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# How do you simplify $\sqrt {504}$?

Last updated date: 13th Jun 2024
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Hint: In this question, we used continued fraction expansion. And fraction is that in mathematics, a fraction is an expression obtained through an iterative process of representing verity because the sum of its integer part and therefore the reciprocal of another number, then writing this other number because the sum of its integer part and another reciprocal, and so on.
${a_0} + \dfrac{1}{{{a_1} + \dfrac{1}{{{a_2} + \dfrac{1}{{}}}}}} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;. \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;. \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;. + \dfrac{1}{{{a_n}}} \\$
It is a finite continued fraction, where n is a non-negative integer, ${a_0}$ is an integer, and ${a_i}$ is a positive integer, for $i = 1,.............n$.
It is generally assumed that the numerator of the entire fraction is $1$. If the arbitrary values and functions are utilized in place of one or more of the numerator or the integer in the denominators, the resulting expression may be a generalized continued fraction.

Complete step by step answer:
The number is$504$.
The factor of $504 = 2 \times 2 \times 2 \times 3 \times 3 \times 7$ has no perfect square factors, so $\sqrt {504}$ can’t be simplified.
Then,
It is an irrational approximation; I will find a continued fraction expansion for $\sqrt {504}$ then truncate it.
To find the simple continued fraction expansion of $\sqrt n$, we use the following algorithm.
${m_0} = 0 \\ {d_0} = 1 \\ {a_0} = \sqrt n \\ {m_{i + 1}} = {d_i}{a_i} - {m_i} \\ {d_{i + 1}} = \dfrac{{n - {m^2}_{i + 1}}}{{{d_i}}} \\ {a_{i + 1}} = \dfrac{{{a_0} + {m_{i + 1}}}}{{{d_{i + 1}}}} \\$
This algorithm stops when ${a_i} = 2{a_0}$, making the end of the repeating part of the continued fraction.
Then, the continued fraction expansion is.
$\left[ {{a_0};\;{a_1},\;{a_2},\;{a_3}........} \right] = {a_0} + \dfrac{1}{{{a_1} + \dfrac{1}{{{a_2} + \dfrac{1}{{{a_3} + ........}}}}}}$
Next, in the question the value of $n = 504$ and $\left[ {\sqrt n } \right] = 22$, since ${22^2} = 484 < 504 < 529 = {23^2}$.
So, by using the continued fraction expansion:
${m_0} = 0 \\ {d_0} = 1 \\ {a_0} = \left[ {\sqrt {504} } \right] = 22 \\ {m_1} = {d_0}{a_0} - {m_0} = 22 \\ \Rightarrow {d_1} = \dfrac{{n - {m_1}^2}}{{{d_0}}} = \dfrac{{504 - {{22}^2}}}{1} = 20 \\$
$\Rightarrow {a_1} = \left[ {\dfrac{{{a_0} + {m_1}}}{{{d_1}}}} \right] = \left[ {\dfrac{{22 + 22}}{{20}}} \right] = 2 \\ {m_2} = {d_1}{a_1} - {m_1} = 40 - 22 = 18 \\$
$\Rightarrow {d_2} = \dfrac{{n - {m_2}^2}}{{{d_1}}} = \dfrac{{504 - 324}}{{20}} = 9 \\ {a_2} = \left[ {\dfrac{{{a_0} + {m_2}}}{{{d_2}}}} \right] = \left[ {\dfrac{{22 + 18}}{9}} \right] = 4 \\ {m_3} = {d_2}{a_2} - {m_2} = 36 - 18 = 18 \\$
$\Rightarrow {d_3} = \dfrac{{n - {m_3}^2}}{{{d_2}}} = \dfrac{{504 - 324}}{9} = 20 \\ {a_3} = \left[ {\dfrac{{{a_0} + {m_3}}}{{{d_3}}}} \right] = \left[ {\dfrac{{22 + 18}}{{20}}} \right] = 2 \\$
$\Rightarrow {m_4} = {d_3}{a_3} - {m_3} = 40 - 18 = 22 \\ {d_4} = \dfrac{{n - {m_4}^2}}{{{d_3}}} = \dfrac{{504 - 484}}{{20}} = 1 \\ {a_4} = \left[ {\dfrac{{{a_0} + {m_4}}}{{{d_4}}}} \right] = \left[ {\dfrac{{22 + 22}}{1}} \right] = 44 \\$
Having reached a value $44$ which is twice the primary value $22$, this is often the top of the repeating pattern of the fraction, and that we have:
$\sqrt {504} = \left[ {22;\;2,\;4,\;2,\;44} \right]$
The first economical approximation for $\sqrt {504}$ is then:
$\sqrt {504} \approx \left[ {22;\;2,\;4,\;2} \right] = 22 + \dfrac{1}{{2 + \dfrac{1}{{4 + \dfrac{1}{2}}}}} \\ = \dfrac{{449}}{{20}} = 22.45 \\$
Then, we again used the repeated value.
$\sqrt {504} = \left[ {22;2,4,2,44,2,4,2} \right] \\ \approx 22.44994432... \\$

Therefore the closer value of $\sqrt {504}$ is:
$\therefore \sqrt {504} \approx 22.44994432...$

Note:
As we know continued fraction is just another way of writing fraction. They have some interesting connections with a jigsaw puzzle problem about splitting a rectangle into squares etc. it is the simple method for finding the square root of a number which has no square factor.