MAT102 Lecture 02B

• Shoe sock b

• movie actress a

• pencil paper b

• highschool college a

• ocean breeze a

• leaf tree a

• computer chip a

• chips salsa b

• phone book b

• 5x a, 4x b

• Quirky

  • $n$ is quirky if $n^2$ is not even
  • $0^2$ is even, so not quirky
  • $1^2$ is not even, so quirky
  • $2^2$ is even, so not quirky
  • $3^2$ is not even, so quirky
  • $4^2$ is even, so not quirky
  • Odd numbers are quirky based on this pattern.
  • $n\text{ is quirky}⟺n=2k+1,k\in \mathbb{Z}$

• Mean Value Theorem

• $\underbrace{\text{If }a\text{ and }b\text{ are both even},then}ab\text{ is also even}$

  • Hypothesis
  • Let $a=2k,k\in \mathbb{Z}$ let $b=2\ell ,\ell \in \mathbb{Z}$
  • Let's show that $ab=2j,j\in \mathbb{Z}$
  • $ab=2j$
  • $(2k)(2\ell )=2j$
  • $4k\ell =2j$
  • Notice we can take out a $2$
  • $2k\ell =j$
  • Since $k,\ell \in \mathbb{Z}$, we can show that $j$ is $2\cdot \text{some integer}$
  • Thus we have proven the above.

• There exists a unique integer $e$, such that for any other integer $n,e\cdot n=n$

  • Assuming $e$ is not euler's number
  • We know that by field axioms that we have additive and multiplicative identities.
  • So we must have some multiplicative identity that should result in the same number multiplying the identity with $n$.
  • We are in the $\mathbb{Z}$ space, so the multiplicative identity is $1$
  • So $e\cdot n=n$ where $e=1$

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  • Proceed by contradiction:
    • Suppose there is a number $e$ such that $e\cdot n=n$, where $e,n\in \mathbb{Z}$
    • Suppose we have another number $d$ where $d\cdot n=n$
    • So $e=n$ and $d=n$
    • Which means $e=d$
    • Contradiction, we have found another number wherein $e\cdot n=n$
    • So there must be a unique number which is the multiplicative identity.