Square Roots | Brilliant Math & Science Wiki (2024)

The square root of a number \(a\) is the answer to the question, \(“\, \)What non-negative number, when squared \((\)raised to the 2\(^\text{nd}\) power\(),\) results in \(a?"\) The symbol for square root is \(“\,\sqrt{\ \ }"\). The square root of the number \(a\) is written as \(\sqrt{a}.\)

What is the square root of \(25?\)

Ask yourself the question, \(“\,\)What non-negative number, when squared, results in \(25?"\)

The answer to that question will be the square root of \(25\).

\(5^2=25\), so \(\sqrt{25}=5\). \(_\square\)

Square roots are crucial in solving quadratic equations and for solving distance problems in geometry.

The square root of a number \(a\), denoted \(\sqrt{a}\), is the number \(b\) such that

\[b^{2} = a\text{ and }b\ge 0.\]

The square root symbol "\(\sqrt{\ }\)" is also sometimes called a radical. The number or expression underneath the top line of the square root symbol is called the radicand. For example, in the expression \(\sqrt{2x+3}\), the radicand is \(2x+3\).

The square root symbol acts as a grouping symbol, which means that all numbers and operations in the radicand are grouped as if they were in parentheses.

Is \(“\,\sqrt{9}+16=\sqrt{9+16}"\) a true statement?

When calculating \(\sqrt{9}+16\), note that only the \(9\) is in the radicand. Thus, \(\sqrt{9}+16=3+16=19\).

When calculating \(\sqrt{9+16}\), note that the sum of \(9\) and \(16\) is in the radicand. Thus, \(\sqrt{9+16}=\sqrt{25}=5\).

These values are not equal, and so "\(\sqrt{9}+16=\sqrt{9+16}\)" is a false statement. \(_\square\)

Also, note that the result of a square root operation is always positive or zero. This fact is often ignored and leads to a common misconception about square roots.

The square root of a positive perfect square is always a positive integer. These square roots can be found by thinking of perfect squares until a match for the radicand is found.

For reference, a table of the first ten perfect squares is listed below:

\[\begin{array}{r|r}n & n^2 \\\hline 0 & 0 \\1 & 1 \\2 & 4 \\3 & 9 \\4 & 16 \\5 & 25 \\6 & 36 \\7 & 49 \\8 & 64 \\9 & 81 \\10 & 100 \\\end{array}\]

Find the value of \(\sqrt{36}\).

The goal is to think of a non-negative number that, when squared, results in \(36\). Since \(36 = 6^2\), it follows that \(\sqrt{36} = 6.\) \(_\square\)

The square root of a fraction can be found in the same way, as long as both numerator and denominator are perfect squares.

Find the value of \(\sqrt{\dfrac{25}{49}}\).

The goal is to think of a non-negative number that, when squared, results in \(\dfrac{25}{49}\). Since \(25 = 5^2\) and \(49=7^2\), it follows that \(\sqrt{\dfrac{25}{49}} = \dfrac{5}{7}\). \(_\square\)

The square root operation is not just defined for perfect squares. The square root operation can also be applied to any non-negative real number (this domain will later be expanded to negative real numbers and complex numbers).

When the square root operation is performed on a positive integer that is not a perfect square, the result is an irrational number. Calculators are often used to find the decimal approximation of such a result. However, a decent approximation can be found without a calculator. First, it will be explored how to find integer bounds for a square root.

Which consecutive integers is \(\sqrt{56}\) between?

First, note that \(56\) is not a perfect square. This means that \(\sqrt{56}\) is an irrational number.

What perfect squares is \(56\) between? Looking at the list of perfect squares, \(56\) is between \(49\) and \(64\).

Because \(49<56<64\), it stands to reason that \(\sqrt{49}<\sqrt{56}<\sqrt{64}\).

Thus, \(7<\sqrt{56}<8\). \(_\square\)

This approximation is confirmed when checking on a calculator: \(\sqrt{56}\approx 7.483315\).

This process can be applied to give an integer approximation to any square root. In order to develop a better approximation, one must consider the concavity of the square root function.

To develop a better approximation of a square root, consider which perfect square the number is closer to.

Approximate \(\sqrt{6}\) to one decimal place.

Note that \(6\) is not a perfect square. Find the integer bounds by considering what perfect squares \(6\) is between.

\(6\) is between \(4\) and \(9\). Thus, \(2<\sqrt{6}<3\).

Now consider which perfect square 6 is closer to: \(4\) or \(9\). \(6\) is nearly halfway between \(4\) and \(9\), but it is slightly closer to \(4\).

Since \(6\) is about halfway between \(4\) and \(9\), it stands to reason that \(\sqrt{6}\) is about halfway between \(2\) and \(3\). \(2.5\) is a good guess to start with.

You can check your guess by squaring \(2.5\). The result should be close to \(6\). Multiplying by hand, \(2.5\times 2.5=6.25\). This appears to be close, but is it the best approximation to one decimal place?

\(2.5\) appears to be a high guess (since \(6.25>6\), so another valid guess would be \(2.4\). Once again, square this number and compare the result to \(6\). Multiplying by hand, \(2.4\times 2.4=5.76\). Note that this is a low guess, so there are no better one-decimal-place approximations for \(\sqrt{6}\) than \(2.4\) and \(2.5\).

Which approximation is better? Note that \(2.4^2=5.76\) is less than \(6\) by \(0.24\), and \(2.5^2=6.25\) is greater than \(6\) by \(0.25\). The closer squared result came from the \(2.4\) approximation. Therefore, \(2.4\) is the best one-decimal-place approximation for \(\sqrt{6}\).

This approximation is confirmed with a calculator: \(\sqrt{6}\approx 2.449490\). \(_\square\)

Below is a graph of the square root function, \(f(x)=\sqrt{x}\):

The points with integer coordinates are shown. These points represent \(x\)-values that are perfect squares. Notice that the points become further apart as the function moves further in the positive \(x\) direction.

Taking a closer look, it is more clear what is happening:

In the above graph, the red dashed segments connect the perfect square points. Note that the square root function is always higher than these straight-line segments. What this means is that the value of a square root will tend to be higher than any linear approximation.

Approximate \(\sqrt{2.5}\) to one decimal place.

\(2.5\) is exactly halfway between the perfect squares \(1\) and \(4\). Using linear thinking, it would stand to reason that the best approximation for \(\sqrt{2.5}\) would be exactly halfway between \(1\) and \(2\). The guess becomes \(1.5\).

Multiplying by hand, \(1.5\times 1.5=2.25\). The guess is low, it is \(0.25\) off from the value of \(2.5\).

Recall that the actual square root tends to be greater than a linear approximation. Now try a guess of \(1.6\).

Multiplying by hand, \(1.6\times 1.6=2.56\). This guess is high, and it is only off by \(0.06\).

Thus, the best one-decimal-place approximation for \(\sqrt{2.5}\) is \(1.6\). It turns out that the best approximation is not halfway between \(1\) and \(2\).

This approximation is confirmed by calculator: \(\sqrt{2.5}\approx 1.581139\). \(_\square\)

Square roots are often used to find the solutions to quadratic equations. The simplest kind of quadratic equation that can be solved with square roots is an equation of the form \(x^2=a\).

Find the solutions to \(x^2=16.\)

The square root can be used to find a solution to this equation. This solution is \(x=\sqrt{16}=4\).

However, there is another solution. This other solution is \(x=-\sqrt{16}=-4\). \(_\square\)

An equation of the form \(x^2=a\), where \(a>0\), will always have two solutions: one positive and one negative.

Let \(a\) be a positive real number. The solutions to the equation \(x^2=a\) are

\[x=\sqrt{a}\quad \text{and} \quad x=-\sqrt{a},\]

or equivalently,

\[x=\pm \sqrt{a}.\]

For these kinds of equations, the positive solution is called the principle square root, while the negative solution is called the negative square root.

Some problems will request, "What are the square roots of a certain number, \(a\)?" Normally, we would consider \(\sqrt{a}\) to have only one possible result. However, in this context, the meaning of the question is to find both solutions of the equation \(x^2=a\).

The square root of a positive integer that is not a perfect square is always an irrational number. The decimal representation of such a number loses precision when it is rounded, and it is time-consuming to compute without the aid of a calculator. Instead of using decimal representation, the standard way to write such a number is to use simplified radical form:

Let \(a\) be a positive non-perfect square integer.

The simplified radical form of the square root of \(a\) is

\[b\sqrt{c}.\]

In this form \(\sqrt{a}=b\sqrt{c}\), both \(b\) and \(c\) are positive integers, and \(c\) contains no perfect square factors other than \(1\).

The process for putting a square root into simplified radical form involves finding perfect square factors, then applying the following identity:

Let \(a\) and \(b\) be positive real numbers. Then,

\[\sqrt{ab}=\sqrt{a}\times\sqrt{b}.\]

Simplify \(\sqrt{72}\).

First, ask yourself, "What is a perfect square factor of \(72\)?"

\(4\) is a perfect square factor of \(72\), and \(9\) is a perfect square factor of \(72\).

For the sake of this process, it is more efficient to find the largest perfect square factor of \(72\). As shown below, \(4\times 9=36\) is the largest perfect square factor of \(72\):

\[\begin{align}\sqrt{72}&=\sqrt{36\times 2} \\&=\sqrt{36}\times\sqrt{2} \\&=6\sqrt{2}.\end{align}\]

Therefore, simplified form of \(\sqrt{72}\) is \(6\sqrt{2}.\ _\square\)

Note: When a number is placed to the left of a square root symbol, multiplication is implied. "\(6\sqrt{2}\)" is read as "\(6\) times the square root of \(2\)."

Rationalizing the denominator is the process of re-writing a rational expression that contains a radical in the denominator as an equivalent rational expression that contains no radical in the denominator.

The following identity is a direct consequence of the definition of a square root, and it is crucial in the process of rationalizing a denominator:

Let \(a\) be a positive real number. Then,

\[\sqrt{a}\times\sqrt{a}=a.\]

Rationalizing the denominator involves applying the identity property of multiplication: the fact that multiplying the numerator and denominator by the same number results in an equivalent rational expression. In this case, the number we choose for this is the square root.

Let \(a\) be a positive real number. In order to rationalize the fraction \(\dfrac{1}{\sqrt{a}}\), both the numerator and the denominator must be multiplied by \(\sqrt{a}\). This is equivalent to multiplying the expression by \(1\), so the resulting expression is equivalent to \(\dfrac{1}{\sqrt{a}}\):

\[\begin{align}\frac{1}{\sqrt{a}}&= \frac{1}{\sqrt{a}} \times \frac{\sqrt{a}}{\sqrt{a}} \\&= \frac{1 \times \sqrt{a}}{\sqrt{a} \times \sqrt{a}} \\&= \frac{\sqrt{a}}{a}.\end{align}\]

Rationalize the following expression:\[\]

\[\frac{1}{\sqrt{2}}.\]

Rationalization will help in calculation as the denominator will be an integer:

\[\begin{align}x&=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \\&=\frac{\sqrt{2}}{2}. \ _\square\end{align}\]

Rationalize the denominator of \(\dfrac{1}{\sqrt{\sqrt{5}}}\).

We have

\[\begin{align}
\dfrac{1}{\sqrt{\sqrt{5}}} & = \dfrac{1 × \sqrt{\sqrt{5}}}{\sqrt{\sqrt{5}} × \sqrt{\sqrt{5}}}\\ \\& = \dfrac{\sqrt{\sqrt{5}}}{\sqrt{5}}\\ \\& = \dfrac{\sqrt{\sqrt{5}} × \sqrt{5}}{\sqrt{5} × \sqrt{5}}\\ \\& = \dfrac{\sqrt{5\sqrt{5}}}{5}.\ _\square\end{align}\]

Sometimes, you will encounter denominators in which an unfavorable term is added to another, making the two seemingly inseparable. In these scenarios, you must multiply the conjugate of the number.

The conjugate of a binomial number \((a+b)\) is the number \((a-b)\). When multiplied to the original number, it will produce \(\displaystyle a^2 - b^2\).

It can be proven rather simply that this always works:

We have

\[\begin{align}(a+b) \times (a-b) &=a^2 - ab + ab - b^2 \\&=a^2 - b^2.\end{align}\]

We can use this technique of multiplying by the conjugate to rationalize denominators with addition or subtraction in them.

Rationalize the following expression:

\[\dfrac{1}{\sqrt{3} - 1}.\]

To rationalize this, multiply by the conjugate:

\[\begin{align}x&=\frac{1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \\&=\frac{\sqrt{3} + 1}{2}. \ _\square\end{align}\]

Rationalize the following expression and find the value of a + b:

\[\dfrac{\sqrt{3} - 1}{\sqrt{3} + 1} = {a + b \sqrt{3}}.\]

To rationalize this, multiply by the conjugate:

\[\begin{align}x&=\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \\&=\frac{(\sqrt{3} - 1)^2}{2}\\&=\frac{4-2\sqrt{3}}{2}\\&= 2-\sqrt{3} \\&= a + b \sqrt{3}\\\\\Rightarrow a+b& =2-1 \\&= 1. \ _\square\end{align}\]

You may have noticed that none of these problems or examples involve finding the square root of a negative number.

Find the value of \(\sqrt{-9}\).

What is a non-negative number that, when squared, results in \(-9?\)

It is not \(3\), because \(3^2\) results in positive \(9\).

Even if we allowed ourselves to answer with a negative number, we still would not find a good result. It cannot be \(-3\), because \((-3)^2\) results in positive \(9\).

It turns out that there is no real number value for \(\sqrt{-9}\).

Evaluating \(\sqrt{-9}\) requires the imaginary unit, \(i\).

Because \(i^2=-1\), \((3i)^2=-9\). Thus, \(\sqrt{-9}=3i\). \(_\square\)

The imaginary unit, \(i\), is a complex number that satisfies the following equation:

\[i^2=-1.\]

With the introduction of complex numbers, the square root of any negative number can be evaluated:

Let \(a\) be a positive real number. Then,

\[\sqrt{-a}=i\sqrt{a}.\]

The square root of a non-negative, non-perfect square real number is always an irrational number. Finding the decimal representation of such a square root will often be accomplished with the help of a calculator. However, you may sometimes be asked to manually calculate the square root of some non-perfect square number.

Find the square root of 39 to 3 decimal places.

We have

\[\begin{array} {c|ccccc|c}6 & \overline{39}. & \overline{00} & \overline{00} & \overline{00} & \overline{00} & \underline{6.2449\ldots} \\& \underline{- 36} & & & & & \\122 & 3 & 00 & & & & \\ & \underline{- 2} & \underline{44} & & & & \\1244 & & 56 & 00 & & & \\ & & \underline{- 49} & \underline{76} & & & \\12484 & & 6 & 24 & 00 & & \\ & & \underline{4} & \underline{99} & \underline{36} & & \\124889 & & 1 & 24 & 64 & 00 & \\ & & \underline{1} & \underline{12} & \underline{40} & \underline{01} & \\ & & & 12 & 23 & 99 & \\\end{array}\]

Thus, the square root of 39 to 3 decimal places is 6.245. \(_\square\)

Checking on a calculator, you find the answer to be correct. But how is this so? What was the method? All it looked like was a bunch of random numbers. Well, the method resembles long division, except you bring two digits down instead of one after subtracting. Here is the method:

1. For \(n\) decimal places, write \(2n+2\) digits after the decimal point, dividing into \(n+1\) pairs of digits (empty digits are considered as 0's, as shown above). For digits to the left of the decimal point, pair the digits from the furthest to the right to the left. Your last pairing will be either 1 or 2 digits.

   If there is 1 digit, the digit will be by itself. If there are 2 digits, the digits can be paired together.

   We now draw 2 columns to the right of the decimal point. This column will be known as the dividend column; the one on the left will be the divisor, and the right one will have an area for the quotient.

2. Now, from the most leftward pair of digit, or pair of digits, find the largest positive integer that has its square less than this 1 or 2-digit number. In the example above, the largest integer satisfying this was 6, since \(6^2 < 39 < 7^2\). So, we put the 6 in the divisor column, and put a 6 in the quotient column.

3. We subtract the square of this number from the first pair of digits, then drop the next pair down. In the example above, we subtracted \(39-36\) to get \(3\), then drop \(00\) from the top of the column to get \(300\).

4. We add the number from the divisor to itself, then shift it across to the left. We now have to find an integer \(a\) such that \(\overline{ka} \times a < \text{New dividend} < \overline{k(a+1)} \times {a+1}\), where \(k\) refers to the number obtained after adding the old divisor to itself and \(\overline{abc}\) refers to concatenating the numbers \(a\), \(b\) and \(c\). Once we find the value for \(a\), we write it down in place, put it as the next number in the quotient. E.g., above, we had \(\overline{12a}\times a = 300\), so \(a=2\) since \(122 \times 2 < 300 < 123 \times 3\).
   Now, we have got the first digit of the square root. Now, we add the first divisor and the first digit. Then we have to find a number \(a\) such that \(\overline{ka} × a \leq\) the present remainder where \(k\) is the second dividend. For example in the example above, the first step is as follows:\[\begin{array} {c|ccccc|c}
   \color{blue}{6} & \overline{39}. & \overline{00} & \overline{00} & \overline{00} & \overline{00} & \underline{\color{blue}{6}} \\& \underline{- 36} & & & & &\\& 3 & 00 & & & & \end{array}\]Now, we add the blue coloured 6's to get \(\color{purple}{12}\).
   Then we have to find a number which fits in the box such that \(12\boxed{\phantom{0}} × \boxed{\phantom{0}} \leq 300\).
   We see that \(12\color{green}{2} × \color{green}{2} = 244 < 300\) and \(12\color{orange}{3} × \color{orange}{3} = 369 > 300\).
   So, we take \(2\) in the box, i.e. \(a = 2\).
   So, we finally have\[\begin{array} {c|ccccc|c}6 & \overline{39}. & \overline{00} & \overline{00} & \overline{00} & \overline{00} & \underline{6.\color{orange}{2}} \\& \underline{- 36} & & & & & \\\color{purple}{12}\color{red}{2} & 3 & 00 & & & & \\& \underline{- 2} & \underline{44} & & & & \\& & 56 & & & & & \end{array}\]

5. We find the \(\overline{ka} \times a\), and then subtract this from the new dividend, then drop the next pair down. E.g., above, we had \(300-244=56\), so the next new dividend became \(5600\).

6. Add the number \(a\) to \(\overline{ka}\), and shift to the left again. Repeat Steps 4-6 until all the pairs have been exhausted. You have now determined the square root manually.

Though the method given above is good, it is also slightly lengthy if many decimal places are required. When the number of significant figures takes priority, we need a better algorithm.

Proposed is the Babylonian method to compute \(x=\sqrt{\alpha}\) :

1. Pick a number \(x_0\) such that \(x_0 > \sqrt{\alpha}\).

2. Define a sequence \(\left \{x_n \right \}_{n=0}^{\infty}\) that is built recursively by the following formula: \[x_{n+1} = \frac{1}{2} \left( x_n + \frac{\alpha}{x_n} \right).\]

3. \(\displaystyle \sqrt{\alpha} = \lim_{n \to \infty} x_n\).

Implementation in Python :

123456

def sqrt(x): EPS = pow(10,-10) ans = x while abs(ans * ans - x) > EPS: ans = 1 / 2.0 * (ans + x / ans) return ans

Code in action :

1234567

20.00000000000000000000000000000010.974999999999999644728632119950 7.264265375854213502293532656040 6.316505985303646042439140728675 6.245402762693970544205512851477 6.244998011514777402908293879591 6.244997998398398308950163482223

Compare with the actual value \(\sqrt{39} = 6.2449979983983982058468...\); this algorithm returns a number that is correct up to 16 significant figures, with only 7 iterations.

We can get the approximate value of \(\sqrt{\alpha}\) to eight to sixteen significant figures in 3 to 4 iterations if we chose \(x_0\) suitably.

This is a result of the following inequality, whose proof is given below:

If \(\varepsilon_n = x_n - \sqrt{\alpha}\) and \(\beta=2\sqrt{\alpha}\), then

\[0< \frac{\varepsilon_n}{\beta} < \left(\frac{\varepsilon_0}{\beta} \right)^{2^n} \quad n = 1,2, \ldots. \]

So, if \(\varepsilon_0 < \beta\) \(\Big(\)prefrerably \(\varepsilon_0 < \frac{\beta}{10}\Big) \), then the error decreases rapidly. Usually, to do this, taking \(x_0=\left\lceil \sqrt{\alpha} \right\rceil\) is sufficient. Here \(\lceil \cdot \rceil\) denotes the ceiling function.

We have\[ x_{n +1} = \frac{1}{2} \left(x_n + \frac{\alpha}{x_n}\right). \]We first show \(x_{n} > \sqrt{\alpha}\) for any \(n \in \mathbb{N} \).

By the AM-GM inequality,\[ x_{n} > \frac{1}{2} \left(2 \sqrt{x_{n - 1} \cdot \frac{\alpha}{x_{n - 1}}}\right) = \sqrt{\alpha}. \]We cannot have equality as \(x_0 > \sqrt{\alpha} \).
From here, it is easy to see that if \(x_0 = \sqrt{\alpha}, \) then \(x_n\) are all equal for all integers \(n\).

Now,\[\begin{align}x_{n + 1} - \sqrt{\alpha} &= \frac{1}{2} \left(x_n + \frac{\alpha}{x_n} - 2\sqrt{\alpha} \right) \\&= \frac{1}{2} \left(\sqrt{x_n} - \sqrt{\frac{\alpha}{x_n}} \right)^{2} \\&= \frac{\left(x_n - \sqrt{\alpha} \right)^{2}}{2 x_n}.\end{align}\]So,\[ \frac{\varepsilon_{n + 1}}{\beta} = \frac{\left(x_n - \sqrt{\alpha} \right)^{2}}{4 x_n \sqrt{\alpha}} < \frac{\left(x_n - \sqrt{\alpha} \right)^{2}}{4 \alpha} = \left(\frac{\varepsilon_{n}}{\beta}\right)^2,\]and by induction, it is easy to see that\[\frac{\varepsilon_{n}}{\beta}< \left(\frac{\varepsilon_{0}}{\beta}\right)^{2^n}\]for \( n > 0 \).

Note that \(\varepsilon_n > 0 \) as \( x_{n} > \sqrt{\alpha} \), then \(\dfrac{\varepsilon_n}{\beta} > 0.\)

This completes the proof. \(_\square\)

This method is illustrated by the following examples:

Compute \(\sqrt{39}\) correct to six decimal places.

Let us take \(x_0 = 7\).

Let's figure out the value of \(\dfrac{\varepsilon_0}{\beta}\): \[\begin{align} \frac{\varepsilon_0}{\beta} &= \frac{7-\sqrt{39}}{2\sqrt{39}} \\ &< \frac{1}{2\sqrt{39}} \\ &< \frac{1}{12} \\\implies 0 &< \frac{\varepsilon_0}{\beta} < \frac{1}{10}. \end{align} \]

Therefore, we need to calculate just three iterates. This is because of the following inequality \((\)note that \(\beta<14)\):

\[\boxed{ 0 < \varepsilon_3 < 14 \times 10^{-8} < 10^{-6} }. \]

Iterating, we get \[\begin{align} x_1 &= 6.28571428\ldots \\ x_2 &= 6.24512987\ldots \\ x_3 &= 6.24499799\ldots. \end{align}\]

Hence the value of \(\sqrt{39}\), correct to six decimals, is 6.244998. \(_\square\)

Note:

• Notice that right after two iterates, we get the value of \(\sqrt{39}\) correct to three decimals, compared to three iterates required in the previous method.
• This method requires only three iterates, compared to 6 iterates required by the previous method for this task.

Compute \(\sqrt{3}\) correct to thirty decimal places.

Let us take \(x_0 = 2\).

Let's calculate \(\dfrac{\varepsilon_0}{\beta}\): \[\begin{align} \frac{\varepsilon_0}{\beta} &= \frac{2-\sqrt3}{2\sqrt3} \\ &= \frac{1}{2\sqrt3(2+\sqrt3)} \\ &= \frac{1}{2(2\sqrt3+3)} \\ &< \frac{1}{2(5)} \\\implies 0&<\frac{\varepsilon_0}{\beta} < \frac{1}{10}. \end{align} \]

This means that \[\boxed{0< \varepsilon_5 < 4 \times 10^{-32}}\] since \(\beta < 4.\)

Hence we need to calculate just 5 iterates to get \(\sqrt3\) correct to thirty decimals!

Iterating, we get (only ten digits are shown here)

\[\sqrt3 \approx x_5 = 1.7320508075\ldots.\ _\square\]

Note: With 20 iterates, the value of \(\sqrt3\) to more than a million digits is known!

For more details in approximating square roots, see here.

The square root of a complex number is somewhat ambiguous. Non-real complex numbers are neither positive nor negative, so it is not well-defined which square root is the principal square root. Therefore, when the square root operation is used on a complex number, the result is interpreted to be all the solutions of an equation:

Let \(z\) be a complex number. Then there are up to two values for \(\sqrt{z}\), and they are equal to the solutions of the equation

\[x^2=z.\]

Note that the square root operation, when used on complex numbers, is not well-defined in the sense that there is only one result.

Find square root of \(i\) using general method.

Assume that \(\sqrt{i}=a+ib\). Then, \(i=(a+ib)^{2}=a^{2}-b^{2}+2abi\).

Equating the real parts of the equation and equating the imaginary parts of the equation gives

\[\begin{array}a^{2}-b^{2}=0 & \text{and} & 2ab=1\end{array}.\]

Solving this system of equations gives

\[\begin{array}a=b=\frac{1}{\sqrt{2}} & \text{or} & a=b=-\frac {1}{\sqrt{2}}.\end{array}\]

Thus, \[\begin{array}\sqrt{i}=\dfrac {1}{\sqrt{2}}+ \dfrac {i}{\sqrt{2}} & \text{or} & \sqrt{i}=-\dfrac {1}{\sqrt{2}}- \dfrac {i}{\sqrt{2}}.\ _\square \end{array}\]

In general, for complex number \(z=a+ib\) such that \(a,b\in\mathbb{R}\),

\[\sqrt{z}=\pm \left[ \sqrt {\frac{\sqrt{a^2+b^2}+a}{2} }+i \sqrt{ \frac{\sqrt{a^2+b^2}-a}{2} } \right]. \]

Another method, using Euler's formula, is to express the complex number \(z\) in the form \(z=re^{i\theta},\) where \(r=|z|\) and \(\theta \in [0, 2\pi) \). Here's an illustration:

Find the square root of \(i\) using Euler's formula.

Let \(z=i\) so that \(r=|z|=|i|=1\). Since \(e^{i\pi/2}=\cos \frac{\pi}{2} + i\sin \frac{\pi}{2} = i \), we have \(\theta= \frac{\pi}{2}. \)

Now,

\[\sqrt{z}=\pm \sqrt{r} e^{i\theta/2}. \]

Hence, with \(z=i\), we get \(\displaystyle\sqrt{i}=\pm e^{i\pi/4}=\pm \left(\frac {1}{\sqrt{2}}+ \frac {i}{\sqrt{2}}\right),\) which is the same answer as before, but much faster to calculate. \(_\square\)

Any complex number \(z=a+bi\) can be written in the polar form as

\[z = r(\cos \theta + i \sin \theta),\]

where \(r=\sqrt{a^2+b^2}\) and \(\theta=\arctan{\frac{b}{a}}\). Consider the square root by raising its power to half and apply De Moivre's theorem:

\[\begin{aligned} z^\frac{1}{2} &= \pm r^\frac{1}{2}(\cos \theta + i \sin \theta)^\frac{1}{2} \\\sqrt{z} &= \pm \sqrt{r}\left(\cos{\frac{\theta}{2}} + i\sin{{\frac{\theta}{2}}}\right).\end{aligned}\]

Find \(a\), where\[\]

\[a = \dfrac{1}{\sqrt{1} + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \dfrac{1}{\sqrt{99} + \sqrt{100}}.\]

Similar to the previous problem, multiply by the conjugate:

\[\begin{align}a=&\dfrac{1}{\sqrt{1} + \sqrt{2}} \times \frac{-\sqrt{1}+\sqrt{2}}{-\sqrt{1}+\sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} \times \frac{-\sqrt{2}+\sqrt{3}}{-\sqrt{2}+\sqrt{3}} \\\\&+ \dfrac{1}{\sqrt{3} + \sqrt{4}} \times \frac{-\sqrt{3}+\sqrt{4}}{-\sqrt{3}+\sqrt{4}} + \cdots + \dfrac{1}{\sqrt{99} + \sqrt{100}} \times \frac{-\sqrt{99}+\sqrt{100}}{-\sqrt{99}+\sqrt{100}}\\\\=&\frac{-\sqrt{1}+\sqrt{2}}{1} + \frac{-\sqrt{2}+\sqrt{3}}{1} + \frac{-\sqrt{3}+\sqrt{4}}{1} + \cdots + \frac{-\sqrt{99}+\sqrt{100}}{1}\\\\=& - \sqrt{1} + \sqrt{2} - \sqrt{2} + \sqrt{3} - \sqrt{3} + \sqrt{4} - \cdots - \sqrt{99} + \sqrt{100}\\\\=& -\sqrt{1} + \sqrt{100}\\\\=&-1 + 10\\\\=&9.\ _\square\end{align}\]