1. Prove that $\sqrt{5} - 3$ is not a rational number.

Proof by contradiction:

Let's assume that $\sqrt{5} - 3$ is a rational number. Then there exists some rational number $r$ such that:

$\sqrt{5} - 3 = r$

Rearranging the equation, we have:

$\sqrt{5} = r + 3$

Now, squaring both sides:

$5 = (r + 3)^2$

Expanding the right-hand side:

$5 = r^2 + 6r + 9$

Rearranging this equation:

$r^2 + 6r + 4 = 0$

This is a quadratic equation. For $r$ to be rational, the discriminant of this quadratic equation must be a perfect square. The discriminant $\Delta$ is given by:

$\Delta = 6^2 - 4 \times 1 \times 4 = 36 - 16 = 20$

Since the discriminant $20$ is not a perfect square, this implies that $r$ cannot be a rational number. Hence, our assumption that $\sqrt{5} - 3$ is rational is incorrect.

Therefore, $\sqrt{5} - 3$ is not a rational number.

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2. Arrange the following in descending order of magnitude: $8^4, 90, 10^6$

First, we need to calculate the values of each term:

$8^4 = 8 \times 8 \times 8 \times 8 = 4096$ $90 = 90$ $10^6 = 1,000,000$

Now we can arrange them in descending order:

$10^6 = 1,000,000 > 4096 = 8^4 > 90$

Thus, the descending order is:

$10^6, 8^4, 90$

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3. Simplify the following expressions:

i. $(4 - 3)(2 + 2)$

Simplifying inside the parentheses:

$(4 - 3) = 1$ $(2 + 2) = 4$

Now, multiplying them:

$1 \times 4 = 4$

So, the simplified result is:

$4$

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ii. $(2 + 3)(3 + 5)$

Simplifying inside the parentheses:

$(2 + 3) = 5$ $(3 + 5) = 8$

Now, multiplying them:

$5 \times 8 = 40$

So, the simplified result is:

$40$

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iii. $\frac{(3 + 2)}{2 - 1}$

Simplifying inside the parentheses:

$(3 + 2) = 5$ $(2 - 1) = 1$

Now, dividing:

$\frac{5}{1} = 5$

So, the simplified result is:

$5$

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iv. $\frac{7(2 - 6) + 11}{7(2 + 11)}$

Simplifying inside the parentheses:

$(2 - 6) = -4$ $(2 + 11) = 13$

Now, simplifying the expression:

$\frac{7(-4) + 11}{7(13)} = \frac{-28 + 11}{91} = \frac{-17}{91}$

So, the simplified result is:

$\frac{-17}{91}$

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4. Rationalize the denominator of the following:

$\frac{2 \sqrt{3} + 6}{3 \sqrt{5} - 2}$

We rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is $3 \sqrt{5} + 2$:

$\frac{(2 \sqrt{3} + 6)(3 \sqrt{5} + 2)}{(3 \sqrt{5} - 2)(3 \sqrt{5} + 2)}$

Using the difference of squares in the denominator:

$(3 \sqrt{5})^2 - 2^2 = 45 - 4 = 41$

Now, expand the numerator:

$(2 \sqrt{3})(3 \sqrt{5}) + (2 \sqrt{3})(2) + 6(3 \sqrt{5}) + 6(2)$ $= 6 \sqrt{15} + 4 \sqrt{3} + 18 \sqrt{5} + 12$

Thus, the expression becomes:

$\frac{6 \sqrt{15} + 4 \sqrt{3} + 18 \sqrt{5} + 12}{41}$

This is the rationalized form of the given expression.

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Do you have any questions or would you like more details on any of these steps?

Here are 5 related questions to explore further:

1. Can you prove that $\sqrt{2}$ is irrational using a similar method?
2. How can you simplify expressions involving cube roots?
3. What is the difference between rational and irrational numbers?
4. How do you rationalize denominators with complex expressions?
5. Can you explain the significance of the discriminant in quadratic equations?

Tip: When simplifying algebraic expressions, always start by resolving parentheses and follow the order of operations (PEMDAS).