Problem 1 - A Gentle Introduction to the Art of Mathematics

Question

A "postage stamp problem" is a problem that (typically) asks us to determine what total postage values can be produced using two sorts of stamps. Suppose that you have \(3 \mathrm{c}\) stamps and \(7 \mathrm{c}\) stamps, show (using strong induction) that any postage value \(12 \mathrm{c}\) or higher can be achieved. That is, $$ \forall n \in \mathbb{N}, n \geq 12 \quad \Longrightarrow \quad \exists x, y \in \mathbb{N}, n=3 x+7 y. $$

Step-by-Step Solution

Answer

Any postage value 12 cents or higher can be made using 3 cent and 7 cent stamps.

Step 1 - Base Cases

To prove the statement using strong induction, start with the smallest value, which is 12 cents. Verify that 12 cents can be formed using the 3 cent and 7 cent stamps. We can write: $$ 12 = 3(4) + 7(0) $$Therefore, 12 cents can be formed.

Step 2 - Prove for the Next Few Values

Now, we systematically show that 13, 14, 15, 16, and 17 cents can also be made:$$ 13 = 3(2) + 7(1) $$$$ 14 = 3(0) + 7(2) $$$$ 15 = 3(5) + 7(0) $$$$ 16 = 3(3) + 7(1) $$$$ 17 = 3(1) + 7(2) $$

Step 3 - Inductive Hypothesis

Assume that for all values of \(k\) such that \(12 \text{c} \text{ or higher (up to } n)\) can be achieved with the given stamps. Here, we need to show that \(n+1\) can also be formed.

Step 4 - Inductive Step

To prove \(n+1\) cents can be formed, consider the values $$\begin{cases}: n+1 - 3 = k \text{ (As we can remove a 3 cent stamp)}\text{Since }12 \text{c or higher up to } n\( can be achieved, \) k \text{ can be formed.}$$

Step 5 - Conclusion

Because \(k = (n+1) - 3 \text{ is } \text{still a valid number}\text{ as stated }. It must be higher than or equal to 12 / previously thought possible.By using previous steps up to complete the proofs for all \) \text{higher values of } n\ \text{ thus induction holds true.}$$Therefore since \( 12 to 13\text{..}.17\text{ will achieved by \) 3 /7 cents stamps - all further stepped out } By induction thus any value can be stated for all n( 12 / higher can be shown same way).$$.

Key Concepts

Strong Induction

Strong induction is a powerful method of proving statements about natural numbers. Unlike regular induction, strong induction assumes the truth of the statement for *all* previous cases rather than just one. In the postage stamp problem, we use strong induction to show that any postage value of 12 cents or higher can be made using 3-cent and 7-cent stamps. This involves:

• Establishing base cases where we verify the initial set of values.
• Assuming the statement is true for all values up to a certain point.
• Proving it holds for the next value based on the assumption that it holds for all previous ones.

The strength of strong induction helps handle situations where direct steps from one case to the next are not visible or intuitive, which is what makes it ideal for this postage stamp scenario.

Base Case

The base case is the first step in induction proofs and serves as the foundation. It verifies that the statement is true for the initial value or set of values. In our problem, the smallest value to check is 12 cents. We verify it by showing that 12 cents can indeed be formed using 3-cent and 7-cent stamps:
\[ 12 = 3 \times 4 + 7 \times 0 \]
Once we validate this initial step, we move on to check subsequent values: 13, 14, 15, 16, and 17 cents. Each of these values can also be constructed using the stamps:

• \[ 13 = 3 \times 2 + 7 \times 1 \]
• \[ 14 = 3 \times 0 + 7 \times 2 \]
• \[ 15 = 3 \times 5 + 7 \times 0 \]
• \[ 16 = 3 \times 3 + 7 \times 1 \]
• \[ 17 = 3 \times 1 + 7 \times 2 \]

These base cases are crucial since they give us a launching point and ensure our hypothesis holds for these specific values.

Inductive Step

The inductive step is the heart of the strong induction method. It shows that if the statement holds for all previous cases, it must also be true for the next case. In the postage stamp problem, after establishing our base cases up to 17 cents, we assume they hold for all values up to some value \(n\). We then need to show it holds for \(n+1\). To do this, we consider:

• If \(n+1 - 3\) can be achieved, which means \(n+1 = 3k + 7m\) once we've removed a 3-cent stamp.

Given that \(k = (n+1) - 3\) is still within the range we previously validated, and since strong induction assumes all preceding values are achievable, \(k\) must be achievable.
By this step, we establish that any amount \(n+1\) can also be formed, assuming \(n\) and all values before it are formable. This completes the inductive argument and shows that all values \( \forall n \text{ such that } n \geq 12\) can be created using the given stamps.