## Math 1190 - Introduction to Sets and Logic

# Quiz 2 - Solutions

### Part (a)

47 divided by 7 leaves 6 remainder 5.

In other words, 47 = 6x7 + 5. Therefore a=5 works.
### Part (b)

There were several ways to approach this. One is to compute all the
values:
- f(0)=3
- f(1)=4
- f(2)=5
- f(3)=6
- f(4)=0 [as 4+3=7 which is congruent to 0 mod 7]
- f(5)=1 [as 5+3=8 which is congruent to 1 mod 7]
- f(6)=2 [as 6+3=9 which is congruent to 2 mod 7]

Since all the numbers A appear once (and only once) in the right hand
column, f is bijective. [You could have also shown this by drawing two
columns of 0's through 6's, and then linking up values and their images.
The crucial point is that you have to connect this with being bijective,
by explicitly stating that each number in the second column corresponds
to EXACTLY one number in the first column.]
Another way to establish that f is bijective is to show that it has
an inverse. In this case, the inverse is the function g(y) which equals
the remainder of y-3 upon division by 7. [Because if y is congruent
to x+3, then x is congruent to y-3. Note that x is NOT equal to y-3; you
have to take remainders, or else you won't have numbers between 0 and 6.]