Series of mathematical questions

Question 1:
Bob is not convinced that an integer "z" is also a rational number because an integer is typically expressed without a denominator. However, every integer can be written as a quotient of two integers, specifically as the quotient of the integer and the number 1. For example, the integer 5 can be expressed as 5/1. Therefore, every integer is a rational number.

Question 4:
To sketch the graph of y = x^2 - 7x + 12, we can start by finding the vertex of the parabola. The vertex can be found using the formula x = -b/2a, where a and b are the coefficients of the quadratic equation. In this case, a = 1 and b = -7, so x = 7/2. Substituting x = 7/2 into the equation, we get y = -1/4.

Therefore, the vertex is (7/2, -1/4), and the minimum value of the function occurs at x = 7/2, with a minimum value of -1/4.

To sketch the graph, we can plot the vertex and the x-intercepts, which can be found by factoring the quadratic equation. Factoring gives us (x - 3)(x - 4), so the x-intercepts are x = 3 and x = 4. We can also find the y-intercept by setting x = 0, which gives us y = 12.

Putting all of this information together, we can sketch the graph as a parabola that opens upward and passes through the points (3, 0), (4, 0), and (0, 12), with a vertex at (7/2, -1/4).

Question 7:
(a) To calculate the average value of the function f(x) = x^2 + x + 1 over the interval [1, 3], we need to find the definite integral of the function over that interval and divide by the length of the interval. The definite integral is:

∫[1,3] (x^2 + x + 1) dx = [x^3/3 + x^2/2 + x] from 1 to 3
= (27/3 + 9/2 + 3) - (1/3 + 1/2 + 1)
= 25

So the average value of the function is:

(1/2) * (25/(3-1)) = 25/4

(b) To prove by contradiction that sqrt(5) is irrational, we assume the opposite, which is that sqrt(5) is rational. This means that sqrt(5) can be expressed as the ratio of two integers, p and q, where q is not equal to 0 and p and q have no common factors.

So we have:

sqrt(5) = p/q

Squaring both sides gives:

5 = p^2/q^2

Multiplying both sides by q^2 gives:

5q^2 = p^2

This means that p^2 is a multiple of 5, so p must be a multiple of 5. Let p = 5k, where k is an integer. Then:

5q^2 = (5k)^2
5q^2 = 25k^2
q^2 = 5k^2

This means that q^2 is a multiple of 5, so q must also be a multiple of