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Question
Which option is representing an irrational number?
Which option is representing an irrational number?
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Solution
The correct option is B 1⋅414213… and 3⋅14159265…
Irrational numbers are real numbers that cannot be represented as a simple fraction. It cannot be expressed in the form of a ratio, such as pq, where p and q are integers, q≠0. Also, the decimal expansion of an irrational number is neither terminating nor repeating.
Check each option one-by-one.
a. 3⋅14159265… and 12, the first number is a decimal expansion that is neither terminating nor repeating, it is an irrational number. But the second number is expressed in the form of a ratio, so it cannot be an irrational number.
b. 1⋅414213… and 3⋅14159265…, both the numbers are decimal expansions that are neither terminating nor repeating, both are irrational numbers.
c. 2.4444 and 23, the first number is decimal expansion is terminating, it cannot be an irrational number. But the second number is expressed in the form of a ratio, so it cannot be an irrational number.
d. 0 and 1⋅414213…, the first number is 0, and 0 is a rational number, since we know, a rational number can be expressed as pq, so 0 is not an irrational number. 1⋅414213…, the decimal expansion of a number is neither terminating nor repeating, it is an irrational number.
So, only the statement in option b is correct.
Irrational numbers are real numbers that cannot be represented as a simple fraction. It cannot be expressed in the form of a ratio, such as pq, where p and q are integers, q≠0. Also, the decimal expansion of an irrational number is neither terminating nor repeating.
Check each option one-by-one.
a. 3⋅14159265… and 12, the first number is a decimal expansion that is neither terminating nor repeating, it is an irrational number. But the second number is expressed in the form of a ratio, so it cannot be an irrational number.
b. 1⋅414213… and 3⋅14159265…, both the numbers are decimal expansions that are neither terminating nor repeating, both are irrational numbers.
c. 2.4444 and 23, the first number is decimal expansion is terminating, it cannot be an irrational number. But the second number is expressed in the form of a ratio, so it cannot be an irrational number.
d. 0 and 1⋅414213…, the first number is 0, and 0 is a rational number, since we know, a rational number can be expressed as pq, so 0 is not an irrational number. 1⋅414213…, the decimal expansion of a number is neither terminating nor repeating, it is an irrational number.
So, only the statement in option b is correct.
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