5 Marks Question

Question 15 Marks

Show that every positive odd integer is of the form (4q + 1) or (4q + 3) for some integer q.

Answer

Let 'a' be a given positive od integer.
On dividing 'a' by 4, let q be the quotient and r be the remainder.
Then, by euclid's algorithm, we have
a = 4q + r, where 0 ≤ r < 4
⇒ a = 4q + r, where r = 0, 1, 2, 3
⇒ a = 4q or a = 4q + 1 or a = 4q + 2 or a = 4q + 3
But, a = 4q and a = 4q + 2 = 2(2q + 1) are clearly even.
Thus, when 'a' is odd, it is of the form:
a = (4q + 1) or (4q + 3) for some integer q.

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Question 25 Marks

Show that $\big(4+3\sqrt2\big)$ is irrational.

Answer

If possible, let $\big(4+3\sqrt2\big)$ be rational.
Then 4 and $3\sqrt2$ are rational.
$\Rightarrow4+3\sqrt2-4$ is rational $\dots(\because$ diference of two rationals is rational$)$
$\Rightarrow3\sqrt2$ is rational
$\Rightarrow\sqrt2$ is rational $\dots(\because$ 3 is rational$)$
This contradicts the fact that $\sqrt2$ is irrational.
The contradiction arises by assuming that $\big(4+3\sqrt2\big)$ is rational.
Hence, $\big(4+3\sqrt2\big)$ is irrational.

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Question 35 Marks

Prove that $\frac{2}{\sqrt7}$ is an irrational number.
HINT: $\frac{2}{\sqrt7}=\Big(\frac{2}{\sqrt7}\times\frac{\sqrt7}{\sqrt7}\Big)=\frac{2}{7}.\sqrt7$

Answer

$\frac{2}{\sqrt7}=\Big(\frac{2}{\sqrt7}\times\frac{\sqrt7}{\sqrt7}\Big)=\frac{2}{7}.\sqrt7$
Let $\frac{2}{7}\sqrt7$ is a rational number.
$\therefore\frac{2}{7}\sqrt7=\frac{\text{p}}{\text{q}},$
where p and q are some integers and
$HCF (p, q) = 1 ...(1)$
$\Rightarrow2\sqrt7\text{q}=\text{7p}$
$\Rightarrow\big(2\sqrt7\text{q}\big)^2=(\text{7p})^2$
$\Rightarrow7(4\text{q}^2)=\text{49p}^2$
$\Rightarrow\text{4q}^2=\text{7p}^2$
$\Rightarrow q^2 $is divisible by $7$
$\Rightarrow q $is divisible by $7 ...(2)$
Let$ q = 7m$, where m is some integer.
$\therefore2\sqrt7\text{q}=\text{2p}$
$\Rightarrow\Big[2\sqrt7\text{q}(7\text{m})\Big]^2=\text{7p}^2$
$\Rightarrow343(4\text{m}^2)=\text{49p}^2$
$\Rightarrow7(\text{4m}^2)=\text{p}^2$
$\Rightarrow p^2$ is divisible by $7$
$\Rightarrow p$ is divisible by $7 ...(3)$
From $(2)$ and $(3), 7$ is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.
Thus, $\frac{2}{\sqrt7}$ is irrational.

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Question 45 Marks

Prove that $\big(4-5\sqrt2\big)$ is an irrational number.

Answer

Let $\text{x}=4-5\sqrt2$ be a rational number.
$\text{x}=4-5\sqrt2$
$\Rightarrow\text{x}^2=\big(4-5\sqrt2\big)^2$
$\Rightarrow\text{x}^2=(4)^2+\big(2\sqrt3\big)^2-2(4)\big(5\sqrt2\big)$
$\Rightarrow\text{x}^2=16+50-40\sqrt2$
$\Rightarrow\text{x}^2-66=-40\sqrt2$
$\Rightarrow\frac{66-\text{x}^2}{40}=\sqrt2$
Since $x$ is a rational number, $x^2$ is also a rational number.
$\Rightarrow 66 − x^2$is a rational number
$\Rightarrow\frac{66-\text{x}^2}{40}$ is a rational number
$\Rightarrow\sqrt2$ is a rational number
But $\sqrt2$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $\big(4-5\sqrt2\big)$ is an irrational number.

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Question 55 Marks

Prove that $\big(5-2\sqrt3\big)$ is an irrational number.

Answer

Let $\text{x}=5-2\sqrt3$ be a rational number.
$\text{x}=5-2\sqrt3$
$\Rightarrow\text{x}^2=\big(5-2\sqrt3\big)^2$
$\Rightarrow\text{x}^2=(5)^2+\big(2\sqrt3\big)^2-2(5)\big(2\sqrt3\big)$
$\Rightarrow\text{x}^2=25+12-20\sqrt3$
$\Rightarrow\text{x}^2-37=-20\sqrt3$
$\Rightarrow\frac{37-\text{x}^2}{20}=\sqrt3$
Since $x$ is a rational number, $x^2$ is also a rational number.
$\Rightarrow 37 − x^2$ is a rational number
$\Rightarrow\frac{37-\text{x}^2}{20}$ is a rational number
$\Rightarrow\sqrt3$ is a rational number
But $\sqrt3$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $\big(5-2\sqrt3\big)$ is an irrational number.

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Question 65 Marks

Show that one and only one out of n, (n + 2) and (n + 4) is divisible by 3, where n is any positive integer.

Answer

on dividing 'n' by 3, let 'q' be the quotient and 'r' be the remainder.
Then, n = 3q + r, where 0 ≤ r < 3
⇒ n = 3q + r, where r = 0, 1 or 2
⇒ n = 3q or n = 3q +1 or n = 3q + 2
Case 1: If n = 3q, then 'n' is clearly divisible by 3
Case 2: If n = 3q +1, then (n + 2) = (3q + 3) = 3(q + 1), which is clearly divisible by 3
In this case, (n + 2) is divisible by 3
Case 3: If n = 3q +2, then (n + 4) = (3q + 6) = 3(q + 2), which is clearly divisible by 3
In this case, (n + 4) is divisible by 3
Hence, one and only one out of n, (n + 2) and (n + 4) is divisible by 3

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Question 75 Marks

Prove that $\sqrt3$ is an irrational number.

Answer

If possible, let $\sqrt{3}$ be rational and let its simplest form be $\frac{a}{b}$
The, a and b are integers having no common factor other than $1$ , and $b \neq 0$
Now, $\sqrt{3}=\frac{ a ^2}{b^2}$
$\Rightarrow 3=\frac{ a ^2}{b^2} \ldots$ (On squaring both sides)
$\Rightarrow 3 b^2=a^2$
$\Rightarrow 3$ divides $a ^2 \ldots\left[\because 3\right.$ divies $\left.3 b^2\right]$
$\Rightarrow 3$ divides a $\ldots\left[\because 3\right.$ is prime and 3 divides $a^2 \Rightarrow 3$ divides $\left.a\right]$
Let $a =3 c$ for some integers c .
Putting $a=3 c$ in (i), we get
$3 b^2=9 c^2$
$\Rightarrow b^2=3 c^2$
$\Rightarrow 3 \text { divides } b^2 \ldots\left[\because 3 \text { divies } 3 c^2\right]$
$\Rightarrow 3 \text { divides } b \ldots\left[\because 3 \text { is prime and } 3 \text { divides } a^2 \Rightarrow 3 \text { divides } a\right]$
Thus, $3$ is a common factor of $a$ and $b$
But, this contradicts the fact that $a$ and $b$ have no common factor other than $1$
The contradiction arises by assuming that $\sqrt{3}$ is rational.
Hence, $\sqrt{3}$ is irrational.

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Question 85 Marks

Prove that $5\sqrt2$ is an irrational number.

Answer

Let $5\sqrt2$ is a rational number.
$\therefore5\sqrt2=\frac{\text{p}}{\text{q}},$ where p and q are some integers and HCF $(p, q) = 1 ...(1)$
$\Rightarrow5\sqrt2\text{q}=\text{p}$
$\Rightarrow\big(5\sqrt2\text{q}\big)^2=\text{p}^2$
$\Rightarrow2(25\text{q}^2)=\text{p}^2$
$\Rightarrow p^2$ is divisible by $2$
$\Rightarrow p$ is divisible by $2 ...(2)$
Let $p = 2m$, where m is some integer.
$\Rightarrow5\sqrt2\text{q}=\text{2m}$
$\Rightarrow\big(5\sqrt2\text{q}\big)^2=\text{2m}^2$
$\Rightarrow2(25\text{q}^2)=\text{4m}^2$
$\Rightarrow\text{25q}^2=\text{2m}^2$
$\Rightarrow q^2$ is divisible by $2$
$\Rightarrow q$ is divisible by $2 ...(3)$
From $(2)$ and $(3)$, $2$ is a common factor of both $p$ and $q$, which contradicts $(1)$.
Hence, our assumption is wrong.
Thus, $5\sqrt2$ is irrational.

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Question 95 Marks

Prove that $\big(2\sqrt3-1\big)$ is an irrational number.

Answer

Let $\text{x}=2\sqrt3-1$ be a rational number.
$\text{x}=2\sqrt3-1$
$\Rightarrow\text{x}^2=\big(2\sqrt3-1\big)^2$
$\Rightarrow\text{x}^2=\big(2\sqrt3\big)^2+1^2-2\big(2\sqrt3\big)(1)$
$\Rightarrow\text{x}^2=12+1-4\sqrt3$
$\Rightarrow\text{x}^2-13=-4\sqrt3$
$\Rightarrow\frac{13-\text{x}^2}{4}=\sqrt3$
Since $x$ is a rational number, $x^2$ is also a rational number.
$\Rightarrow 13 − x^2$ is a rational number
$\Rightarrow\frac{13-\text{x}^2}{4}$ is a rational number
$\Rightarrow\sqrt3$ is a rational number
But $\sqrt3$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.
Thus, $\big(2\sqrt3-1\big)$ is an irrational number.

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