Every irrational number is
$
(a){\text{ surd}} \\
(b){\text{ a prime number}} \\
(c){\text{ not a surd}} \\
(d){\text{ none}} \\
$
Answer
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Hint – In this question the most important thing is the basic definition of irrational number. If a number can be represented in the form of $\dfrac{p}{q},q \ne 0$ can be categorized as an irrational number. Then use basic definitions of surd, prime numbers to get the right option.
Complete step-by-step answer:
An irrational number is a real number that cannot be represented as a ratio or a simple fraction.
Or an irrational number cannot be represented in the form $\dfrac{p}{q},q \ne 0$.
By definition, a surd is an irrational root of a rational number. So we know that surds are always irrational and they are always roots.
For example:
$\sqrt 2 $ is a surd since 2 is a rational number as 2 is written as $\left( {\dfrac{2}{1}} \right)$ and $\sqrt 2 $ is irrational number as $\sqrt 2 $ cannot represented in the form $\dfrac{p}{q},q \ne 0$.
Similarly, the cube root of 9 is also a surd since 9 is rational and the cube root of 9 is irrational.
$ \Rightarrow \sqrt[3]{9}$ Is an irrational number and 9 is a rational number.
On the other hand, $\sqrt \pi $ is not a surd because $\pi $ is not a rational number it is an irrational number as $\pi $ cannot be represented in the form$\dfrac{p}{q},q \ne 0$.
Thus, to answer the question, every surd is an irrational number.
But every irrational number may or may not be a surd for example $\sqrt \pi $
We know that every prime number is integer so it can’t be irrational. So option (B) is already rejected.
So this is the required answer.
Hence option (C) is correct.
Note – The trick concept here was the knowledge of surd and prime numbers. A surd is a number left in ‘square root form’ or even in a ‘cube root form etc’. They are therefore irrational numbers, but not all of them. The reason that we leave numbers as surd is because in decimal form they can go on forever hence to represent them in a more convenient way. A prime number is one that is divisible by one and itself only.
Complete step-by-step answer:
An irrational number is a real number that cannot be represented as a ratio or a simple fraction.
Or an irrational number cannot be represented in the form $\dfrac{p}{q},q \ne 0$.
By definition, a surd is an irrational root of a rational number. So we know that surds are always irrational and they are always roots.
For example:
$\sqrt 2 $ is a surd since 2 is a rational number as 2 is written as $\left( {\dfrac{2}{1}} \right)$ and $\sqrt 2 $ is irrational number as $\sqrt 2 $ cannot represented in the form $\dfrac{p}{q},q \ne 0$.
Similarly, the cube root of 9 is also a surd since 9 is rational and the cube root of 9 is irrational.
$ \Rightarrow \sqrt[3]{9}$ Is an irrational number and 9 is a rational number.
On the other hand, $\sqrt \pi $ is not a surd because $\pi $ is not a rational number it is an irrational number as $\pi $ cannot be represented in the form$\dfrac{p}{q},q \ne 0$.
Thus, to answer the question, every surd is an irrational number.
But every irrational number may or may not be a surd for example $\sqrt \pi $
We know that every prime number is integer so it can’t be irrational. So option (B) is already rejected.
So this is the required answer.
Hence option (C) is correct.
Note – The trick concept here was the knowledge of surd and prime numbers. A surd is a number left in ‘square root form’ or even in a ‘cube root form etc’. They are therefore irrational numbers, but not all of them. The reason that we leave numbers as surd is because in decimal form they can go on forever hence to represent them in a more convenient way. A prime number is one that is divisible by one and itself only.
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