MAT102 Tutorial W6

• Assume $n\in \mathbb{N}$
• 1
  • Show that $n^2$ contains even powers of all its prime factors.
  • $n=p_1^{a_1}{p}_2^{a_2}\dots p_n^{a_n}$
  • Now we need to show that $a_i=2k$ for some $k\in \mathbb{Z}$.
  • $n^2=(p_1^{a_1}{p}_2^{a_2}\dots p_n^{a_n}{)}^2$
  • This means that:
    • $n^2=p_1^{2a_1}{p}_2^{2a_2}\dots p_n^{2a_n}$
    • We can see that we have $2a_i$. This means that we can factor out a two, $2(a_i)$. This means that the $a_i$ is even.
• 2
  • Show that $n$ is a perfect square if and only if all its prime factors have even power.
  • A perfect square is $n^2$ as $n\in \mathbb{N}$.
  • Proof
    • $⟹$ Get by question 1
    • $\Leftarrow$
      • Let $n=p_1^{2a_1}{p}_2^{2a_2}\dots p_n^{2a_n}$
      • $n=(p_1^{a_1}{p}_2^{a_2}\dots p_n^{a_n}{)}^2$
      • This means $a^2$
• 3
  • Show that $\sqrt{n}$ is either an $\mathbb{Z}$ or $\mathbb{R}\setminus \mathbb{Q}$ (Irrationals)
    • Proof
      • Case one: $n$ is a perfect square
        • $n$ is a perfect square.
        • $n=a^2,a\in \mathbb{N}$
        • $\sqrt{n}=a$
        • $a\in \mathbb{Z}$
      • Case two: $\sqrt{n}$ will be irrational
        • $n$ is not a perfect square.
        • Contradiction:
          • Assume $\sqrt{n}$ is rational.
          • $\sqrt{n}=\frac{p}{q},p,q\in \mathbb{Z}$
          • $gcd(p,q)=1,q\ne 0$
          • $n=\frac{p^2}{q^2}$
          • $n{q}^2=p^2$
          • Notice if $n$ isn't a perfect square. Then at least one of it's prime factors has an odd power.
            • $n=p_1^{2a_1+1}{p}_2^{a_2}\dots p_n^{a_n}$
            • Like this
          • If $n{q}^2=p^2$ and $n$ has an odd power, then it's a contradiction. Because $p^2$ is a perfect square.
• True or false
  • For $a,b\in \mathbb{Z}$ if $a^2\mid b^2$ then $a\mid b$
  • Proof
    • $a^2\mid b^2$ means $a^2=k{b}^2$ for some $k\in \mathbb{Z}$.
    • $a\mid b$ means $a=jb$ for some $j\in \mathbb{Z}$.
    • We want to show that if $a^2\mid b^2$ then $a\mid b$.
    • $a^2=k{b}^2$
    • $a=\sqrt{k{b}^2}$
    • $\sqrt{k{b}^2}\mid b$
    • $\sqrt{k{b}^2}=jb$
    • $k{b}^2=(jb{)}^2$
    • $k{b}^2=j^2{b}^2$
    • $k=j^2$
    • This shows us that the multiplier for our division statement is squared if $a$ and $b$ are both squared.
    • But so far I've only proved that the multiplier is squared.
  • Notes
    • $a\mid b⟺\{\begin{array}{l}a=p_1^{a_1}{p}_2^{a_2}\dots p_n^{a_n}\\ b=p_1^{b_1}{p}_2^{b_2}\dots p_n^{b_n}\\ \text{where }{a}_i\le b_i\end{array}$
    • Let $\{\begin{array}{l}{a}^2=p_1^{2a_1}{p}_2^{2a_2}\dots p_n^{2a_n}\\ b^2=p_1^{2b_1}{p}_2^{2b_2}\dots p_n^{2b_n}\end{array}$
      Notice that $2a_i\le 2b_i⟹a_i\le b_i$
    • This tells us that $a\mid b$
• Find all possible value of $gcd(n+1,2-n)$
  • $d=gcd(n+1,2-n)$
  • This means $d\mid n+1$ and $d\mid 2-n$.
  • $d\mid gcd(n+1,(n+1)+(2-n))$
  • $d\mid gcd(n+1,(n+1)-(2-n))$
  • $(n+1)-(2-n)$
    • $n+1-2+n$
    • $2n-1$
    • doesn't simplify
  • $(n+1)+(2-n)$
    • $n+1+2-n$
    • $+1+2$
    • $3$
  • $d=gcd(n+1,3)$
  • $n+1=d(3)$
  • $\frac{n+1}{3}=d$

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  • Notes
    • $gcd(a,b)=gcd(a,b-a)$
    • $gcd(a,b)=d=gcd(d(a,-b))$
  • $gcd(n+1,1-2n)$
  • $gcd(n+1,2n-1)$
    • Negate the second term
  • $gcd(n+1,n-2)$
  • $gcd(n+1,(n-2-n-1))$
  • $gcd(n+1,-3)$
  • $gcd(n+1,3)$
  • Since $3$ is prime, it can be $1$ or $3$.