Tag: existence of irrational numbers

Quizzes (0) Questions (127)

Questions Related to existence of irrational numbers

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

$\pi$ is a(n) ________ while $\dfrac{22}{7}$ is rational.

  1. Integer

  2. Whole Number

  3. <p>Rational Number
    </p>

  4. <p>Irrational Number
    </p>
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The value $\dfrac{22}7$ is a rational number, as it can be expressed in the form $\dfrac pq$. 

We consider it as an approximate value of $\pi$ because $\pi$ is close to $\dfrac{22}7$. 
But actually its value is $3.14159....$, which is neither terminating nor repeating. 
Thus, $\pi $ is irrational, but $\dfrac{22}7$ is rational.

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

Check whether following statement is true or false.
$7\sqrt{5}$ is a rational number.

  1. True

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

An irrational number is any real number that cannot be expressed as a ratio a/b, where a and b are integers and b is non-zero.
$7\sqrt5$ is irrational as it can never be expressed in the form a/b

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

$3+2\sqrt{5}$ a rational number.

  1. True

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

Let's assume that $3+2\sqrt5$ is rational..... 

then 

$3+2\sqrt5 = p/q $

$\sqrt5 =( p-3q)/(2q) $ 

now take $p-3q$ to be P and $2q$ to be Q........where P and Q are integers 

which means, $\sqrt5= P/Q$...... 

But this contradicts the fact that $\sqrt5$ is rational 

So our assumption is wrong and $3+2\sqrt5$ is irrational.

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

$\sqrt { 2 } ,\sqrt { 3 }$ are

  1. Whole numbers

  2. Rational numbers

  3. Irrational numbers

  4. Integers
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A rational number is any number that can be expressed as a fraction $\dfrac pq$ of two integers with $q$ not equal to zero.
As in the case of $\sqrt2$ and $\sqrt3$, it cannot be expressed as a fraction $\dfrac pq$.

Hence, option $A$ is the correct answer.

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 irrational and if $a, b$ are two odd prime numbers, then $a^2 -b^2$ is composite. As per the above passage mark the correct answer to the following question.
$\sqrt{7}$ is:

  1. a rational number

  2. an irrational number

  3. not a real number

  4. terminating decimal
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The basic definition for a rational number is that it can be represented in the form of $p/q$, where p and q are integers and q is a non-zero integer. Here, $\sqrt7$ is not a perfect square and thus cannot be expressed in the form of $p/q$, thus it is an irrational number.

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

Consider the given statements:
I. All surds are irrational numbers.
II. All irrationals numbers are surds.
Which of the following is true.

  1. Only I

  2. Only II

  3. Both I and II

  4. Neither I nor II
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A surd, by its very definition is an irrational number.

However, not every irrational number can be expressed as a surd.
Hence, only statement 1 is true.

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

If $a\neq 1$ and $ln{ a }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 3 }+........=3\left( lna+{ \left( ln{ a } \right)  }^{ 2 }+{ \left( ln{ a } \right)  }^{ 3 }+{ \left( ln{ a } \right)  }^{ 4 }+...... \right)$ then $a$ is

  1. $an\ irrational\ number$
  2. $a\ transcendental\ number$
  3. $an\ algeberaic\ number$
  4. $a\ surd$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, for $a\ne 1$,

$ln{ a }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 2 }+{ \left( ln{ a }^{ 2 } \right)  }^{ 3 }+........=3\left( lna+{ \left( ln{ a } \right)  }^{ 2 }+{ \left( ln{ a } \right)  }^{ 3 }+{ \left( ln{ a } \right)  }^{ 4 }+...... \right)$
or, $\dfrac{\ln a^2}{1-\ln a^2}=3\times \dfrac{\ln a}{1-\ln a}$
or, $\dfrac{2\ln a}{1-2\ln a}=3\times \dfrac{\ln a}{1-\ln a}$
or, $2(1-\ln a)=3(1-2\ln a)$ [Since $a\ne 1\Rightarrow \ln a \ne 0$ ]
or, $4\ln a =1$
or, $a=\sqrt[4]{e}$.
So clearly $a$ is an irrational number.