Tag: comparison of irrational numbers

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Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Let x and y be rational and irrational numbers, respectively, then x + y necessarily an irrational number.


State True or False.

  1. True

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

Yes.

Let x $= 21, y =\sqrt{2}$ be a rational number
Now $x+y=21 +\sqrt{2}=21+1.4142....=22.4142....$ , which is non-terminating and non-recurring. Hence x+y is irrational.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

The descending order of the surds $\sqrt[3]{2} , \sqrt[6]{3} , \sqrt[9]{4}$ is _________.

  1. $\sqrt[9]{4} , \sqrt[6]{3} , \sqrt[3]{2}$
  2. $\sqrt[9]{4} , \sqrt[3]{2} , \sqrt[6]{3}$
  3. $\sqrt[3]{2} , \sqrt[6]{3} , \sqrt[9]{4}$
  4. $\sqrt[6]{3} , \sqrt[9]{4} , \sqrt[3]{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\sqrt[3]{2} \approx 1.26$

$\sqrt[6]{3} \approx 1.201$
$\sqrt[9]{4} \approx 1.166$

$\therefore$ Ascending order is $\sqrt[9]{4} < \sqrt[6]{3} < \sqrt[3]{2}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Identify the irrational number(s) between $2\sqrt{3}$ and $3\sqrt{3}$

  1. $\sqrt{19}$
  2. $\sqrt{29}$
  3. $\cfrac { 4\sqrt { 3 } }{ \sqrt { 3 } } $
  4. $\sqrt{17}$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation
$2\sqrt{3}=\sqrt{12}$
$3\sqrt{3}=\sqrt{27}$
$\therefore \sqrt{176}\sqrt{19}$ are irrational no between them $\sqrt{29}$ lie out of it.
As $\dfrac{4\sqrt{3}}{\sqrt{3}}=4$ (Rational)

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Compare the following pairs of surds. $\sqrt[4]{64}, \sqrt[6]{128}$    

  1. $\sqrt[4]{64} > \sqrt[6]{128}$
  2. $\sqrt[4]{64} < \sqrt[6]{128}$
  3. $\sqrt[4]{64} \neq \sqrt[6]{128}$
  4. $\sqrt[4]{64} = \sqrt[6]{128}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\sqrt[4]{64}=\sqrt[4]{2^6}=2\sqrt[4]{2^2}=2\sqrt{2}=2\sqrt[6]{2^3}=2\sqrt[6]{8}$
$\sqrt[6]{128}=\sqrt[6]{2^7}=2\sqrt[6]{2}$
$\sqrt[4]{64}>\sqrt[6]{128}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

The smallest of $\sqrt[3]{4},    \sqrt[4]{5},     \sqrt[4]{6},    \sqrt[3]{8}$ is:

  1. $\sqrt[3]{8}$
  2. $\sqrt[4]{5}$
  3. $\sqrt[3]{4}$
  4. $\sqrt[4]{6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

(B) $\sqrt[3]{4}, \sqrt[4]{5},  \sqrt[4]{6}, \sqrt[3]{8}$

$=4^{1/3}, 5^{1/4}, 6^{1/4}, 8^{1/3}$

L.C.M of 3 & 4 $=12$

So, the given surds can be written as,

$=4^{4/12}, 5^{3/12}, 6^{3/12}, 8^{4/12}$

$=(4^{4})^{1/12}, (5^{3})^{1/12}, (6^{3})^{1/12}, (8^{4})^{1/12}$

$=(256)^{1/12}, (125)^{1/12}, (216)^{1/12}, (4096)^{1/12}$

$\therefore $ The smallest one is $\sqrt[4]{5}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

$\sqrt{11}-\sqrt{10} .... \sqrt{12}-\sqrt{11}$,use appropriate inequality to fill the gap.

  1. <

  2. >
  3. $=$

  4. cannot determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We first consider $\sqrt { 11 } -\sqrt { 10 }$ as follows:


$\sqrt { 11 } -\sqrt { 10 } =3.317-3.162=0.156$

Now we find the value of $\sqrt { 12 } -\sqrt { 11 }$ as follows:

$\sqrt { 12 } -\sqrt { 11 } =3.464-3.317=0.147$


Since $0.156>0.147$

Hence, $\sqrt { 11 } -\sqrt { 10 }>\sqrt {12} -\sqrt {11}$