The Number System — Irrational Numbers

1. a) $\sqrt{49} = 7$. Rational — it simplifies to a whole number, which can be written as $\frac{7}{1}$.

b) $\sqrt{10}$. Irrational — $10$ is not a perfect square, so $\sqrt{10}$ is a non-terminating, non-repeating decimal.

c) $0.333... = 0.\overline{3} = \frac{1}{3}$. Rational — it is a repeating decimal, which can be expressed as a fraction.

d) $\pi$. Irrational — $\pi$ is a non-terminating, non-repeating decimal ($3.14159...$). It cannot be expressed as a fraction of two integers.

e) $\frac{7}{11}$. Rational — it is already expressed as a fraction of two integers.

2. $\sqrt{2}$ is irrational because $2$ is not a perfect square. Its decimal expansion ($1.41421356...$) neither terminates nor repeats.

To locate it between whole numbers: $1^2 = 1$ and $2^2 = 4$. Since $1 < 2 < 4$, we have $\sqrt{1} < \sqrt{2} < \sqrt{4}$, so:

$1 < \sqrt{2} < 2$

Since $\sqrt{2} \approx 1.414$, it is closer to $1$ (distance $\approx 0.414$) than to $2$ (distance $\approx 0.586$).

3. Identify the perfect squares around $50$:

$7^2 = 49 \quad \text{and} \quad 8^2 = 64$

So $\sqrt{50}$ is between $7$ and $8$, closer to $7$ since $50$ is just above $49$.

Try $7.0^2 = 49.00$ and $7.1^2 = 50.41$.

Since $49.00 < 50 < 50.41$, we know $7.0 < \sqrt{50} < 7.1$.

Try $7.07^2 = 49.9849$ — very close to $50$.

Try $7.08^2 = 50.1264$ — just above $50$.

Since $49.9849$ is closer to $50$ than $50.1264$ is, $\sqrt{50} \approx 7.1$ (to the nearest tenth).

4. Identify the perfect squares to locate each value:

$\sqrt{3}$: $1^2 = 1$ and $2^2 = 4$, so $1 < \sqrt{3} < 2$. Since $1.7^2 = 2.89$ and $1.8^2 = 3.24$, we get $\sqrt{3} \approx 1.73$.

$\sqrt{8}$: $2^2 = 4$ and $3^2 = 9$, so $2 < \sqrt{8} < 3$. Since $2.8^2 = 7.84$ and $2.9^2 = 8.41$, we get $\sqrt{8} \approx 2.83$.

$\pi$: By definition, $\pi \approx 3.14$.

$\sqrt{17}$: $4^2 = 16$ and $5^2 = 25$, so $4 < \sqrt{17} < 5$. Since $4.1^2 = 16.81$ and $4.2^2 = 17.64$, we get $\sqrt{17} \approx 4.12$.

In order on the number line: $\sqrt{3} \approx 1.73$, $\sqrt{8} \approx 2.83$, $\pi \approx 3.14$, $\sqrt{17} \approx 4.12$.

5. The side length of a square with area $A$ is $s = \sqrt{A}$:

$s = \sqrt{75} \text{ ft}$

This is the exact side length. To approximate:

$8^2 = 64, \quad 9^2 = 81$

So $8 < \sqrt{75} < 9$. Refining:

$8.6^2 = 73.96, \quad 8.7^2 = 75.69$

So $8.6 < \sqrt{75} < 8.7$. Further:

$8.66^2 = 74.9956, \quad 8.67^2 = 75.1689$

Therefore $\sqrt{75} \approx 8.66$ ft (to the nearest hundredth).

Since $75$ is not a perfect square, $\sqrt{75}$ is irrational.