Exercise 1.4

1. \(2-\sqrt{5}\)

   Solution:

   Please note: if Irrational Number is added or subtracted from a Rational Number result will be Irrational Number

   Here in this problem:

   \(\Rightarrow\text{2 is a rational number}~\\\&~\sqrt{5}\text{ is Irrational} \\ \therefore\ \color{blue}{2-\sqrt{5}\ \text{is Irrational}}\)
2. \((3+\sqrt{23})-\sqrt{23}\)

   Solution:

   \[ \require{cancel} (3+\sqrt{23})-\sqrt{23}\\ =3+{(\sqrt{23}-\sqrt{23})}\\ =3+\cancelto{0}{(\sqrt{23}-\sqrt{23})}\\ =3 \\ \Rightarrow\color{blue}{\text{Rational Number}} \]
3. \(\dfrac{2\sqrt{7}}{7\sqrt{7}}\)

   Solution:

   \[ \require{cancel} \dfrac{2\sqrt{7}}{7\sqrt{7}} = \dfrac{2\cancel{\sqrt{7}}}{7\cancel{\sqrt{7}}} = \dfrac{2}{7} \Rightarrow \dfrac{p}{q} \mid q \neq 0\\ \Rightarrow \color{blue}{\text{Rational Number}} \]
4. \(\dfrac{1}{\sqrt{2}}\)

   Solution:

   \(\sqrt{2}\) is Irrational

   Note: If any rational number is divided or multiplied by an Irrational number result is Irrational Number

   Hence, \(\dfrac{1}{\sqrt{2}}\Rightarrow\) Irrational Number

5. \(2\pi\)

   Solution:

   \(\pi\) is Irrational

   Note: If any rational number is divided or multiplied by an Irrational number result is Irrational Number

   Hence, \(2\pi\Rightarrow\) Irrational Number

1. \((3 + \sqrt{3})(2 + \sqrt{2})\)

   Solution:

   \[ (3 + \sqrt{3})(2 + \sqrt{2}) \\ = 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{3}\sqrt{2} \\ = 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6} \]
2. \((3 + \sqrt{3})(3 - \sqrt{3})\)

   Solution:

   \[ (3 + \sqrt{3})(3 - \sqrt{3}) \\ = 9 - (\sqrt{3})^2 \\ = 9 - 3 \\ = 6 \]
3. \((\sqrt{5} + \sqrt{2})^2\)

   Solution:

   \[ (a+b)^2 = a^2 + 2ab + b^2\\ (\sqrt{5} + \sqrt{2})^2 \\ = (\sqrt{5})^2 + 2\sqrt{5}\sqrt{2} + (\sqrt{2})^2 \\ = 5 + 2\sqrt{10} + 2 \\ = 7 + 2\sqrt{10} \]
4. \((\sqrt{5} - \sqrt{2})^2 (\sqrt{5} + \sqrt{2})^2\)

   Solution:

   \[ ((\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}))^2 \\ = (\sqrt{5})^2 - (\sqrt{2})^2 \\ = 5 - 2 \\= 3 \]

Solution:

There is no contradiction. \(\pi\) is defined as the ratio of circumference to diameter, and that ratio simply happens to be an irrational real number.

Rational vs. ratio of lengths

• “Rational” means a number that can be written as \(\dfrac{p}{q}\), with both integers, \(q \ne 0\).
• But any “ratio of lengths” can be rational or irrational, it's just a real number.
• For example, the ratio of the diagonal to the side of a square is \(\sqrt{2}\), which is irrational.

No contradiction

• The confusion comes from mixing up “ratio” with “rational.”
• Every rational number is a ratio, but not every ratio is rational.
• \(\pi\) is a ratio, which happens to be irrational.

Represent \(\sqrt{9.3}\) on the number line:

Solution:

1. Draw a number line. Mark point O as 0.
2. Mark point A at 9.3 units from O.
3. Mark point B at 1 unit beyond A (OB = 10.3)
4. Find the midpoint M of OB (midpoint is at \(\frac{10.3}{2} = 5.15\))
5. With M as center and radius MB = 5.15, draw a semicircle on OB
6. At point A (9.3 units), draw a perpendicular that intersects the semicircle at P
7. The length AP equals \(\sqrt{9.3}\)
8. Transfer AP onto the number line starting from O; mark point C so OC = AP
9. Point C represents \(\sqrt{9.3}\)

Rationalise the denominators of the following:

1. \(\dfrac{1}{\sqrt{7}}\)

   Solution:

   Multiply numerator & denominator by \(\sqrt{7}\): \[ \dfrac{1}{\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{\sqrt{7}}{7} \]

2. \(\dfrac{1}{\sqrt{7}-\sqrt{6}}\)

   Solution:

   Multiply numerator & denominator by \(\sqrt{7}+\sqrt{6}\): \[ \frac{1}{\sqrt{7}-\sqrt{6}}\times\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}} \\\\ = \frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} \\\\ = \frac{\sqrt{7}+\sqrt{6}}{7-6}\\ \\= \sqrt{7}+\sqrt{6} \]

3. \(\dfrac{1}{\sqrt{5}+\sqrt{2}}\)

   Solution:

   Multiply numerator & denominator by \(\sqrt{5}-\sqrt{2}\): \[ \dfrac{1}{\sqrt{5}+\sqrt{2}} \times \dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\\\ = \dfrac{\sqrt{5}-\sqrt{2}}{(\sqrt{5})^2-(\sqrt{2})^2} \\\\ = \dfrac{\sqrt{5}-\sqrt{2}}{5-2} \\\\= \dfrac{\sqrt{5}-\sqrt{2}}{3} \]

4. \(\dfrac{1}{\sqrt{7}-2}\)

   Solution:

   Multiply numerator & denominator by \(\sqrt{7}+2\): \[ \frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2} \\\\ = \frac{\sqrt{7}+2}{(\sqrt{7})^2 - (2)^2} \\\\ = \frac{\sqrt{7}+2}{7-4} \\= \frac{\sqrt{7}+2}{3} \]