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Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Which one of the following is not true?

  1. $\sqrt{2}$ is an irrational number
  2. If a is a rational number and $\sqrt{b}$ is an irrational number then $a\sqrt{b}$ is irrational number

  3. Every surd is an irrational number

  4. The square root of every positive integer is always irrational
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

(a) All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction. An irrational number has endless non-repeating digits to the right of the decimal point. Here are some irrational numbers:


$π = 3.141592…$
$\sqrt {2} = 1.414213…$

Therefore, $\sqrt {2}$ is an irrational number.

(b) Let us take a rational number $a=\dfrac {2}{1}$ and an irrational number $b=\sqrt {2}$, then their product can be determined as:

$a\times b=2\times \sqrt { 2 } =2\sqrt { 2 }$ which is also an irrational number.

Therefore, if $a$ is a rational number and $\sqrt {b}$ is an irrational number than $a\sqrt {b}$ is an irrational number.

(c) By definition, a surd is a irrational root of a rational number. So we know that surds are always irrational and they are always roots.

For eg, $\sqrt {2}$ is a surd since $2$ is rational and $\sqrt {2}$ is irrational.

Surds are numbers left in root form $\sqrt {}$ to express its exact value. It has an infinite number of non-recurring decimals. 

Therefore, every surd is an irrational number.

(d) Let us take a positive integer $4$, now square root of $4$ will be:

$\sqrt {4}=2$ which is not an irrational number 

Hence, the square root of every positive integer is not always irrational.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Which one of the following is not true?

  1. When x is not a perfect square, $\sqrt{x}$ is an irrational number
  2. The index form of $\sqrt[m]{x^n}$ is $x^{\frac{n}{m}}$
  3. The radical form of $\left(x^{\frac{1}{n}}\right)^{\frac{1}{m}}$ is $\sqrt[m]{x^n}$

  4. Every real number is an irrational number
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(D)\,\, Real = Rational + Irrational$

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

If $ x = ( 2 + \sqrt3)^n , n \epsilon N $ and $ f = x - [x],$ then $ \dfrac {f^2}{1-f} $ is :

  1. An irrational number

  2. A non-integer rational number

  3. An odd number

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

Let x = (2 + sqrt(3))^n. Let y = (2 - sqrt(3))^n. Since 0 < 2 - sqrt(3) < 1, y is between 0 and 1. x + y is an integer, so f = x - [x] = 1 - y. Then f^2 / (1 - f) = (1 - y)^2 / y. This simplifies to an even integer.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

The product of two irrational numbers is 

  1. Always irrational

  2. Always rational

  3. Can be both rational and irrational

  4. always an integer
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $p=\sqrt 3$ and $q=\sqrt 3$ be two irrational numbers 
$pq=\sqrt 3\times \sqrt 3=3$
which is rational
Now let $p=\sqrt 3$ and $q=\sqrt 2$
$pq=\sqrt 3\times \sqrt 2=\sqrt 6$
which is an irrational number
So the product can be both rational and irrational .
Option $C$ is correct.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers
State whether the given statement is True or False :

$4-5\sqrt { 2 } $ is an irrational number.

  1. True

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

$\text{Here 4 is a rational number and }$$5\sqrt2$ is a $irrational$ number


And $\text{the difference of rational and irrational is always an irrational number}$

So that $(4-5\sqrt2)$ is an irrational number.

hence option A is correct.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers
State whether the given statement is True or False :

$5-2\sqrt { 3 } $ is an irrational number.

  1. True

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

$\text{Here 5 is a rational number and }$$2\sqrt3$ is a $irrational$ number


And $\text{the difference of rational and irrational is always an irrational number}$

So that $(5 - 2\sqrt3)$ is an irrational number.

hence option A is correct.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers
State whether the given statement is True or False :

$3+\sqrt { 2 } $ is an irrational number.

  1. True

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

$3 \text{ is a  rational number and }$$\sqrt2$ is a $irrational$ number


And $\text{the addition of rational and irrational is always an irrational number}$

So that $(3+\sqrt2)$ is an irrational number.

hence option A is correct

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

The equation $\sqrt{x+4}$- $\sqrt{x-3}$+ 1=0 has:

  1. no root

  2. one real root

  3. one real root and one imaginary root

  4. two imaginary roots

  5. two real roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Longrightarrow \sqrt { x+4 } -\sqrt { x-3 } +1=0\ \Longrightarrow \sqrt { x+4 } +1=\sqrt { x-3 } \ \Longrightarrow x+4+1+2\sqrt { x+4 } =x-3\ \Longrightarrow 2\sqrt { x+4 } =-8\ \Longrightarrow x+4=16\ \therefore x=12$

But x = 12 will not satisfy given equation.
$\therefore$ No roots for given equation.