Show that 5|a + b when (a,b) ∈ S

Question

I've been trying this problem for a bit now and I'm confused about what the notation even is. The previous problem was about recursive definitions and our teacher provides with powerpoints but I can't seem to find anything about this notation in them. I searched online for help but nothing comes up for me. Please, any help is greatly appreciated.

Thank you,

-jjleahy

EDIT:

Let S be the subset of the ordered pairs of integers defined recursively by:

Basis Step: (0,0) ∈ S

Recursive step: If (a,b) ∈ S, then (a + 2, b + 3) ∈ S and (a + 3, b + 2) ∈ S.

Answer 1

I think this can be done by induction on number of steps it takes to get to the tuple $(a,b)$ from $(0,0)$ by performing the two operations mentioned successively .

Base case: no operation is performed on $(0,0)$ and $5\mid (0+0)$. Now assume that all tuple you get by applying n successive mentioned operations to get to $(a,b)$, $5\mid a+b$. And you choose a tuple $(a,b)$ obtained from $n+1$ successive application of operations. Then either $(a-2,b-3) $ is in $S$ or $(a-3,b-2)$ is in $S$. Either way, $5\mid a-2+b-3$ and $5\mid 5$ implies $5\mid a+b$ or $5\mid a-3+b-2$ and $5\mid 5$ implies $5\mid a+b$.

Answer 2

Hints:

1. $5\mid a+b $ means that the sum of $a$ and $b$ is divisible by 5. For example, if $(a,b)=(4,6)$, then $5\mid a+b$ is true because $a+b=10$ and 10 is divisible by 5.
2. $S$ is a subset of the ordered pairs of integers means that the elements of $S$ have the form $(a,b)$ where $a$ and $b$ are integers. For example $(2,3)$, $(5,5)$, and $(4,6)$ are elements of $S$.
3. $S$ is defined recursively by the "Basis step:" $(0,0)\in S$ and the "Recursive Step:" if $(a,b)\in S$ then $(a+2, b+3)\in S$ and $(a+3, b+2)\in S$ means two things:

a) $(0,0)$ is an elment of $S$, and

b) if $(a,b)$ is an element of $S$ then $(a+2, b+3)$ and $(a+3, b+2)$ are both elements of $S$. For example, if you knew that $(14,16)$ was an element of $S$, you could conclude that $(16,19)$ and $(17,18)$ were both elements of $S$.

4. In order to show that $5\mid a+b$ when $(a,b)\in S$, you have to show two things

a) if $(a,b)=(0,0)$ then $a+b$ is divisible by 5, and

b) if $(a,b)$ is in $S$ and $a+b$ is divisible by five, then the two "new" element generated by the "Recursive Step" from $(a,b)$ (namely $(a+2, b+3)$ and $(a+3, b+2)$) will also have the property that the sums of their coordinates are divisible by 5.

Hope that helps.