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 \(ab \in F\)

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