8. Just for this question, pretend that 1 as a prime number. It is not actually a prime
number, but if we pretend it is for a minute, it makes the following statement easier
to write.
Bertrand’s postulate is a theorem in number theory which states that for every positive
integer n, there is a prime number p such that 1
2
n < p ≤ n. Use Bertrand’s postulate
along with strong induction to prove that every integer n ≥ 1 can be written as a sum
of distinct primes.
“Distinct” means that no prime shows up twice: 9 = 5 + 2 + 2 would not be valid since
2 is repeated, but 9 = 5 + 3 + 1 is valid. A sum with only one term, like 2 = 2, is valid.
You do not need to prove Bertrand’s postulat

To prove that every integer \( n \geq 1 \) can be written as a sum of distinct primes using strong induction and Bertrand's postulate (with the assumption that 1 is a prime number), we'll proceed as follows.

### Step 1: **Base Case**
For \( n = 1 \):
- \( n = 1 \) is trivially written as a sum of distinct primes because we are pretending that 1 is a prime number. Thus, \( n = 1 \) can be written as the sum of the distinct prime \( 1 \) alone.
  
Thus, the base case holds.

### Step 2: **Inductive Hypothesis**
Assume that for all integers \( k \), where \( 1 \leq k \leq n \), every integer \( k \) can be written as a sum of distinct primes. This is the strong induction hypothesis.

### Step 3: **Inductive Step**
We need to prove that \( n + 1 \) can also be written as a sum of distinct primes.

#### Case 1: \( n+1 \) is prime
- If \( n+1 \) is prime, then \( n+1 \) itself is a valid sum of distinct primes (with only one term). This satisfies the condition that \( n+1 \) can be written as a sum of distinct primes.

#### Case 2: \( n+1 \) is not prime
- If \( n+1 \) is not prime, then we can use the inductive hypothesis. Since Bertrand’s postulate guarantees that between any number \( n/2 \) and \( n \), there is a prime \( p \), we can use this prime number to break down \( n+1 \) into two smaller distinct primes.
  
  - Let \( p \) be a prime such that \( 1/2(n+1) < p \leq n+1 \) (by Bertrand’s postulate). Now, subtract this prime \( p \) from \( n+1 \). That is:
    \[
    n+1 - p = m
    \]
    where \( m \) is a smaller number than \( n+1 \).

  - By the inductive hypothesis, \( m \) can be written as a sum of distinct primes, say:
    \[
    m = p_1 + p_2 + \dots + p_k
    \]
  
  - Hence, \( n+1 \) can be written as:
    \[
    n+1 = p + p_1 + p_2 + \dots + p_k
    \]
    Since \( p \) and the primes \( p_1, p_2, \dots, p_k \) are distinct, this is a valid sum of distinct primes.

### Conclusion:
By strong induction, we have proven that every integer \( n \geq 1 \) can be written as a sum of distinct primes. Bertrand’s postulate ensures that we can always find a prime within the appropriate range to apply the inductive step.

Would you like any clarifications or more details on this proof?

### Related Questions:
1. How does Bertrand's postulate work for large values of \( n \)?
2. Can you give more examples of writing integers as sums of distinct primes?
3. What would change in this proof if we do not treat 1 as a prime number?
4. How does strong induction differ from simple induction?
5. Can this method be extended to non-prime numbers?

### Tip:
When applying strong induction, always carefully verify the base case and ensure that the inductive hypothesis covers all relevant cases up to \( n \).

Just for this question, pretend that 1 is a prime number. It is not actually a prime number, but if we pretend it is for a minute, it makes the following statement easier to write. Bertrand’s postulate is a theorem in number theory which states that for every positive integer n, there is a prime number p such that 1/2 n < p ≤ n. Use Bertrand’s postulate along with strong induction to prove that every integer n ≥ 1 can be written as a sum of distinct primes. 'Distinct' means that no prime shows up twice: 9 = 5 + 2 + 2 would not be valid since 2 is repeated, but 9 = 5 + 3 + 1 is valid. A sum with only one term, like 2 = 2, is valid.

Explore how to prove that every integer greater than or equal to 1 can be written as a sum of distinct primes using Bertrand's Postulate and strong induction. This in-depth solution covers key concepts in number theory and is suitable for advanced high school or college-level students.

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