# We can use an unlimited supply of 4-cent and 7-cent postage stamps to make (exactly) any amount of postage that is 18 cents or more.

$P(n)$: postage of exactly n cents can be made using only 4 cents and 7 cents

Basis ($n=18$)

$1 \cdot 4 + 2\cdot7 = 18$, thus $P(18)$ holds

Induction step:

Let $i$ be an arbitrary number such that $i \geq18$ and suppose that $k, L \in \mathbb N$ such that $i = 4\cdot k + 7 \cdot L$ k representing the # of 4 stamps and L representing the # of 7 stamp

I was told I have to prove 2 cases with them being $L > 0$ and $L =0$. I don't get how they got the two cases. I'm only aware that I have to prove that any number $i \geq 18$ can be achieved with 4 stamps and 7 stamps.

Could someone explain this proof?

• If $L>0$, you can take away one 7-cent stamp and replace it with two 4-cent stamps to get $i+1$ cents. If $L=0$, you can use that you have at least five 4-cent stamps (since $i\ge18$) to make $i+1$ cents. – user84413 Sep 10 '17 at 23:19
• The two cases are a hint from your instructor (or book) to help you solve the problem more easily. As is often the case with exercises like this, it is counterproductive to try to figure out why your instructor gave you a hint, at least until after you have actually followed the hint and have seen where it led you. Try it and see. – David K Sep 10 '17 at 23:29
• This is a duplicate. I'm on my phone so it's hard for me to look it up but someone please look it up so we can close this thing down. – marty cohen Sep 10 '17 at 23:30
• @martycohen I found several other versions of 4-cent and 7-cent stamps, but none with the "two cases" hint: math.stackexchange.com/questions/1766227 math.stackexchange.com/questions/697897 math.stackexchange.com/questions/1743231 math.stackexchange.com/questions/1576870 -- I don't know if that helps. – David K Sep 10 '17 at 23:37

Remember that in general, the structure of the inductive step would be like this:

Let $i\in\mathbb N$ and $i \geq 18,$ and suppose $P(i)$ is true. We wish to show that $P(i+1)$ is true.

Since $P(i),$ ...
...
... and therefore $P(i+1).$ This proves the inductive step.

Your problem is to fill in the "..." parts with a valid mathematical argument.

Your instructor's hint about cases $L > 0$ and $L=0$ gives you a little more detail about how the inductive step can be structured. The hint says to try this:

Let $i\in\mathbb N$ and $i \geq 18,$ and suppose $P(i)$ is true. We wish to show that $P(i+1)$ is true.

Since $P(i),$ there exist $k, L \in \mathbb N$ such that $i = 4\cdot k + 7 \cdot L.$ Consider the following two cases, which cover all possible values of $L$:

Case 1: Suppose $L>0.$ Then ...
...
... and therefore $P(i+1).$

Case 2: Suppose $L=0.$ Then ...
...
... and therefore $P(i+1).$

Hence in either case we have shown $P(i+1).$ This proves the inductive step.

You still have to fill in the "..." parts. With two separate arguments. But the only way to learn how to do this is to start trying.

By the way, the details of your question seem to imply that your book and/or instructor consider the least element of $\mathbb N$ to be $0.$ In some other contexts, people consider the least element of $\mathbb N$ to be $1$ (and therefore $0 \not\in\mathbb N$). I think what you have written so far is probably perfectly OK, but in general, when dealing with $\mathbb N$ make sure you know which way it has been defined.

• Case 1: L > 0. We have $k' = k+2$ and $L' = L - 1$. $$4 \cdot k' + 7 \cdot L' = 4(k+2) + 7(L-1) = 4k + 7L + 1 = i + 1 \text{ By I.H and definition of i, k', and L'}$$ – jimmy jimmy Sep 11 '17 at 1:10
• Case 2: k' = k-5, L' = 3 works. Thanks I get it – jimmy jimmy Sep 11 '17 at 1:21
• Case 2: If L= 0, then $18 \le N = 4k + 0*L$ so $k \ge 4.5$ so $k \ge 5$. – fleablood Sep 11 '17 at 1:30

INDUCTION STEP:

Suppose you can make $N$ by $4m + 7n$.

Then you can make $N+1$ by doing $4(m+2) + 7(n-1) = 4m +8 + 7n -7 = 4m + 7n + 1 = N + 1$.

That'd a good induction step.... except what if $n < 1$?

Well then you could do $N = 4m + 7n = 4(m -5) + 7(n+3)=4m -20 + 7n + 21 = 4m + 7n + 1$.

That'd a good alternative induction step... except what if $m < 5$?

Well,... then you'd do the first one. Any any case you can either do the first induction step, or the second induction step, or take your choice of which one to use.

Okay, what if both $m < 5$ and $n < 1$. Well, ... then we'd be screwed.

But If that were the case $m \le 4;n=0$ and $N \le 4*4 + 7*0 = 16$.

So, so long as $N > 16$ we will have our choice of induction steps and we can always get $N+1$.

... Provided our base case is $N > 16$.

BASE CASE

If $N=17$ we .... can't do it.

If $N = 18$ we can do it with $m = 1$ and $n= 2$ and so $N = 4m + 7n = 4*1 + 7*2 = 4+14 = 18$.

Hence we are done by induction.

[Note: We can do $16 = 4*4 + 7*0$ but to get $17$ we have to do $4*(4-5) + 7*(0+3)= 4(-1) + 7*3$. And we can also do $14$ to $15$ to $16$ by $14 = 4*0 + 7*2; 15 = 4*(0+2) + 7*(2-1) = 4*2 + 7*1; 16 = 4*(2+2)+7*(1-1) = 4*4 + 7*0$. But then we hit a wall because we have both $m < 5$ and $n < 1$ so we can't do the induction step.]

\begin{eqnarray*} 18=1 \times 4 + 2 \times 7 \\ 19=3 \times 4 + 1 \times 7 \\ 20=5 \times 4 + 0 \times 7 \\ 21=0 \times 4 + 3 \times 7 \\ \end{eqnarray*} Now add $4$ to these.

• Thats strong induction I'm aware of how to do it this way. The question is in simple induction with the given hint – jimmy jimmy Sep 11 '17 at 1:11
• If $N \ge 18$ then either you have at least $5$ 4 cent stamps OR you will have at least $1$ 7 cent stamps If you you have $5$ 4 cents stamps trade them in for $3$ 7 cents stamps. If you don't have $5$ 4 cents stamps but you have a $7$ cent stamp, trade in in ofr $2$ 4 cent stamps. If you don't have either $5$ 4 cent stamps nor $1$ seven cents stamps, then all you have is $4$ 4 cents stamps = $16$ cents. – fleablood Sep 11 '17 at 1:24
• @jimmyjimmy Technically, it's possible (though maybe a bit complicated) to build a simple-induction proof on these four statements. However, this answer has already been given for previous 4,7-cent stamp questions, and it doesn't explain the two cases you were asking about. Also, I think for what you want, the way you're doing it is easier. – David K Sep 11 '17 at 2:22

For the record, why don't we prove that if $a$, $b$ are relatively prime positive integers then every integer $> ab$ is of the form $u a + v b$, with $u$,$v$ integers $>0$. Indeed, take $k> ab$ integer. Then (exactly) one of the integers $a$, $2 a$, $\ldots$, $b a$ will be $\equiv k \mod b$, let that be $u a$. So $k - u a = v b$, with $v > 0$.