Tag: existence of irrational numbers
Questions Related to existence of irrational numbers
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
$0.\overline{35}$ is equal to
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$\displaystyle\frac{35}{66}$
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$\displaystyle\frac{35}{77}$
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$\displaystyle\frac{35}{99}$
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none of these
$3.\overline{25}$ is equal to
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$\displaystyle\frac{320}{99}$
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$\displaystyle\frac{321}{99}$
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$\displaystyle\frac{322}{99}$
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$\displaystyle\frac{323}{99}$
Given that,$3.\overline{25}$.
Let,
$x=3.\overline{25}$
$x=3.252525.....$
Multiply by 100 both sides,
$ 100x=100\times 3.252525..... $
$ 100x=325.2525..... $
$ 100x=322+3.2525..... $
$ 100x=322+x $
$ 99x=322 $
$ x=\dfrac{322}{99} $
Hence, this is the answer.
$0.\overline{05}$ is equal to
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$\displaystyle\frac{3}{99}$
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$\displaystyle\frac{4}{99}$
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$\displaystyle\frac{5}{99}$
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none of these
Given that,$0.\overline{05}$
Let,
$ x=0.\overline{05} $
$ x=0.05050505..... $
Multiply by $100$ both sides,
$ 100x=100\times 0.05050505..... $
$ 100x=5.050505..... $
$ 100x=5+0.050505..... $
$ 100x=5+x $
$ 99x=5 $
$ x=\dfrac{5}{99} $
Hence, this is the answer.
Which statement is true?
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$ \displaystyle \frac{-8}{12} $= $ \displaystyle \frac{10}{-15} $
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$ \displaystyle \sqrt{3} $ is not a real number
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Additive identity of 5 is -5
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$ \displaystyle \frac{2}{5} $>$ \displaystyle \frac{4}{5} $
Irrational number is defined as
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a real number that cannot be made by dividing two integers.
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a real number that can be made by dividing two integer.
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a number that can be made derived after multiplying two integers.
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a real number that can be written as whole number.