3 Marks Question

Question 13 Marks

Prove that $\frac{1}{{\sqrt 2 }}$ is irrational.

Answer

We can prove $\frac{1}{{\sqrt 2 }}$ irrational by contradiction.
Let us suppose that $\frac{1}{{\sqrt 2 }}$is rational.
It means we have some co-prime integers $a$ and $b (b ≠ 0)$
Such that
$\frac{1}{{\sqrt 2 }}$= $\frac ab$
$\Rightarrow \sqrt 2 = \frac{b}{a}..........(1)$
$R.H.S$ of $(1)$ is rational but we know that is$\sqrt 2 $ irrational.
It is not possible which means our supposition is wrong.
Therefore,$\frac{1}{{\sqrt 2 }}$can not be rational.
Hence, it is irrational.

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Question 23 Marks

Prove that $3+2\sqrt { 5 }$ is irrational.

Answer

Let us assume, to the contrary, that is $3 + 2 \sqrt { 5 }$ rational.
That is, we can find coprime integers $a$ and $b ( b \neq 0 )$ such that
$3 + 2 \sqrt { 5 } = \frac { a } { b } \text { Therefore, } \frac { a } { b } - 3 = 2 \sqrt { 5 }$
$\Rightarrow \frac { a - 3 b } { b } = 2 \sqrt { 5 }$
$\Rightarrow \frac { a - 3 b } { 2 b } = \sqrt { 5 } \Rightarrow \frac { a } { 2 b } - \frac { 3 } { 2 }$
Since a and b are integers,
We get $\frac { a } { 2 b } - \frac { 3 } { 2 }$ is rational, also so $\sqrt { 5 }$ is rational.
But this contradicts the fact that $\sqrt { 5 }$ is irrational.
This contradiction arose because of our incorrect
assumption that $3 + 2 \sqrt { 5 }$ is rational.
So, we conclude that $3 + 2 \sqrt { 5 }$ is irrational.

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Question 33 Marks

Prove that $\sqrt 5 $ is irrational.

Answer

Let us prove $\sqrt 5 $ irrational by contradiction.
Let us suppose that $\sqrt 5 $ is rational. It means that we have co-prime integers a and b $(b ≠ 0)$
Such that $\sqrt 5 = \frac{a}{b}$
$\Rightarrow $ b $\sqrt 5 $=a
Squaring both sides, we get
$\Rightarrow 5 b^2=a^2 \ldots(1)$
It means that $5$ is factor of $a^2$
Hence, $5$ is also factor of a by Theorem. $... (2)$
If, $5$ is factor of $a ,$ it means that we can write $a = 5c$ for some integer $c .$
Substituting value of $a$ in $(1) ,$
$5b^2= 25c^2$
$⇒ b^2=5c^2$
It means that $5$ is factor of $b^2$ .
Hence, $5$ is also factor of b by Theorem. $... (3)$
From $(2)$ and $(3) $, we can say that $5$ is factor of both $a$ and $b .$
But, $a$ and $b$ are co-prime .
Therefore, our assumption was wrong. $\sqrt 5 $ cannot be rational. Hence, it is irrational.

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Question 43 Marks

There is a circular path around a sports field. Sonia takes $18$ minutes to drive one round of the field, while Ravi takes $12$ minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?

Answer

By taking $LCM$ of time taken $($in minutes$)$ by Sonia and Ravi, We can get the actual number of minutes after which they meet again at the starting point after both start at the same point and at the same time, and go in the same direction.
$18 = 2 \times 3 \times 3 = 2 \times 3 ^ { 2 }$

$12 = 2 \times 2 \times 3 = 2^2 \times 3$

$LCM ( 18,12 ) = 2 ^ { 2 } \times 3 ^ { 2 } = 36$
Therefore, both Sonia and Ravi will meet again at the starting point after $36$ minutes.

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Question 53 Marks

Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers.

Answer

Numbers are of two types $-$ prime and composite.
Prime numbers can be divided by $1$ and only itself, whereas composite numbers have factors other than $1$ and itself.
It can be observed that
$7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)$
$= 13 × (77 + 1)= 13 × 78= 13 ×13 × 6$
The given expression has $6$ and $13$ as its factors.
Therefore, it is a composite number.
$7 × 6 × 5 × 4 × 3 × 2 × 1 + 5$
$= 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)$
$= 5 × (1008 + 1)= 5 ×1009$
$1009$ cannot be factorized further
Therefore, the given expression has $5$ and $1009$ as its factors.
Hence, it is a composite number.

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Question 63 Marks

Show that $5-\sqrt{3}$ is irrational.

Answer

Let us assume, to the contrary, that $5-\sqrt{3}$ is rational.
That is, we can find coprime numbers $a$ and $b\ (b ≠ 0)$ such that $5-\sqrt{3}=\frac{a}{b}$
Therefore, $5-\frac{a}{b}=\sqrt{3}$
Rearranging this equation, we get $\sqrt{3}=5-\frac{a}{b}=\frac{5 b-a}{b}$
Since $a$ and $b$ are integers, we get $5-\frac{a}{b}$ is rational, and so $\sqrt{3}$ is rational.
But this contradicts the fact that $\sqrt{3}$ is irrational
This contradiction has arisen because of our incorrect assumption that $5-\sqrt{3}$ is rational.
So, we conclude that $5-\sqrt{3}$ is irrational.

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Question 73 Marks

Prove that $\sqrt { 3 }$ is irrational.

Answer

Let us assume $\sqrt3$ be a rational, then as every rational can be represented in the form $p/q$ where $q\neq 0$
Let $\sqrt3=p/q$ where $p,q$ have no common factor.
Now squaring on both sides we get $3=p^2 / q^2 \Rightarrow 3 \times q^2=p^2$
Which means $3$ divides $p^2$ which implies $3$ divides $p$
Hence we can write $p=3\times k,$ where $k$ is some constant.
This gives $3 \times q^2=9 \times k^2 q^2=3 \times k^2$
Which means $3$ divides $q^2$ which implies $3$ divides $q.$
$3$ divides $p$ and $q$ which means $3$ is a common factor for $p$ and $q.$
And this is a contradiction for our assumption that $p$ and $q$ have no common factor…
Hence we can say our assumption that $\sqrt3$ is rational is wrong… And
therefore $\sqrt3$ is an irrational…

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