Number-Theory theory class

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Multiplication of surds

(i)

$$\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{ab}$$

if $a > 0, \ b > 0, \ n$ is a +ve rational number

(ii)

$$\frac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\frac{a}{b}}$$

(iii)

$$\sqrt[n]{\sqrt[m]{a}} = \sqrt[n]{\sqrt[m]{a}} = \sqrt[mn]{a}$$

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(i)

$$\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{ab}$$

(ii)

$$\sqrt[n]{a} \times \sqrt[n]{b} = a^{\frac{1}{mn}} \times b^{\frac{1}{mn}}$$ $$(a^m)^{\frac{1}{mn}} \times (b^n)^{\frac{1}{mn}} \Rightarrow \sqrt[mn]{a^m} \times \sqrt[mn]{b^n} \Rightarrow \sqrt[mn]{a^m b^n}$$

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Ex1:

$$\sqrt[3]{3} \times \sqrt[4]{2}$$

$$3^{\frac{1 \times 4}{3 \times 4}} \times 2^{\frac{1 \times 3}{4 \times 3}}$$ $$= (3^4)^{\frac{1}{12}} \times (2^3)^{\frac{1}{12}}$$ $$= \sqrt[12]{81} \times \sqrt[12]{8}$$ $$= \sqrt[12]{81 \times 8} = \sqrt[12]{648}$$

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Ex2:

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Q:

$$(\sqrt{10} + \sqrt{11} + \sqrt{12}) \times (\sqrt{10} + \sqrt{11} - \sqrt{12}) \times (\sqrt{10} - \sqrt{11} + \sqrt{12}) \times (\sqrt{10} - \sqrt{11} - \sqrt{12})$$

Let:

• $a = \sqrt{10} + \sqrt{11}$

• $b = \sqrt{12}$

$$[(\sqrt{10} + \sqrt{11})^2 - (\sqrt{12})^2][(\sqrt{10} - \sqrt{11})^2 - (\sqrt{12})^2]$$ $$(10 + 11 + 2\sqrt{110} - 12)(10 + 11 - 2\sqrt{110} - 12)$$ $$(9 + 2\sqrt{110})(9 - 2\sqrt{110})$$

Let:

• $a + b = 9 + 2\sqrt{110}$

• $a - b = 9 - 2\sqrt{110}$

$$= 9^2 - (2\sqrt{110})^2$$ $$=81-4\times 110=81-440=\boxed{{\displaystyle -359}}$$

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Rationalizing factor ⇒

If the product of two surds is rational number, then each of the two is R.F of each other.

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Note:

(i)
R.F of $\sqrt[n]{a} \iff \sqrt[n]{a^{n-1}}$

$$\sqrt[n]{a} \times \sqrt[n]{a^{n-1}} = \sqrt[n]{a^n} = a$$

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(ii)
R.F of $\sqrt[n]{a^m} \iff \sqrt[n]{a^{n-m}}$

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(iii)
R.F of $\sqrt{a} + \sqrt{b} \iff \sqrt{a} - \sqrt{b}$

$$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})$$ $$a - b \Rightarrow a$$

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(iv)

$$\left( \sqrt[3]{a} + \sqrt[3]{b} \right) \Rightarrow \left( \sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2} \right)$$

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(v)
R.F of $\left( \sqrt[3]{a} - \sqrt[3]{b} \right) \iff \left( \sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2} \right)$

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$$(a + b) = \left( a^{\frac{1}{3}} \right)^2 + \left( b^{\frac{1}{3}} \right)^2 = \left( a^{\frac{2}{3}} + b^{\frac{2}{3}} \right) \left( \sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2} \right)$$ $$a + b = \sqrt[3]{a + \sqrt[3]{b}} \left( \sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2} \right)$$

Ex3:

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$$x = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}}$$ $$y = \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}}$$ $$(x + y)^4 + (x^4 + y^4) = ?$$

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Solution:

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$$x + y = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}} + \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}}$$ $$= \frac{(\sqrt{7} + \sqrt{3})^2 + (\sqrt{7} - \sqrt{3})^2}{7 - 3}$$ $$= \frac{7 + 3 + 2\sqrt{21} + 7 + 3 - 2\sqrt{21}}{4} = \frac{20}{4} = 5$$ $$\boxed{x + y = 5}, \quad \boxed{xy = 1}$$

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$$x^2 + y^2 + 2(xy) = 25 \Rightarrow x^2 + y^2 = 23$$

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$$x^4 + y^4 + 2(xy)^2 = 529 \Rightarrow \boxed{x^4 + y^4 = 527}$$

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$$(x + y)^4 + (x^4 + y^4)$$ $$5^4 + 527 = 625 + 527 = \boxed{1152}$$

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Ex4:

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Q.

$$x = 7 + 4\sqrt{3}, \quad xy = 1$$ $$\frac{1}{x^2} + \frac{1}{y^2} = ?$$

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Solution:

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$$\frac{y^2 + x^2}{(xy)^2} = x^2 + y^2 \quad \text{(since } xy = 1 \text{)}$$ $$y = \frac{1}{x} = \frac{1}{7 + 4\sqrt{3}} \times \frac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} = \frac{7 - 4\sqrt{3}}{49 - 16 \times 3} = 7 - 4\sqrt{3}$$ $$x^2 + y^2 = (7 + 4\sqrt{3})^2 + (7 - 4\sqrt{3})^2$$ $$= 49 + 48 + 56\sqrt{3} + 49 + 48 - 56\sqrt{3}$$

(Since the irrational parts cancel out)

$$= 97 + 97 = \boxed{194}$$

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Ex5:

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Q.

$$3\sqrt[3]{3} \cdot \left( \sqrt[3]{\frac{4}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{1}{9}} \right)^{-1} = 1 + a$$ $$a^3 = ?$$

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Solution:

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$$3\sqrt[3]{3} \left( \frac{\sqrt[3]{4} - \sqrt[3]{2} + 1}{\sqrt[3]{9}} \right)^{-1} = 1 + a$$ $$3\sqrt[3]{3} \left( \frac{\sqrt[3]{2}^2 - \sqrt[3]{2} + 1}{\sqrt[3]{9}} \right)^{-1} = 1 + a$$ $$= \frac{3\sqrt[3]{3} \cdot \sqrt[3]{9}}{(\sqrt[3]{2})^2 - \sqrt[3]{2} + 1} \times \frac{\sqrt[3]{2} + 1}{\sqrt[3]{2} + 1}$$ $$= \frac{3\sqrt[3]{27} \cdot (\sqrt[3]{2} + 1)}{(\sqrt[3]{2})^3 + 1}$$ $$= 3 \cdot (\sqrt[3]{2} + 1) = 1 + a \Rightarrow a = \sqrt[3]{2} \Rightarrow a^3 = 2$$

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Home work:

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85. Simplify

$$\sqrt[3]{a + \frac{a + 8}{3} \sqrt[3]{\frac{a - 1}{3}}} + \sqrt[3]{a - \frac{a + 8}{3} \sqrt[3]{\frac{a - 1}{3}}}$$

Options:
(1) 1 (2) 2 (3) -2 (4) 4

Solution:

Answer: 2

Let

$$x=\sqrt[3]{\,a+\frac{a+8}{3}\sqrt{\frac{a-1}{3}}\,}+\sqrt[3]{\,a-\frac{a+8}{3}\sqrt{\frac{a-1}{3}}\,}=u+v .$$

Then

$$u^3=a+\frac{a+8}{3}\sqrt{\frac{a-1}{3}},\qquad v^3=a-\frac{a+8}{3}\sqrt{\frac{a-1}{3}}.$$

So

$$u^3+v^3=2a,\qquad uv=\sqrt[3]{a^2-\left(\frac{a+8}{3}\right)^2\!\left(\frac{a-1}{3}\right)} =\sqrt[3]{\frac{(4-a)^3}{27}}=\frac{4-a}{3}.$$

Cubing $x=u+v$:

$$x^3=u^3+v^3+3uv(u+v)=2a+(4-a)x.$$

Thus $x$ satisfies

$$x^3-(4-a)x-2a=0.$$

Or write it as: x^3  -4x + a(x - 2) = 0
For the coefficient of a to be 0, x = 2 and that also makes x^3 - 4x = 0.
So answer = 2.

86. Find the value of

$$\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}} - \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + ...}}}}$$

Options:
(1) 1 (2) 2 (3) 0 (4) cannot be found
Solution:
Let the first term be x.
Then: x = sqrt(2.x) => x^2 = 2x => x = 0,2. But x > 0 => x = 2.
Similarly for the second term y = sqrt(2 + y) => y^2 - y - 2 = 0 => y = 2,-1 but y > 0 => y = 2
So answer = 2 - 2 = 0.

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87. Simplify

$$\sqrt{2 + \sqrt{3}} + \sqrt{2 - \sqrt{3}}$$

Options:
(1) $\sqrt{2}$ (2) $\sqrt{3}$ (3) $\sqrt{5}$ (4) $\sqrt{6}$
Solution:
$$\sqrt{2+\sqrt3}+\sqrt{2-\sqrt3}=x$$

Square both sides:

$$x^2=(2+\sqrt3)+(2-\sqrt3)+2\sqrt{(2+\sqrt3)(2-\sqrt3)} =4+2\sqrt{4-3}=4+2=6.$$

Since the expression is positive, $x=\sqrt6$.

Answer: $\boxed{\sqrt6}$.

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88. Evaluate

$$\sqrt{14 + 6\sqrt{5}} - \sqrt{14 - 6\sqrt{5}}$$

Options:
(1) $\sqrt{5}$ (2) $-\sqrt{5}$ (3) $2\sqrt{5}$ (4) $3\sqrt{5}$
Solution:
$$x=\sqrt{14+6\sqrt5}-\sqrt{14-6\sqrt5}$$

Square it:

$$x^2=(14+6\sqrt5)+(14-6\sqrt5)-2\sqrt{(14+6\sqrt5)(14-6\sqrt5)} =28-2\sqrt{14^2-(6\sqrt5)^2}.$$ $$14^2-(6\sqrt5)^2=196-180=16 \Rightarrow x^2=28-2\cdot4=20.$$

Since the first radical is larger, $x>0$. Hence

$$x=\sqrt{20}=2\sqrt5.$$

Answer: $2\sqrt{5}$.

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89. Evaluate

$$\sqrt{8 + \sqrt{63}} - \sqrt{8 - \sqrt{63}}$$

Options:
(1) $\sqrt{7}$ (2) $\sqrt{14}$ (3) $\sqrt{2}$ (4) $\sqrt{28}$

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$$a=\sqrt{\,8+\sqrt{63}\,}-\sqrt{\,8-\sqrt{63}\,}$$

Square it:

$$a^2=(8+\sqrt{63})+(8-\sqrt{63})-2\sqrt{(8+\sqrt{63})(8-\sqrt{63})} =16-2\sqrt{64-63}=16-2\cdot1=14.$$

Since $\sqrt{8+\sqrt{63}}>\sqrt{8-\sqrt{63}}$, $a>0$. Thus

$$a=\sqrt{14}.$$

Answer: $\boxed{\sqrt{14}}$.