1. Say whether the following is true or false and support your answer by a proof
Answer: The statement is false. There do not exist natural numbers and such that .

Proof:
Assume for contradiction that (where ) satisfying . Rewrite the equation as:

Let and . Then . Since , we derive:

Suppose and for some integer . Substituting back:

  1. .
  2. .

For , must be even and a multiple of , so is a multiple of .
Testing :

  • (invalid).

For :

  • (invalid).

No valid exists. Conclusion: The equation has no solutions in .

2. Say whether the following is true or false and support your answer by a proof: The sum of any five consecutive integers is divisible by 5 (without remainder).
Answer: The statement is true.

Proof by Induction:

Base Case (): Consider the five consecutive integers starting at : Since is divisible by (), the base case holds.

Inductive Hypothesis: Assume that for some integer , the sum of five consecutive integers starting at is divisible by . That is: , where is an integer.

Inductive Step (): We need to show that the sum of the next five consecutive integers, starting at , is also divisible by . Consider: Simplify the sum: From the inductive hypothesis, we know: Thus, the sum becomes: Since is an integer, is divisible by .

Conclusion: By induction, the sum of any five consecutive integers is divisible by .

3. For any integer n, consider the expression
Answer: The statement is true.
Proof by Induction (for non-negative integers ):

  1. Base Case: :
  2. Inductive Step: Assume is odd for some integer . For : By hypothesis, is odd, and is even.
    Odd + Even = Odd, so is odd.
    By induction, the statement holds for all . For negative integers, parity (even/odd) is preserved, so the case analysis above suffices.

Conclusion: The statement is true for all integers .

4. Prove that every odd natural number is of one of the forms 4n + 1 or 4n + 3, where n is an integer.
Proof: We aim to show that every odd natural number is of one of the forms or , where n is an integer.

By the Division Algorithm (or Remainder Theorem), for any integer a and a positive integer b = 4, there exist unique integers q and r such that:

Here, r is the remainder when a is divided by 4, and r can take one of the values 0, 1, 2, or 3.

Now, consider the possible cases for r:

  1. Case r = 0:
    This is an even number because it is divisible by 2.

  2. Case r = 1:
    This is an odd number because it is not divisible by 2.

  3. Case r = 2:
    This is an even number because it is divisible by 2.

  4. Case r = 3:
    This is an odd number because it is not divisible by 2.

Since a is an odd natural number, it cannot be of the form or (as these are even). Therefore, a must be of one of the forms:

Letting n = q, we conclude that every odd natural number is of one of the forms or , where n is an integer.

QED

5. Prove that for any integer n, at least one of the integers n, n + 2, n + 4 is divisible by 3.
Proof: We aim to prove that for any integer n, at least one of the integers n, n+2, or n+4 is divisible by 3.

By the Division Algorithm, any integer n can be expressed as:

  • If , then n is divisible by 3.
  • If , then: so n+2 is divisible by 3.
  • If , then: so n+4 is divisible by 3.

In each case, at least one of the integers n, n+2, or n+4 is divisible by 3.

Conclusion: For any integer n, at least one of the integers n, n+2, or n+4 is divisible by 3.

QED

6. A classic unsolved problem in number theory asks if there are infinitely many pairs of ‘twin primes’, pairs of primes separated by 2, such as 3 and 5, 11 and 13, or 71 and 73. Prove that the only prime triple (i.e. three primes, each 2 from the next) is 3, 5, 7.
Proof: Assume, for contradiction, that there exists an integer n>3 such that n, n + 2, and n+4 are all prime.

For any integer n, it must satisfy one of the following modulo 3:

Analyzing each case:

  1. If , then n is divisible by 3.
  2. If , then , so n + 2 is divisible by 3.
  3. If , then , so n + 4 is divisible by 3.

Thus, one of n, n + 2, or n+4 must be divisible by 3. Since n>3, this number would be greater than 3 and therefore composite (as primes divisible by 3 must equal 3). This contradicts the assumption that all three numbers are prime.

Hence, no such n>3 exists. The only valid prime triple is (3, 5, 7).

QED

7. Prove that for any natural number,
Proof by Mathematical Induction:
We aim to prove that for any natural number n:

For n = 1:

Thus, the formula holds for n = 1.

Assume that the formula holds for some arbitrary k:

We need to show that the formula holds for k + 1, i.e.,

Using the inductive hypothesis:

Simplifying: which is exactly the desired formula for k + 1.

Conclusion

By the principle of mathematical induction, the formula holds for all natural numbers n.

QED

8. Prove (from the definition of a limit of a sequence) that if the sequence tends to limit L as , then for any fixed number M>0, the sequence tends to the limit .
Proof:

Using the definition of a limit, we will show that if , then for any fixed constant .

By the definition of convergence, for every , there exists a natural number such that for all :

To prove , we need to show that for every , there exists such that for all :

Factor out :

Let (if ).
From the assumption, there exists such that for all :

Multiply both sides by :

Thus, for , .

For any fixed , the sequence converges to . Therefore:

QED

9. Given an infinite collection , of intervals of the real line, their intersection is defined to be Give an example of a family of intervals , (n = 1, 2, ), such that for all n and Prove that your example has the stated property.
Proof: We consider the sequence of nested open intervals:

for .

We need to show that for all .

Since , we have:

Since for all , every element in is also in . Thus, .

By definition,

For any , we must have for all . This means:

for all .

Taking the limit as , we see that , so any in the intersection must satisfy , which is impossible.

Thus,

The given sequence of intervals satisfies the required conditions, proving that their intersection is empty.

10. Give an example of a family of intervals , , such that for all and consists of a single real number. Prove that your example has the stated property.
Proof: Consider the sequence of closed intervals:

for .

We first verify that . Since

and , we have implies ,

so every element in is also in , proving .

Now, we compute the intersection:

For any , we must have

for all .

Taking the limit as , we see that , meaning that the only possible value for is 0.

Thus,

This proves that the intersection consists of the single real number 0.