$$ \begin{array}{l} ext { 16. Write down the values of } \ \begin{array}{llll} ext { (i) }\left|\frac{3}{4} ight| & ext { (ii) }|-0.28| & ext { (iii) }|3-11| & ext { (iv) }\left|\frac{5}{7}-\frac{3}{8} ight|\end{array} \ ext { (v) }\left|\frac{|-2|-3}{|-6|-|-8|} ight|\end{array} $$ 17. Given that $$ \sqrt{2} $$ and $$ \sqrt{3} $$ are irrational numbers, show that the following are irrational $$ ] $$ numbers: $$ \begin{array}{llll} ext { (i) } \frac{\sqrt{2}}{2} & ext { (ii) } \sqrt{2}+\sqrt{3} & ext { (iii) } \sqrt{2}+1 & ext { (iv) } \sqrt{3}+3\end{array} $$

Question

$$ \begin{array}{l}	ext { 16. Write down the values of } \ \begin{array}{llll}	ext { (i) }\left|\frac{3}{4}ight| & 	ext { (ii) }|-0.28| & 	ext { (iii) }|3-11| & 	ext { (iv) }\left|\frac{5}{7}-\frac{3}{8}ight|\end{array} \ 	ext { (v) }\left|\frac{|-2|-3}{|-6|-|-8|}ight|\end{array} $$ 17. Given that $$ \sqrt{2} $$ and $$ \sqrt{3} $$ are irrational numbers, show that the following are irrational $$ ] $$ numbers: $$ \begin{array}{llll}	ext { (i) } \frac{\sqrt{2}}{2} & 	ext { (ii) } \sqrt{2}+\sqrt{3} & 	ext { (iii) } \sqrt{2}+1 & 	ext { (iv) } \sqrt{3}+3\end{array} $$

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### Problem 16: Absolute Values

We need to find the absolute values for the following expressions:

1. $$ \left|\frac{3}{4}\right| $$
2. $$ |-0.28| $$
3. $$ |3-11| $$
4. $$ \left|\frac{5}{7}-\frac{3}{8}\right| $$
5. $$ \left|\frac{|-2|-3}{|-6|-|-8|}\right| $$

#### (i) $$ \left|\frac{3}{4}\right| $$

The absolute value of a positive number is the number itself:
$$
\left|\frac{3}{4}\right| = \frac{3}{4}
$$

#### (ii) $$ |-0.28| $$

The absolute value of a negative number is its positive counterpart:
$$
|-0.28| = 0.28
$$

#### (iii) $$ |3-11| $$

First, calculate $$ 3 - 11 $$:
$$
3 - 11 = -8
$$
Now, take the absolute value:
$$
|-8| = 8
$$

#### (iv) $$ \left|\frac{5}{7}-\frac{3}{8}\right| $$

First, we need to find a common denominator to subtract the fractions. The least common multiple of 7 and 8 is 56.

Convert each fraction:
$$
\frac{5}{7} = \frac{5 \times 8}{7 \times 8} = \frac{40}{56}
$$
$$
\frac{3}{8} = \frac{3 \times 7}{8 \times 7} = \frac{21}{56}
$$

Now, subtract:
$$
\frac{40}{56} - \frac{21}{56} = \frac{40 - 21}{56} = \frac{19}{56}
$$

Taking the absolute value:
$$
\left|\frac{19}{56}\right| = \frac{19}{56}
$$

#### (v) $$ \left|\frac{|-2|-3}{|-6|-|-8|}\right| $$

Calculate the absolute values:
$$
|-2| = 2 \quad \text{and} \quad |-6| = 6, \quad |-8| = 8
$$

Now substitute:
$$
\left|\frac{2-3}{6-8}\right| = \left|\frac{-1}{-2}\right| = \left|\frac{1}{2}\right| = \frac{1}{2}
$$

### Summary of Absolute Values
1. $$ \left|\frac{3}{4}\right| = \frac{3}{4} $$
2. $$ |-0.28| = 0.28 $$
3. $$ |3-11| = 8 $$
4. $$ \left|\frac{5}{7}-\frac{3}{8}\right| = \frac{19}{56} $$
5. $$ \left|\frac{|-2|-3}{|-6|-|-8|}\right| = \frac{1}{2} $$

---

### Problem 17: Showing Irrationality

We need to show that the following numbers are irrational:

1. $$ \frac{\sqrt{2}}{2} $$
2. $$ \sqrt{2}+\sqrt{3} $$
3. $$ \sqrt{2}+1 $$
4. $$ \sqrt{3}+3 $$

#### (i) $$ \frac{\sqrt{2}}{2} $$

Assume $$ \frac{\sqrt{2}}{2} $$ is rational, then it can be expressed as $$ \frac{a}{b} $$ where $$ a $$ and $$ b $$ are integers. This implies:
$$
\sqrt{2} = \frac{2a}{b}
$$
Squaring both sides gives:
$$
2 = \frac{4a^2}{b^2} \implies 2b^2 = 4a^2 \implies b^2 = 2a^2
$$
This means $$ b^2 $$ is even, hence $$ b $$ is even. Let $$ b = 2k $$:
$$
(2k)^2 = 2a^2 \implies 4k^2 = 2a^2 \implies 2k^2 = a^2
$$
Thus, $$ a^2 $$ is even, so $$ a $$ is even. This contradicts the assumption that $$ a $$ and $$ b $$ are coprime. Therefore, $$ \frac{\sqrt{2}}{2} $$ is irrational.

#### (ii) $$ \sqrt{2}+\sqrt{3} $$

Assume $$ \sqrt{2}+\sqrt{3} $$ is rational, then:
$$
\sqrt{2} = r - \sqrt{3} \quad (r \text{ is rational})
$$
Squaring both sides:
$$
2 = r^2 - 2r\sqrt{3} + 3 \implies 2r\sqrt{3} = r^2 + 1 \implies \sqrt{3} = \frac{r^2 + 1}{2r}
$$
Since $$ r $$ is rational, $$ \sqrt{3} $$ must be rational, which is a contradiction. Thus, $$ \sqrt{2}+\sqrt{3} $$ is irrational.

#### (iii) $$ \sqrt{2}+1 $$

Assume $$ \sqrt{2}+1 $$ is rational:
$$
\sqrt{2} = r - 1
$$
Squaring gives:
$$
2 = r^2 - 2r + 1 \implies r^2 - 2r - 1 = 0
$$
The discriminant $$ (-2)^2 - 4(1)(-1) = 4 + 4 = 8 $$ is not a perfect square, hence $$ r $$ is irrational. Therefore, $$ \sqrt{2}+1 $$ is irrational.

#### (iv) $$ \sqrt{3}+3 $$

Assume $$ \sqrt{3}+3 $$ is rational:
$$
\sqrt{3} = r - 3
$$
Squaring gives:
$$
3 = r^2 - 6r + 9 \implies r^2 - 6r + 6 = 0
$$
The discriminant $$ (-6)^2 - 4(1)(6) = 36 - 24 = 12 $$ is not a perfect square, hence $$ r $$ is irrational. Therefore, $$ \sqrt{3}+3 $$ is irrational.

### Summary of Irrational Numbers
1. $$ \frac{\sqrt{2}}{2} $$ is irrational.

1. $$ \left|\frac{3}{4}\right| = \frac{3}{4} $$ 2. $$ |-0.28| = 0.28 $$ 3. $$ |3-11| = 8 $$ 4. $$ \left|\frac{5}{7}-\frac{3}{8}\right| = \frac{19}{56} $$ 5. $$ \left|\frac{|-2|-3}{|-6|-|-8|}\right| = \frac{1}{2} $$ --- 1. $$ \frac{\sqrt{2}}{2} $$ is irrational. 2. $$ \sqrt{2}+\sqrt{3} $$ is irrational. 3. $$ \sqrt{2}+1 $$ is irrational. 4. $$ \sqrt{3}+3 $$ is irrational.

17. Given that and are irrational numbers, show that the following are irrational numbers:

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Solución

Problem 16: Absolute Values

(i)

(ii)

(iii)

(iv)

(v)

Summary of Absolute Values

Problem 17: Showing Irrationality

(i)

(ii)

(iii)

(iv)

Summary of Irrational Numbers

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