**Problem 1.**

Prove or give counterexample to the following statement: The sum of any even and odd integer is odd

**Solution.**

The statement is true. We take any even number, which can be written as 2a, then we take any odd number, which can be written as 2b+1. The sum of these two numbers is 2a + 2b+1 = 2(a + b) + 1, which is an odd number.

**Problem 2.**

Show that 6.3215215215 is a rational number.

**Solution.**

Let n = 6.3215215215 Then 10n = 63.215215215 and 10000n = 63215.215215 Subtracting 10n from 10000n, we obtain: 9990n = 63152, thus n = 63152 / 9990, which shows that n is rational. It is true that the fraction representing n could be simplified further, but what we have done is enough to establish that n is rational as required.

**Problem 3.**

Two athletes run a circular track at a steady pace so that
the first completes one round in eight minutes and the second in 10 minutes. If
they both start from the same spot at

**Solution.**

We just need to find the least common multiplier of 8 and 10
and this will tell us when two athletes will meet again. Hence, the answer is

**Problem 4.**

Show that the square of any integer can never have remainder 2 when divided by 3.

**Solution.**

Any number n can have one of the following three representations:

1) n = 3q

2) n = 3q + 1

3) n = 3q + 2

First case gives: n^{2} = 9q^{2} = 3 * (3q^{2})
+ 0

Second case gives: n^{2} = 9q^{2} + 6q + 1 =
3 * (3q^{2} + 2q) + 1

Third case gives: n^{2} = 9q^{2} + 12q + 4 =
3 * (3q^{2} + 4q + 1) + 1

As we can see in all three cases we never get a remainder which equals 2.

**Problem 5.**

Is it true or false that for all real numbers x: λx^{2}ϋ = λxϋ^{2}?

**Solution.**

The statement is false: Counter-example: x = 1.5

λ1.5^{2}ϋ = λ2.25ϋ = 2

λ1.5ϋ^{2}
= 1^{2} = 1

**Problem 6.**

Prove or give counterexample to the following statement: The product of any two irrational numbers is irrational.

**Solution.**

The statement is false: Counter example: sqrt(2) * sqrt(2) = 2, where sqrt is a square root

**Problem 7.**

Given that a, b, and c are odd integers, can the following
equation have rational solutions: ax^{2} + bx
+ c = 0?

**Solution.**

Suppose that there exists rational solution to the equation x = p/q. We note that since gcd(p, q) = 1, we conclude that p and q both cannot be even numbers.

a(p/q)^{2} + b(p/q) + c = 0
or ap^{2} + bpq + cq^{2} = 0

Lets consider three possible cases:

1) p
is odd, q is odd: ap^{2} is odd, bpq is odd, cq^{2}
is odd, some of three odd numbers is odd, but 0 is even

2) p
is odd, q is even: ap^{2} is odd, bpq is even,
cq^{2} is even, some of one odd number and two even numbers is odd, but
0 is even

3) p
is even, q is odd: ap^{2} is even, bpq is even,
cq^{2} is odd, some of one odd number and two even numbers is odd, but
0 is even

Thus, we conclude that all possibilities lead to a contradiction, hence the assumption about the existence of a rational solution is false.

**Problem 8.**

Find lcm(24, 36).

**Solution.**

72

**Problem 9.**

Simplify: n!/(n-2)!

**Solution.**

n(n 1)

**Problem 10.**

Prove using mathematical induction that for all integers n greater than 1:

(1 1/2^{2})(1 1/3^{2})
(1
1/n^{2}) = (n + 1) / 2n

**Solution.**

Base case n = 2: left hand side = (1 1/2^{2}) = Ύ,
right hand side = (2 + 1) / (2 * 2 ) = Ύ

Assumption is that (1 1/2^{2})(1
1/3^{2})
(1 1/n^{2}) = (n + 1) / 2n

We need to show that (1 1/2^{2})(1
1/3^{2})
(1 1/n^{2}) (1 1/(n+1)^{2}) = (n + 2) /
2(n+1)

(1 1/2^{2})(1 1/3^{2})
(1
1/n^{2}) (1 1/(n+1)^{2}) = ((n + 1) / 2n) * (1 1/(n+1)^{2})
=

((n + 1) / 2n) * (n^{2} + 2n) / (n+1)^{2}) =
(n + 2) / 2(n+1) as needed

Prove any way possible that for all integers greater than 1:

sqrt(n) < 1 / sqrt(1) + 1 / sqrt(2) + + 1 / sqrt(n), where sqrt stands for the square root.

**Solution 1
(Mathematical Induction)**

Base case n = 2: left hand side = sqrt(2), right hand side = 1 + 1 /sqrt(2) = (sqrt(2) + 1) / sqrt(2) > (1 + 1) / sqrt(2) = sqrt(2) = left hand side

Assumption is that sqrt(n) < 1 / sqrt(1) + 1 / sqrt(2) + + 1/ sqrt(n)

We need to show that sqrt(n + 1) < 1 / sqrt(1) + 1 / sqrt(2) + + 1 / sqrt(n) + 1 / sqrt(n + 1)

1 / sqrt(1) + 1 / sqrt(2) + + 1 / sqrt(n) + 1 / sqrt(n + 1) > sqrt(n) + 1 / sqrt(n + 1)

Thus, it is enough for us to show that sqrt(n) + 1 / sqrt(n + 1) > sqrt(n + 1) or equivalently sqrt(n + 1) - sqrt(n) < 1 / sqrt(n + 1)

sqrt(n + 1) - sqrt(n) = (sqrt(n + 1) - sqrt(n)) * (sqrt(n + 1) + sqrt(n)) / (sqrt(n + 1) + sqrt(n)) = 1 / (sqrt(n + 1) + sqrt(n)) < 1 / sqrt(n + 1)

**Solution
2.**

Simple algebra leads to a much faster proof:

1 / sqrt(1) + 1 / sqrt(2) + + 1 / sqrt(n) > 1 / sqrt(n) + 1 / sqrt(n) + + 1 / sqrt(n) = n / sqrt(n) = sqrt(n)

**Problem 12.**

Explain why the following cannot be a valid principal of a mathematical induction:

We are given predicate P(n)

- Base case: P(0), P(1), P(2) are all true
- Given that P(k) is true, one can show that P(3k) is true as well for all positive k

Conclusion: P(n) is true for all n

**Solution.**

The proposed above is not a valid principal of mathematical induction, because there is no way to show that P(4) is true.