5 Marks Questions

Question 15 Marks

Define irrational number and prove that $3+\sqrt[2]{5}$ is an irrational number.

Answer

Irrational numbers: are those numbers that cannot be written in form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. In other words, these are the numbers whose decimal expansion is non-terminating and non-repeating.
Let $3+\sqrt[2]{5}$ be a rational number
$\therefore$ We can find two integers $a, b(b \neq 0)$ such that
$\begin{array}{l}3+\sqrt[2]{5}=\frac{a}{b} \\\sqrt[2]{5}=\frac{a}{b}-3\end{array}$
Since $a$ and $b$ are integers, $\frac{a}{b}-3$ is also a rational number and hence $\sqrt[2]{5}$ should be rational.
This contradicts the fact that $\sqrt[2]{5}$ is irrational.
Therefore, our assumption s wrong and hence, $3+\sqrt[2]{5}$ is an irrational number.

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Question 25 Marks

Use Euclid's Division Lemma to show that the square of any positive integer is either of the form $3 m$ or $3 m+1$ for some integer m.

Answer

We know, By Euclid's Division Lemma
If a and b are two positive integers, then
$a= bq+r$ where $0 \leq r<{b} \ldots\ldots (1)$
Let $a$ be any positive integer and $b=3,$ using equation 1, we get,
$a=3q+r$ where $0\leq r <3$
We know can be either 0, 1 or 2

If r = 0	If r = 1	If r = 2
The equation becomes, $a = 3q+0$ $\Rightarrow$ a = 3q	The equation becomes, $a = 3q + 1$	The equation becomes, $a = 3q + 2$
Squaring both sides, $a^2=(3 q)^2$	Squaring both sides, $a^2=(3 q+1)^2$	Squaring both sides, $a^2=(3 q+2)^2$
$\Rightarrow$ $a^2=9 q^2$	$\Rightarrow$ $a^2=9 q^2+6 q+1$	$\Rightarrow$ $a^2=9 q^2+12 q+4$
$\Rightarrow a^2=3\left(3 q^2\right)$	$\Rightarrow a^2=3\left(3 q^2+2 q\right)+1$	$\Rightarrow a^2=9 q^2+12 q+3+1$ $\Rightarrow a^2=3\left(3 q^2+4 q+1\right)+1$
Let $m=3 q^2$	Let $m=3 q^2+2 q$	Let $m=3 q^2+4 q+1$
$\Rightarrow a^2=3 m$	$\Rightarrow a^2=3 m+1$	$\Rightarrow a^2=3 m+1$

Hence, square of any positive number can be expressed of the form $3m$ or $3m+1$ for some integer m.
Hence proved.

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Question 35 Marks

Prove that $\sqrt{5}$ is an irrational number.

Answer

Let $\sqrt{5}$ be a rational number.
Then it is of the form $\frac{a}{b}$, where a and b are co-prime.
Now, $\sqrt{5}=\frac{a}{b}$
$\Rightarrow \quad 5=\frac{a^2}{b^2}$
$\Rightarrow \quad a^2=5 b^2\ldots\ldots \text {(i)}$
$\Rightarrow \quad 5$ divides $a^2$
$\Rightarrow \quad 5$ divides $a$
Let $a=5 c$ for some integer $c$
Putting $a=5 c$ in $(i)$,
$\begin{aligned}& 5 b^2=25 c^2 \\\Rightarrow & b^2=5 c^2 \\\Rightarrow & 5 \text { divides } b^2 \\\Rightarrow & 5 \text { divides } b\end{aligned}$
Thus 5 is a common factor of a and b.
But this is not possible as $a$ and $b$ are co-primes.
$\Rightarrow$ Our assumption that $\sqrt{5}$ is rational is wrong.
$\Rightarrow \sqrt{5}$ is an irrational number.

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