MAT102 Lecture 13

• No elementary antiderivative for $\int e^{-x^2}dx$
• You can prove or disprove this with field theory.
• 1
  • If we have a field: $\mathbb{F}=\mathbb{Z}\cup \{\frac{1}{n}\mid n\in \mathbb{Z}\setminus \{0\}\}$
  • Let's have $\frac{1}{3}\in \mathbb{F}$ and $\frac{1}{3}\in \mathbb{F}$. But $\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\notin \mathbb{F}$
• 2
  • ${\mathbb{Z}}_n$ is a field if and only if $n$ is prime.
  • ${\mathbb{Z}}_n$ satisfied all field axioms except multiplicative inverse.
    • Fermat's theorem
    • If $p$ is prime and $p\nmid a$ then $a^{p-1}\equiv 1(\mathrm{mod}p)$
    • $(a^{p-2})(a)\equiv 1(\mathrm{mod}p)$
    • Example:
      • ${\mathbb{Z}}_5=\{[0],[1],[2],[3],[4]\}$
      • $2^4\equiv 1(\mathrm{mod}5)$
      • $3^4\equiv 1(\mathrm{mod}5)$
      • $4^4\equiv 1(\mathrm{mod}5)$
  • ${\mathbb{Z}}_{39}$
  • If $n$ is not prime, then ${\mathbb{Z}}_n$ isn't a field.
  • An element $k$ is a zero divisor if $\mathrm{\exists }m$ such that $k\ne 0$ and $m\ne 0$ but $km=0$.
    • ${\mathbb{Z}}_6,[2][3]=[0]$
      • Notice that $2\cdot 3=6$ which is our $n$
    • These zero divisors can't have multiplicative inverses.
  • If $n$ is not prime, write $n=rs$
  • Now $[r][s]=[0]$ so then ${\mathbb{Z}}_n$ is not a field.
• 3
  • Example of a multiplicative inverse.
  • $(1,2)\cdot (1,\frac{1}{2})$
  • Now to break this, we need a multiplicative inverse where it breaks that it should equal $(1,1)$
  • We need to somehow find that $(0,1)\cdot (a,b)=(a\cdot 0,b\cdot 1)$ there aren't any real numbers that result in $(1,1)$ here.
• 4
  • Fermat's Little Theorem:
    • if $a\in \mathbb{Z},p\in \mathbb{N}$ are such that $p$ is prime and $p\nmid a$, then $a^{p-1}\equiv 1(\mathrm{mod}p)$.
  • $3^7\equiv 1(\mathrm{mod}8)$
  • This is not congruent
  • $3^7\not\equiv 1(\mathrm{mod}8)$
    • $3^7=(3^2{)}^3\cdot 3=3(\mathrm{mod}8)$
    • $3^6\equiv 1$
    • We need to show $3^7$ doesn't have a remainder of $3$ when divided by $8$.
  • $3^7=3^{2\cdot 3+1}=(3^2{)}^3\cdot 3^1$
  • $3^7\equiv 1\cdot 3(\mathrm{mod}8)$ $3^7\equiv 3(\mathrm{mod}8)$

• 1
  • a)
    • $\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}:a,b\in \mathbb{Q}\}$
    • Show that $\mathbb{Q}(\sqrt{2})$ is closed under addition and multiplication
    • If $x,y\in \mathbb{Q}(\sqrt{2})$
    • Does $x+y\in \mathbb{Q}(\sqrt{2})$ and does $xy\in \mathbb{Q}(\sqrt{2})$
    • Addition:
      • So we know $x=a+b\sqrt{2}$ but $y=c+d\sqrt{2}$
      • $x+y=a+b\sqrt{2}+c+d\sqrt{2}$
      • $x+y=(a+c)+(b\sqrt{2}+d\sqrt{2})$
      • $x+y=(a+c)+\sqrt{2}(b+d)$
      • Let $A=a+c$ and $B=c+d$
      • Then we have $x+y=A+B\sqrt{2}$
      • So we have shown that it still takes in the form as defined by $\mathbb{Q}\sqrt{2}$.
    • Multiplication:
      • $\{\begin{array}{l}x=a+b\sqrt{2}\\ y=c+d\sqrt{2}\end{array}$
      • Now $x+y=(a+b\sqrt{2})(c+d\sqrt{2})$
      • $x+y=ac+ad\sqrt{2}+bc\sqrt{2}+2bd$
        • $x+y=ac+\sqrt{2}(ad+bc)+2bd$
        • $x+y=(ac+2bd)+\sqrt{2}(ad+bc)$
        • Let $A=ac+2bd$ because $ac+2bd\in \mathbb{Q}$
        • Let $B=ad+bc$ because $ad+bc\in \mathbb{Q}$.
        • So we have $x+y=A+B\sqrt{2}$ which is in $\mathbb{Q}(\sqrt{2})$
    • Solution:
      • Let $x=a+b\sqrt{2}$ and $y=c+b\sqrt{2}$
      • $\dots$
      • Above solution is correct.
  • b)
    • Show that $\mathbb{Q}(\sqrt{2})$ is a field.
    • Since $\mathbb{Q}$ has most of the axioms true, such as commutativity, associativity, etc. We just need to prove the additive inverse and the multiplicative inverse.
    • So let's show that there is $x+n=0$ and $x\cdot r=1$
    • Let $x=a+b\sqrt{2}$ and $n=c+d\sqrt{2}$
    • now we have
      • $(a+b\sqrt{2})=-(c+d\sqrt{2})$
      • $(a+b\sqrt{2})=-c-d\sqrt{2}$
      • $a+b\sqrt{2}+c+d\sqrt{2}=0$
    • Solution
      • $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt{2})\subseteq \mathbb{R}$
      • Commutativity is inherited down from $\mathbb{R}$ and so is associativity, etc.
      • Multiplicative Inverse:
        • Let $x=a+b\sqrt{2}$
        • So $x^{-1}=\frac{1}{a+b\sqrt{2}}$, this doesn't seem to be of the correct form for being in $\mathbb{Q}(\sqrt{2})$.
        • How do we show that this is part of the set.
        • Conjugate
        • $x^{-1}=\frac{1}{a+b\sqrt{2}}\frac{a-b\sqrt{2}}{a-b\sqrt{2}}$
        • $x^{-1}=\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}$
        • How do we show that $a^2-2b^2\ne 0$
        • For the sake of contradiction, …
          • No rational solutions.