Tag: rational numbers

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Questions Related to rational numbers

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

State whether the following statement is true or not:
$7-\sqrt { 2 } $ is irrational.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let us suppose $7-\sqrt 2$ is rational.

$=>7-\sqrt 2$ is in the form of $\dfrac pq$ where $p$ and $q$ are integers and $q\neq0$

$=>\sqrt2=-\dfrac pq+7$

​$=>\sqrt2=\dfrac{-p+7q}{q}$

as $p, q$ and $7$ are integers $\dfrac{-p+7q}{q}$ is a rational number.
$=>\sqrt 2$ is a rational number.

But we know that $\sqrt 2$ is an irrational number.

So this is a contradiction.

This contradiction has arisen because of our wrong assumption that $7-\sqrt 2$ is a rational number.

Hence, $7- \sqrt2$  is an irrational number.
Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Assuming  that x,y,z  are positive real numbers,simplify the following :


$ (\sqrt{x})^{-2/3}\sqrt{y^{4}}\div \sqrt{xy^{-1/2}} $

  1. $ \dfrac{y^{9/4}}{x^{5}} $
  2. $ \dfrac{y^{9/4}}{x^{5/6}} $
  3. $ \dfrac{y^{9/4}}{x^{-5/6}} $
  4. $ \dfrac{y^{-9/4}}{x^{5/6}} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

$(\sqrt{x})^{-\frac{2}{3}}\sqrt{y^4}\div \sqrt{xy^{-\frac{1}{2}}}$

$=\dfrac{x^{-\frac{1}{3}}y^2}{x^{\frac{1}{2}}y^{-\frac{1}{4}}}$

$=x^{-\frac{1}{3}-\frac{1}{2}}y^{2+\frac{1}{4}}$

$=x^{-\frac{5}{6}}y^{\frac{9}{4}}$

$=\dfrac{y^{\frac{9}{4}}}{x^{\frac{5}{6}}}$
Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

The multiplicative inverse of $-1 + \sqrt{2}$ is

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

Multiplicative inverse of $\sqrt2-1$ is $\dfrac{1}{\sqrt2-1}$
Now multiply and divide with $\sqrt2+1$
We get $\dfrac{\sqrt2+1}{(\sqrt2+1)(\sqrt2-1)} $

$=\dfrac {\sqrt 2+1}{(\sqrt 2)^2-1^2}$
$= \sqrt2+1$

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

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