Tag: rational and irrational numbers
Questions Related to rational and irrational numbers
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
$3.\overline{25}$ 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.
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$\dfrac{11}{2}$
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$\sqrt{16}$
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$\sqrt{9}$
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$\sqrt{11}$
The square root of any prime number is
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rational
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irrational
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co-prime
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composite
The square root of any prime number is irrational.
Example: $\sqrt{2}$ is a irrational number.
$\dfrac {7}{9}$ is a/an _______ number.
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rational
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composite
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irrational
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prime
$\dfrac{7}{9}$ is of the form $\dfrac {p}{q}$ form , hence it is rational no.
$\sqrt {23}$ is not a ...... number.
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irrational
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co-prime
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composite
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rational
As per the theorem, the square root of any prime number is irrational. $\sqrt {23}$ is a prime number, so is not a rational number. It is irrational.
Therefore, $D$ is the correct answer.