Real Numbers (LAQs)

LAQ-1 : Prove that √5 + √7 is an irrational number

$$\text{Let us assume that } \sqrt{5} + \sqrt{7} \text{ is a rational number.}$$

Overview:

$$\text{i.e., } \sqrt{5} + \sqrt{7} = \frac{a}{b}; \text{ where } a, b \text{ are integers and } b \neq 0$$

$$(\sqrt{5} + \sqrt{7})^2 = \left(\frac{a}{b}\right)^2$$

$$5 + 7 + 2\sqrt{5}\sqrt{7} = \frac{a^2}{b^2}$$

$$12 + 2\sqrt{35} = \frac{a^2}{b^2}$$

$$2\sqrt{35} = \frac{a^2}{b^2} – 12$$

$$2\sqrt{35} = \frac{a^2 – 12b^2}{b^2}$$

$$\text{Since } a, b \text{ are integers, } \frac{a^2 – 12b^2}{2b^2} \text{ is rational and so } \sqrt{35} \text{ is a rational number.}$$

$$\text{But this contradicts the fact } \sqrt{35} \text{ is an irrational number.}$$

$$\text{This contradiction has arisen because of our wrong assumption that } \sqrt{5} + \sqrt{7} \text{ is a rational number.}$$

$$\text{Hence, } \sqrt{5} + \sqrt{7} \text{ is an irrational number.}$$

LAQ-2 : Use division algorithm to show that the square of any positive integer is of the form 5m or 5m + 1 or 5m + 4, where ‘m’ is a whole number

$$\text{According to Euclid’s division algorithm, any positive integer can be of the form } a = bq + r, \text{ where } 0 \leq r < b$$

$$\text{In this case, } b = 5 \text{ then } r = 0, 1, 2, 3, 4$$

$$\text{Thus, any positive integer can be of the form } 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4$$

$$(5q)^2 = 5(5q^2) = 5m, \text{ where } m = 5q^2$$

$$(5q + 1)^2 = 25q^2 + 10q + 1 = 5(5q^2 + 2q) + 1 = 5m + 1, \text{ where } m = 5q^2 + 2q$$

$$(5q + 2)^2 = 5m + 2$$

$$(5q + 3)^2 = 5m + 3$$

$$(5q + 4)^2 = 5m + 4$$

$$\text{Thus, the square of any positive integer is of the form } 5m \text{ or } 5m + 1 \text{ or } 5m + 4, \text{ where ‘m’ is a whole number}$$

LAQ-3 : Show that √5 – √3 is an irrational number

$$\text{Let us assume that } \sqrt{5} – \sqrt{3} \text{ is a rational number.}$$

$$\text{i.e., } \sqrt{5} – \sqrt{3} = \frac{a}{b}; \text{ where } a, b \text{ are co-prime integers and } b \neq 0$$

$$(\sqrt{5} – \sqrt{3})^2 = \left(\frac{a}{b}\right)^2$$

$$\sqrt{5}^2 + \sqrt{3}^2 – 2\sqrt{5}\sqrt{3} = \frac{a^2}{b^2}$$

$$5 + 3 – 2\sqrt{15} = \frac{a^2}{b^2}$$

$$8 – 2\sqrt{15} = \frac{a^2}{b^2}$$

$$-2\sqrt{15} = \frac{a^2}{b^2} – 8$$

$$\sqrt{15} = \frac{8b^2 – a^2}{2b^2}$$

$$\text{Since } a \text{ and } b \text{ are integers, } \frac{8b^2 – a^2}{2b^2} \text{ is rational and so } \sqrt{15} \text{ is rational.}$$

$$\text{But, this contradicts the fact that } \sqrt{15} \text{ is irrational.}$$

$$\text{This contradiction has arisen because of our wrong assumption that } \sqrt{5} – \sqrt{3} \text{ is a rational number.}$$

$$\text{Hence, } \sqrt{5} – \sqrt{3} \text{ is an irrational number.}$$

LAQ-4 : Show that cube of any positive integer will be in the form of 8m or 8m + 1 or 8m + 3 or 8m + 5 or 8m + 7, where m is a whole number

$$\text{According to Euclid division lemma, any positive integer can be of the form } a = bq + r, \text{ where } 0 \leq r < b$$

$$\text{In this case, } b = 8 \text{ then } r = 0, 1, 2, 3, 4, 5, 6, 7. \text{ Thus, any positive integer can be of the form } 8q, 8q + 1, 8q + 2, \ldots, 8q + 7$$

$$(8q)^3 = 512q^3 = 8(64q^3) = 8m, \text{ where } m = 64q^3$$

$$(8q + 1)^3 = 512q^3 + 192q^2 + 24q + 1 = 8(64q^3 + 24q^2 + 3q) + 1 = 8m + 1, \text{ where } m = 64q^3 + 24q^2 + 3q$$

$$(8q + 2)^3 = 512q^3 + 384q^2 + 96q + 8 = 8(64q^3 + 48q^2 + 12q + 1) = 8m, \text{ where } m = 64q^3 + 48q^2 + 12q + 1$$

$$(8q + 3)^3 = 512q^3 + 576q^2 + 216q + 27 = 512q^3 + 576q^2 + 216q + 24 + 3 = 8(64q^3 + 72q^2 + 27q + 3) + 3 = 8m + 3, \text{ where } m = 64q^3 + 72q^2 + 27q + 3$$

$$(8q + 4)^3 = 8m \
(8q + 5)^3 = 8m + 5 \
(8q + 6)^3 = 8m \
(8q + 7)^3 = 8m + 7$$

$$\text{Thus, the cube of any positive integer is of the form } 8m \text{ or } 8m + 1 \text{ or } 8m + 3 \text{ or } 8m + 5 \text{ or } 8m + 7 \text{ where } m \text{ is a whole number.}$$

LAQ-5 : If x² + y² = 27xy, then show that, log(x – y/5) = 1/2 [logx + logy]

$$x^2 + y^2 = 27xy$$

$$x^2 + y^2 – 2xy = 27xy – 2xy$$

$$(x – y)^2 = 25xy$$

$$\log((x – y)^2) = \log(25xy)$$

$$2 \log(x – y) = \log 25 + \log x + \log y$$

$$2 \log(x – y) = \log 5^2 + \log x + \log y$$

$$2 \log(x – y) = 2 \log 5 + \log x + \log y$$

$$2 \log(x – y) – 2 \log 5 = \log x + \log y$$

$$2[\log(x – y) – \log 5] = \log x + \log y$$

$$2[\log\left(\frac{x – y}{5}\right)] = \log x + \log y$$

$$\log\left(\frac{x – y}{5}\right) = \frac{1}{2} (\log x + \log y)$$

LAQ-6 : Prove that √p + √q is an irrational where p, q are primes

$$\text{Let us assume that } \sqrt{p} + \sqrt{q} \text{ is a rational number.}$$

$$\text{i.e., } \sqrt{p} + \sqrt{q} = \frac{a}{b}; \text{ where } a \text{ and } b \text{ are coprimes and } a \neq 0; b \neq 0$$

$$(\sqrt{p} + \sqrt{q})^2 = \left(\frac{a}{b}\right)^2$$

$$p + q + 2\sqrt{pq} = \frac{a^2}{b^2}$$

$$2\sqrt{pq} = \frac{a^2}{b^2} – p – q$$

$$2\sqrt{pq} = \frac{a^2 – pb^2 – qb^2}{b^2}$$

$$\sqrt{pq} = \frac{a^2 – pb^2 – qb^2}{2b^2}$$

$$\text{We know that, the root of the product of different prime numbers is irrational.}$$

$$\text{Since } a^2, p, b^2, q \text{ are integers, } \frac{a^2 – pb^2 – qb^2}{2b^2} \text{ must be a rational number and so } \sqrt{pq} \text{ must be a rational number.}$$

$$\text{This contradicts the fact that } \sqrt{pq} \text{ is an irrational number. This contradiction has arisen due to our wrong assumption that } \sqrt{p} + \sqrt{q} \text{ is a rational number.}$$

$$\text{Hence, } \sqrt{p} + \sqrt{q} \text{ is an irrational number.}$$

LAQ-7 : Prove that √2 + √7 is an irrational number

$$\text{Let us assume that } \sqrt{2} + \sqrt{7} \text{ is a rational.}$$

$$\text{i.e., } \sqrt{2} + \sqrt{7} = \frac{a}{b}; \text{ where } a, b \text{ are co-prime integers and } b \neq 0$$

$$(\sqrt{2} + \sqrt{7})^2 = \left(\frac{a}{b}\right)^2$$

$$\sqrt{2}^2 + \sqrt{7}^2 + 2\sqrt{14} = \frac{a^2}{b^2}$$

$$2 + 7 + 2\sqrt{14} = \frac{a^2}{b^2}$$

$$9 + 2\sqrt{14} = \frac{a^2}{b^2}$$

$$2\sqrt{14} = \frac{a^2}{b^2} – 9$$

$$2\sqrt{14} = \frac{a^2 – 9b^2}{b^2}$$

$$\sqrt{14} = \frac{a^2 – 9b^2}{2b^2}$$

$$\text{Since } a \text{ and } b \text{ are integers, } \frac{a^2 – 9b^2}{2b^2} \text{ is rational and so } \sqrt{14} \text{ is rational.}$$

$$\text{But this contradicts the fact that } \sqrt{14} \text{ is rational. This contradiction has arisen because of our wrong assumption that } \sqrt{2} + \sqrt{7} \text{ is a rational.}$$

$$\text{Hence, } \sqrt{2} + \sqrt{7} \text{ is an irrational number.}$$

LAQ-8 : If (2.3)^x = (0.23)^y = 1000, then find the value of 1/x – 1/y

$$\text{Given } (2.3)^x = (0.23)^y = 1000$$

$$(2.3)^x = 1000$$

$$(2.3)^x = 10^3$$

$$[(2.3)^x]^{1/x} = [10^3]^{1/x} \quad \text{(1)}$$

$$(0.23)^y = 1000$$

$$(0.23)^y = 10^3$$

$$[(0.23)^y]^{1/y} = [10^3]^{1/y} \quad \text{(2)}$$

$$0.23 = \frac{2.3}{10}$$

$$10^{\frac{3}{y}} = \frac{10^{\frac{3}{x}}}{10}$$

$$10^{\frac{3}{y}} = 10^{\left(\frac{3}{x} – 1\right)}$$

$$\frac{3}{y} = \frac{3}{x} – 1$$

$$\frac{3}{y} – \frac{3}{x} = -1$$

$$\frac{1}{x} – \frac{1}{y} = \frac{1}{3}$$