How do you simplify square roots that are irrational?
Answer
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Hint: The irrational numbers are those numbers which cannot be expressed as the ratio of two integers. They have a decimal expansion which is neither terminating nor repeating. In the case of square roots, the square root of an imperfect square is an irrational number. These square roots are simplified into the decimal expansions by using the long division method.
Complete step-by-step answer:
We know that the irrational numbers are the numbers having the decimal expansions which are neither terminating nor repeating. Talking about the square roots, all the square roots of the numbers which are imperfect squares are irrational numbers. The examples of the imperfect squares are $2,5,11,22,34...$ etc. So the square roots of these numbers $\sqrt{2},\sqrt{5},\sqrt{11},\sqrt{34},.....$ are all irrational.
Now, the simplification of the irrational square roots means obtaining the square root in the form of decimal expansion. This is done with the help of the long division method. Let us take an example of the irrational square root $\sqrt{107}$. For obtaining its decimal expansion by the long division, we first place a bar over the pairs starting from the unit place. In the number $107$ the pair of numbers from the unit place is $07$. So it is written as $1\overline{07}$. Then we place a bar on the left over digit, in this case $1$. So the number is now written as $\overline{1}\text{ }\overline{07}$. Now, we have to choose the divisor whose square is less than or equal to the leftmost digit. The leftmost digit is $1$, so we choose the divisor as $1$. Then choosing the leftmost digit as the dividend, we divide it as below
\[1\overset{1}{\overline{\left){\begin{align}
& \overline{1}\text{ }\overline{07} \\
& \underline{1} \\
& 0 \\
\end{align}}\right.}}\]
Now, we take the whole pair down, and treating it as the dividend, we divide it as below
\[1\overset{1}{\overline{\left){\begin{align}
& \underline{\begin{align}
& \overline{1}\text{ }\overline{07} \\
& 1 \\
\end{align}} \\
& 0\text{ }\overline{07} \\
\end{align}}\right.}}\]
For the next divisor, we double the previous divisor and write it one the tens place with the empty ones place as below
\[1\overset{1}{\overline{\left){\begin{align}
& \underline{\begin{align}
& \overline{1}\text{ }\overline{07} \\
& 1 \\
\end{align}} \\
& 1\underline{{}}\left| \!{\overline {\,
0\text{ }\overline{07} \,}} \right. \\
\end{align}}\right.}}\]
For the one place, we follow the same procedure as for choosing the first divisor. So the ones place is choose as $0$ and also place it as the next digit on the quotient
\[1\overset{10}{\overline{\left){\begin{align}
& \underline{\begin{align}
& \overline{1}\text{ }\overline{07} \\
& 1 \\
\end{align}} \\
& 1\underline{0}\left| \!{\overline {\,
\begin{align}
& \underline{\begin{align}
& 0\text{ }\overline{07} \\
& 0 \\
\end{align}} \\
& 7 \\
\end{align} \,}} \right. \\
\end{align}}\right.}}\]
Following the same procedure, we perform the long division and obtain the result as $10.34$.
Note: Do not make the pairs of the digits starting from the leftmost digit. They are made from the rightmost digit. When we are finished with the digits, then we extract the pair of digits after the decimal point.
Complete step-by-step answer:
We know that the irrational numbers are the numbers having the decimal expansions which are neither terminating nor repeating. Talking about the square roots, all the square roots of the numbers which are imperfect squares are irrational numbers. The examples of the imperfect squares are $2,5,11,22,34...$ etc. So the square roots of these numbers $\sqrt{2},\sqrt{5},\sqrt{11},\sqrt{34},.....$ are all irrational.
Now, the simplification of the irrational square roots means obtaining the square root in the form of decimal expansion. This is done with the help of the long division method. Let us take an example of the irrational square root $\sqrt{107}$. For obtaining its decimal expansion by the long division, we first place a bar over the pairs starting from the unit place. In the number $107$ the pair of numbers from the unit place is $07$. So it is written as $1\overline{07}$. Then we place a bar on the left over digit, in this case $1$. So the number is now written as $\overline{1}\text{ }\overline{07}$. Now, we have to choose the divisor whose square is less than or equal to the leftmost digit. The leftmost digit is $1$, so we choose the divisor as $1$. Then choosing the leftmost digit as the dividend, we divide it as below
\[1\overset{1}{\overline{\left){\begin{align}
& \overline{1}\text{ }\overline{07} \\
& \underline{1} \\
& 0 \\
\end{align}}\right.}}\]
Now, we take the whole pair down, and treating it as the dividend, we divide it as below
\[1\overset{1}{\overline{\left){\begin{align}
& \underline{\begin{align}
& \overline{1}\text{ }\overline{07} \\
& 1 \\
\end{align}} \\
& 0\text{ }\overline{07} \\
\end{align}}\right.}}\]
For the next divisor, we double the previous divisor and write it one the tens place with the empty ones place as below
\[1\overset{1}{\overline{\left){\begin{align}
& \underline{\begin{align}
& \overline{1}\text{ }\overline{07} \\
& 1 \\
\end{align}} \\
& 1\underline{{}}\left| \!{\overline {\,
0\text{ }\overline{07} \,}} \right. \\
\end{align}}\right.}}\]
For the one place, we follow the same procedure as for choosing the first divisor. So the ones place is choose as $0$ and also place it as the next digit on the quotient
\[1\overset{10}{\overline{\left){\begin{align}
& \underline{\begin{align}
& \overline{1}\text{ }\overline{07} \\
& 1 \\
\end{align}} \\
& 1\underline{0}\left| \!{\overline {\,
\begin{align}
& \underline{\begin{align}
& 0\text{ }\overline{07} \\
& 0 \\
\end{align}} \\
& 7 \\
\end{align} \,}} \right. \\
\end{align}}\right.}}\]
Following the same procedure, we perform the long division and obtain the result as $10.34$.
Note: Do not make the pairs of the digits starting from the leftmost digit. They are made from the rightmost digit. When we are finished with the digits, then we extract the pair of digits after the decimal point.
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