Using equivalence classes to simplify problems

- Operations on integers mod
*m*are well defined- [
*a*] + [*b*], − [*a*], [*a*] ⋅ [*b*], [*a*]^{n}

- [
- Inverses
- units

- Exponentiation
*n*^{[a]}not well defined- definition of totient (
*ϕ*(*m*))

Definition: [*a*] = [*b*] if *a* ≡ *b*(*m**o**d**m*) which means *m*∣(*a* − *b*) which means there is some *k* with *a* − *b* = *k**m*.

Equivalently, [*a*] = [*b*] if and only if *a* = *b* + *k**m* for some *k*. We'll use this form several times.

Claim: defining [*a*] + [*b*] as [*a* + *b*] is well-defined. That is, if [*a*] = [*a*ʹ] and [*b*] = [*b*ʹ] then [*a* + *b*] = [*a*ʹ + *b*ʹ].

Proof: to start, we'll just choose a different representative of *a*.

Suppose [*a*] = [*a*ʹ]. Then *a* = *a*ʹ + *k**m* for some *k*. Thus *a* + *b* = *a*ʹ + *k**m* + *b* = (*a*ʹ + *b*) + *k**m*. Thus [*a* + *b*] = [*a*ʹ + *b*].

Now, suppose [*a*] = [*a*ʹ] and [*b*] = [*b*ʹ]. Then we can use the above twice to conclude [*a* + *b*] = [*a*ʹ + *b*ʹ]:

[*a* + *b*] = [*a*ʹ + *b*] = [*b* + *a*ʹ] = [*b*ʹ + *a*ʹ] = [*a*ʹ + *b*ʹ]

Claim: defining [*a*][*b*] as [*a**b*] is well defined.

Proof: as above, we'll just choose a different representative for [*a*] and then use commutativity.

Suppose [*a*] = [*a*ʹ]. Then *a* = *a*ʹ + *k**m* for some *k*. So *a**b* = (*a*ʹ + *k**m*)*b* = *a*ʹ*b* + (*k**b*)*m*

Thus [*a**b*] = [*a*ʹ*b*].

Claim: defining − [*a*] as [ − *a*] is well defined.

Proof: suppose [*a*] = [*a*ʹ]. Then *a* = *a*ʹ + *k**m*. Thus − *a* = − *a*ʹ + ( − *k*)*m*. Thus [ − *a*] = [ − *a*ʹ].

Claim: if *n* is an integer, then defining [*a*]^{n} as [*a*^{n}] is well defined.

Proof: use induction on *n* and the fact that multiplication is well defined.

Compute 11

^{734}*m**o**d*10. Use the fact that [11] = [1]. Answer: [1]Compute 9

^{735}*m**o**d*10. Use the fact that [9] = [ − 1]. Answer: [9]Compute 7

^{735}*m**o**d*10. Use the fact that 7^{735}= 7^{2 ⋅ 367 + 1}= 49^{367}⋅ 7 and [49] = [ − 1]. Answer: [3].

a

**ring**is a set of things that you can add and multiply, and addition and multiplication obey the usual rules (e.g. commutativity, distributivity, etc.)in a ring, a

**unit**is an element with a multiplicative inverse. That is,*x*is a unit of*R*if there exists a*y*∈*R*with*x**y*= 1.- Examples:
- units of Z are 1 (inverse is 1) and -1 (inverse is -1)
- units of Q are everything other than 0. Inverse of
*p*/*q*is*q*/*p*. - units of R are everything other than 0.

You proved in homework that units of Z

_{m}are [*a*] where*a*and*m*are relatively prime.- Examples:
- units of Z
_{7}are {[1], [2], [3], [4], [5], [6]}. Inverses:- inverse of [1] is [1] because [1][1] = [1]
- inverse of [2] is [4] because [2][4] = [8] = [1]
- inverse of [3] is [5] because [3][5] = [15] = [1]
- inverse of [6] is [6] because [6][6] = [36] = [1]
- [0] not a unit because [0] times anything is [0] ≠ [1]

- units of Z
_{4}are {[1], [3]}.- inverse of [1] is [1]
- inverse of [3] is [3]
- [2] has no inverse: [2][0] = [0] ≠ [1], [2][1] = [2] ≠ [1], [2][2] = [0] ≠ [1] and [2][3] = [2] ≠ [1].

- units of Z

Defining *n*^{[a]} as [*n*^{a}] doesn't work. Counterexample: let *m* = 5, *n* = 2, *a* = 2. Then [*a*] = [2] = [7], but [2^{2}] = [4] and [2^{7}] = [128] = [3] ≠ [4].

The **totient** of *m*, written *ϕ*(*m*) (pronounced "phi" of *m*, LaTeX `\phi`

) is the number of units of Z_{m}.

Examples: *ϕ*(7) = 6, *ϕ*(4) = 2.

We'll use totient in next lecture to define exponentiation.