Tag: properties of irrational numbers

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Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

Simplify by combining similar terms :$\displaystyle 3\sqrt{147}-\frac{7}{3}\sqrt{\frac{1}{3}}+7\sqrt{\frac{1}{3}}$

  1. $\displaystyle \frac{189}{3\sqrt{3}}$
  2. $\displaystyle \frac{175}{3\sqrt{3}}$
  3. $\displaystyle \frac{208\sqrt{3}}{3}$
  4. $\displaystyle \frac{203}{3\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Simplify the expression: 3*sqrt(147) = 3*sqrt(49*3) = 21*sqrt(3). The other terms are -(7/3)(1/sqrt(3)) + 7(1/sqrt(3)) = (14/3)*(1/sqrt(3)) = 14/(3*sqrt(3)). Combining these requires a common denominator. The result is (21*3*sqrt(3) + 14)/(3*sqrt(3)) = (189+14)/(3*sqrt(3)) = 203/(3*sqrt(3)).

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

$\sqrt {5}$ is a\an ......... number.

  1. rational

  2. whole

  3. integer

  4. irrational
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\sqrt {5} = \dfrac {a}{b}$

$b\sqrt {5} = a$ $(a$ and $b$ are co-prime i.e. they have no common factors$) ...(1)$ 
$5b^{2} = a^{2}$ (squaring both sides)
Therefore $5$ divides $a^{2}$
As per Fundamental Theorem of Arithmetic, $5$ divides $a.$
Let's take it as $a = 5c,$
$5b^{2} = 25 c^{2}$
$b^{2} = 5c^{2}$
As per Fundamental Theorem of Arithmetic, $5$ divides $a.$
So $a$ and $b$ have $5$ as a common factor but $a$ and $b$ have only $1$ common factor $1$ from equation $(1),$ so it is not rational.
So, we conclude that $\sqrt {5}$ is irrational.
Therefore, $D$ is the correct answer.

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

$\sqrt{21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}}=$...........

  1. $\sqrt{5}-2+2\sqrt{3}$
  2. $\sqrt{5}-\sqrt{4}-\sqrt{12}$
  3. $-\sqrt{5}+\sqrt{4}+\sqrt{12}$
  4. $-\sqrt{5}-\sqrt{4}+\sqrt{12}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression is sqrt(21 - 4*sqrt(5) + 8*sqrt(3) - 4*sqrt(15)). This is of the form sqrt((a+b+c)^2) = |a+b+c|. Expanding (sqrt(5) - 2 - 2*sqrt(3))^2 gives 5 + 4 + 12 - 4*sqrt(5) - 4*sqrt(15) + 8*sqrt(3) = 21 - 4*sqrt(5) + 8*sqrt(3) - 4*sqrt(15). Thus the square root is |sqrt(5) - 2 - 2*sqrt(3)|, which equals -sqrt(5) + 2 + 2*sqrt(3).

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

State whether the following statements are true or false. 
$\sqrt {n}$ is not irrational if n is a perfect square

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

False ,

$\sqrt{4}=2$ where 2 is a rational number.Here n is perfect square the  $\sqrt{n}$ is rational number 
$\sqrt{5}=2.236..$ is not rational  number But it is irrational number . here n is not a perfect square the  $\sqrt{n}$ is  irrational  number
So $\sqrt{n}$ is not irrational number if n is perfect square

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

If $p$ is prime, then $\sqrt {p}$ is:

  1. Composite number

  2. Rational number

  3. Positive integer

  4. Irrational number
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

SInce, we know that prime numbers are those which are never perfect square and not divisible by any other number except by itself.
which are $2,3,5,7,...$
Clearly, if $p$ is prime then $\sqrt p $ is irrational number.
Option $D$ is correct. 

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

$6+\sqrt{2}$ is a rational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let's assume that $6+\sqrt2$ is rational..... 

then 

$6+\sqrt2 = p/q $

$\sqrt2 =( p-6q)/(q) $ 

now take $p-6q$ to be P and $q$ to be Q........where P and Q are integers 

which means, $\sqrt2= P/Q$...... 

But this contradicts the fact that $\sqrt2$ is rational 

So our assumption is wrong and $6+\sqrt2$ is irrational.