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

The simplified form of the expression $\sqrt { \sqrt [ 3 ]{ 729{ x }^{ 12 } }  } -\dfrac { { x }^{ -2 }-{ x }^{ -3 } }{ { x }^{ -4 }-{ x }^{ -5 } } $ is

  1. ${ 3x }^{ 2 }$
  2. ${ 3x }^{ 3 }$
  3. ${ 2x }^{ 2 }$
  4. ${ 4x }^{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\sqrt{(729x^{12})^{\frac{1}{3}}}-\cfrac{x^{-2}-x^{-3}}{x^{-4}-x^{-5}}$


$=\sqrt{(3^6x^{12})^{\frac{1}{3}}}-\cfrac{x^{-2}-x^{-3}}{x^{-4}-x^{-5}}$


$=\sqrt{(3^2x^4)}-\cfrac{x^{-2}(1-x^{-1})}{x^{-4}(1-x^{-1})}$

$=3x^2-x^2$

$\=2x^2$

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)} $

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