Tag: irrational numbers
Questions Related to irrational numbers
State True or False.
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True
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False
$\ { (\sqrt { 5 } -2) }\ \sqrt { 5 } =2.2360679775........\ \ \sqrt { 5 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Subtraction\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\sqrt { 5 } -2) }\quad is\quad an\quad irrational\quad number.\ \quad $
State True or False.
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True
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False
$\ { \frac { -2 }{ 5 } \sqrt { 8 } }={ \frac { -4 }{ 5 } \sqrt { 2 } }=-0.8*\sqrt { 2 } \ \sqrt { 2 } =1.41421356237........\ \ \sqrt { 2 } is\quad an\quad irrational\quad number,\quad since\quad its\quad decimal\quad representaion\quad is\quad non\quad terminating\quad non\quad repeating.\ Multiplication\quad of\quad rational\quad with\quad irrational\quad is\quad irrational.\ Hence,\quad { (\frac { -2 }{ 5 } \sqrt { 8 } ) }\quad is\quad an\quad irrational\quad number.\ \quad $
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True
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False
$\displaystyle \frac { (2+\sqrt { 2 } )(3-\sqrt { 5 } ) }{ (3+\sqrt { 5 } )(2-\sqrt { 2 } ) } =\frac { { (2+\sqrt { 2 } ) }^{ 2 }{ (3-\sqrt { 5 } ) }^{ 2 } }{ (9-5)(4-2) } =\frac { (4+2+4\sqrt { 2 } )(9+5-6\sqrt { 5 } ) }{ 8 } \$
The given statement is false.
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True
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False
${ (2+\sqrt { 3 } ) }({ 2-\sqrt { 3 } ) }=4-3=1\ Hnece\quad rational.\ $
State TRUE or FALSE
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True
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False
${ (2+\sqrt { 3 } ) }^{ 2 }=4+3+4\sqrt { 3 } =7+4\sqrt { 3 } \ The\quad above\quad given\quad expression\quad consists\quad of\quad an\quad algebric\quad equation\quad \quad \ consisting\quad of\quad irrational\quad terms,\quad hence\quad it\quad is\quad an\quad irrational\quad expression.\ $
Value of $\pi$ is equal to (approximately)
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$3.41$
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$3.14$
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$\displaystyle \frac{23}{7}$
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$\displaystyle \frac{21}{7}$
$\displaystyle 3.14=\frac{22}{7}=\pi(pi)$
$\sqrt3$ is
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rational number
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irrational number
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natural number
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None
Let $\sqrt3$ is a rational number
$\therefore \sqrt3 = \displaystyle \frac{a}{b}$ [Where a & b are co-primes]
$a^2=3b^2$ .......(i)
$\Rightarrow$ 3 divides $a^2$
$\Rightarrow$ 3 also divides a
$\Rightarrow$ a=3c
[Where c is any non-zero positive integer]
$\Rightarrow a^2 = 9c^2$
From equation (i)
$3b^2=9c^2$
$\Rightarrow b^2 = 3c^2 \Rightarrow$ 3 divides $b^2$
$\Rightarrow$ 3 also divides b
So, 3 is a common factor of a and b.
Our assumption is wrong, because a and b are not co - primes.
It means $\sqrt3$ is an irrational number.
$\sqrt2 + \sqrt3$ is
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irrational
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rational
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natural
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None
$\cfrac{m}{n} = \sqrt{2} + \sqrt{3} $
Square both sides:
$\cfrac{m^2}{ n^2} = 5 + 2\sqrt{6} $
$\sqrt{6} = \cfrac{\left(m^{2} - 5n^{2}\right)}{\left(2n^{2}\right)} $
so if $\sqrt{2} + \sqrt{3} $ is rational, then so is $ \sqrt{6}$
Let a and b be the integers with gcd(a,b) = 1 such that
$\cfrac{a}{b} = \sqrt{6}$
Square both sides and multiply by $b^2$:
$a^2 = 6b^2 $
then implies that a is divisible by 2 (since 2 is prime).
Therefore we can write a=2k for some integer k:
$4k^{2} = \left(2k \right)^{2} = 6b^{2} $
Divide by 2:
$2k^{2} = 3b^{2} $
Now the left side is divisible by 2, so $3b^{2}$ is divisible by 2, from which it follows that b is divisible by 2.
However, this would mean that 2 divides gcd(a,b) = 1. Contradiction.
$\therefore \sqrt{6} $ is irrational
, and $\therefore \sqrt{2} + \sqrt{3} $ is also irrational.
Every surd is
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a natural number
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an irrational number
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a whole number
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a rational number
Which of the following is irrational?
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$\displaystyle\frac{1}{3}$
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$\displaystyle\frac{48}{5}$
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$0.7777\dots$
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$1.73202002\dots$
$1.73202002$ is the irrational number because it can not be expressed as a fraction