha + d_{2} \alpha^2 =0\]

Which is entirely in terms of integers.

Now we note that \(\alpha = \sqrt[3]{2}\) implying \(\alpha^3 = 2\). So if we multiply through this equation by alpha, we see that \[2d_{2} + d_{0} \alpha + d_{1} \alpha^2 = 0\]

and then if we repeat the process again: \[2d_{1} + 2d_{2} \alpha + d_{0} \alpha^2 = 0\]

which gives us a system of equations we can represent with a matrix as \[\left( \begin{array}{ccc}d_{0} & d_{1} & d_{2} \\2a_{2} & a_{0} & a_{1} \\2a_{1} & 2a_{2} & a_{0} \end{array} \right)\] which is solvable only if the determinant equals 0: \[0 = \left| \begin{array}{ccc}d_{0} & d_{1} & d_{2} \\2d_{2} & d_{0} & d_{1} \\2d_{1} & 2d_{2} & d_{0} \end{array} \right| =\] \[=d_{0}(d_{0}^2 - 2d_{2}d_{1}) - d_{1}(2d_{2}d_{0} - 2d_{1}^2) + d_{2}(4d_{2}^2 - 2d_{1} d_{0}) =\] \[= d_{0}^3 + 2d_{1}^3 + 4d_{2}^3 - 6d_{0}d_{1}d_{2}\]. Now we wish to explore the truthfulness of this statement by analyzing it in terms of evens and odds. Since 0 is even and every factor on the right contains a multiple of 2, \(d_{0}\) must be even. Now let us assume \(d_{1}\) is odd. We can explore this assumption by substituting \(d_{0}\) with a multiple of 2, as \(d_{0} = 2e_{0}, e_{0} \in \mathbb {Z}\), since we have proved it is even. Inserting this into our equation we get \[8e_{0}^3 + 2d_{1}^3 + 4d_{2}^3 - 12e_{0}d_{1}d_{2} = 0\] \[\Rightarrow 4e_{0}^3 + d_{1}^3 + 2d_{2}^3 - 6e_{0}d_{1}d_{2} = 0\] Where in the last step we divided through by 2. Now zero is even, and all the terms on the left hand side contain a multiple of 2, making them even, except for \(d_{1}\). Therefore \(d_{1}\) is also even. We will repeat this process one more time, defining \(d_{1}\) with a multiple of 2, as \(d_{1} = 2e_{1}, e_{1} \in \mathbb {Z}\) and inserting into our equation to get \[8e_{0}^3 + 16e_{1}^3 + 4d_{2}^3 - 24e_{0}e_{1}d_{2} = 0\] \[\Rightarrow 2e_{0}^3 + 4e_{1}^3 + d_{2}^3 - 6e_{0}e_{1}d_{2} = 0\] where in the last step we divided through by 4. Zero is even and all of the terms on the left hand side are even because they are multiples of 2, so \(d_{2}\) must also be even. So we have proven that in order for this equation to have a nonzero solution, all of the \(d_{i}\)’s must be even. Therefore if we can prove one of them must be odd, we are done. We will start by analyzing our original equation containing integer coefficients \[d_{0} + d_{1} \alpha + d_{2} \alpha^2 = 0\] If any of these coefficients are odd, the proof is done and it is linearly independent. So let us assume they are all even. Then by the definition of even numbers, we can redefine these coefficients as multiples of two: \[d_{0} = 2f_{0}, \text{ where } f_{0} \in \mathbb {Z}\], \[d_{1} = 2f_{1}, \text{ where } f_{1} \in \mathbb {Z}\], \[d_{2} = 2f_{2}, \text{ where } f_{2} \in \mathbb {Z}\] We can then divide by two to gain a new set of coefficients \[f_{0} + f_{1} \alpha + f_{2} \alpha^2 = 0\] and continue this process inductively until one of them is odd. We are guaranteed this is not an infinite process by the fundamental theorem of algebra, which states that all integers are composed of a product of primes. Since primes are odd, and products of odd numbers are also odd, we will eventually obtain a set of coefficients where at least one of them is odd. We can redefine our \(d_{i}\)s as this set of coefficients and we arrive at a contradiction, for at least one of them is odd but they must all be even in order for \[\left| \begin{array}{ccc}d_{0} & d_{1} & d_{2} \\2d_{2} & d_{0} & d_{1} \\2d_{1} & 2d_{2} & d_{0} \end{array} \right| = 0\] Since all of the \(d_{i}\)s must equal zero for the solution to equal zero, all of the \(a_{i}\)s must also equal zero. Therefore if \(a_{0} + a_{1} \alpha + a_{2} \alpha^2 =0\), where \(a_{i} \in \mathbb {Q}\), then all \(a_{i} = 0\), as we desired.

We wish to show F is a field. To do this, we must show that F satisfies the four axioms of a field: 1) closure under multiplication and addition 2) Every element contains an inverse 3) For every element, there is a corresponding negative element, or additive inverse, such that when added to the element it sums to zero 4) There exists a multiplicative identity element.

In part (A), we proved that \(\forall a,b \in \mathbb {f}\), it can be shown that \(ab \in \mathbb {F}\), and \(a+b \in \mathbb{F}\). Therefore F is closed under multiplication and addition.

The inverse of a polynomial is given by changing the variables of the equation and solving. Specifically, for \(y_{1}=a_{0} + a_{1} \alpha + a_{2} \alpha^2\), we can switch the order and solve using the quadratic formula to obtain \[\alpha=a_{0} + a_{1} y_{2} + a_{2} y_{2}^2\] \[\Rightarrow a_{2} y_{2}^2 + a_{1} y_{2} + (a_{0} - \alpha) = 0\] \[\Rightarrow y_{2}= \frac{-a_{1} \pm \sqrt{a_{1}^2-4a_{2}(a_{0}-\alpha)}}{2a_{2}}\] To prove this is in fact the inv

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