Square root Did You Mean Square?

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In mathematics, the principal square root of a non-negative real number $x\,\!$ is denoted $\sqrt x$ and represents the non-negative real number whose square (the result of multiplying the number by itself) is $x\,\!$ .

For example, $\sqrt 9 = 3$ since 3² = 3 × 3 = 9.

This example suggests how square roots can arise when solving quadratic equations such as $x^2=9 \,$ or, more generally

$ax^2+bx+c=0. \,$

There are two solutions to the square root of a non-zero number. For a positive real number, the two square roots are the principle square root and the negative square root. For negative real numbers, the concept of imaginary and complex numbers has been developed to provide a mathematical framework to deal with the results.

Square roots of positive integers are often irrational numbers, i.e., numbers not expressible as a quotient of two integers. For example, $\sqrt 2$ cannot be written exactly as m/n, where n and m are integers. Nonetheless, it is exactly the length of the diagonal of a square with side length 1.

The discovery that $\sqrt 2$ is irrational is attributed to Hippasus, a disciple of Pythagoras.

The square root symbol (√) was first used during the 16th century. It has been suggested that it originated as an altered form of lowercase r, representing the Latin radix (meaning "root").

Have to see

1 Properties
2 Common errors involving equations with the principal square root function
3 Computation
4 Square roots of complex numbers
5 Square roots of matrices and operators
6 Infinitely nested square roots
7 Square roots of the first 20 positive integers
8 See also
9 External links

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Properties

The principal square root function $\sqrt{x}$ is a function which maps the non-negative real domain $\mathbb{R}^+ \cup \{0\}$ into the non-negative real codomain $\mathbb{R}^+ \cup \{0\}$ .

The principal square root function $\sqrt{x}$ always returns a single unique value.

There are only two solutions to the equation $x = \sqrt{x}$ The solution set is { 0,1 }.

To obtain both roots of a positive number, take the value given by the principal square root function as the first root (root₁) and obtain the second root (root₂) by subtracting the first root from zero (ie root₂ = 0 - root₁).

The following important properties of the square root functions are valid for all positive real numbers $x$ and $y$ :

$\sqrt{xy} = \sqrt x \sqrt y \qquad \Rightarrow \qquad \sqrt{\left( 100y \right) } \, = \, 10 \cdot \sqrt y$

$\sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}} \qquad \Rightarrow \qquad \sqrt{\left ( \frac{x}{100}\right ) } = \frac{\sqrt{x}}{10}$

$\sqrt{x^2} = \left|x\right|$ for every real number

x

(see absolute value)

$\sqrt x = x^{1/2}$

The square root function generally maps rational numbers to algebraic numbers; $\sqrt x$ is rational if and only if $x$ is a rational number which, after cancelling, is a ratio of two perfect squares. In particular, $\sqrt 2$ is irrational.

In geometrical terms, the square root function maps the area of a square to its side length.

Suppose that $x$ and $a$ are reals, and that $x^2 = a \,$ , and we want to find $x$ . A common mistake is to "take the square root" and deduce that $x = \sqrt a$ . This is incorrect, because the principal square root of $x 2$ is not $x$ , but the absolute value $\left| x \right|$ , one of our above rules. Thus, all we can conclude is that $\left| x \right| = \sqrt a$ , or equivalently $x = \pm\sqrt a$ .

In calculus, for instance when proving that the square root function is continuous or differentiable or when computing certain limits, the following identity often comes handy:

$\sqrt x - \sqrt y = \frac{x-y}{\sqrt x + \sqrt y}$

It is valid for all non-negative numbers

x

and

y

which are not both zero.

The function $f(x) = \sqrt x$ has the following graph, made up of half a parabola lying on its side:

The function is continuous for all non-negative $x$ , and differentiable for all positive $x$ (it is not differentiable for $x = 0$ since the slope of the tangent there is ∞). Its derivative is given by

$f'(x) = \frac{1}{2\sqrt x}$

Its Taylor series about $x = 0$ can be found using the binomial theorem:

$\sqrt{x+1}=1 + \sum_{n=1}^\infty { (-1)^{n+1} (2n-2)! \over n! (n-1)! 2^{2n-1} }x^n$

$= 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16} x^3 - \frac{5}{128} x^4 + \dots$

for $\left| x \right| AND1lt; 1$ .

Common errors involving equations with the principal square root function

Not recognising $x 1 / 2$ as the principal square root of $x$ in a squaring step followed by principal square root step.

$-1 = -1^1 = -1^{2/2} = (-1^2)^{1/2} = 1^{1/2} = 1 \,$

In details we have:

$-1=-1^1 \,$

$-1^1=-1^{2/2} \,$

$-1^{2/2}=\sqrt{-1^2}\qquad\mathbf{Error!} \,$

This is because $-1^{2/2}=-1 \,$ and $\sqrt{-1^2}=\sqrt{1}$ so

$-1\ne\sqrt{1}$

The error comes in taking the principal square root of a square of −1.
Not taking into account false root(s) when squaring equations with the principal square root function

$x - 1 = \sqrt{x} \,$

$x^2 - 2x + 1 = x \,$ squaring both sides which introduces the false root x=0.38197

$x^2 - 3x + 1 = 0 \,$ move x from the right hand side to the left hand side

Using the quadratic equation formula, we get the solutions

$x=2.618 \,$ or $x=0.38197 \,$

But when we substitute back $x=0.38197 \,$

$0.38197 - 1 = \sqrt{0.38197} \,$

$-0.61803 = 0.61803 \,$ Error!

Computation

Pocket calculators typically implement good routines to compute the exponential function and the natural logarithm, and then compute the square root of $x$ using the identity

$\sqrt{x} = e^{\frac{1}{2}\ln x}$

The same identity is exploited when computing square roots with logarithm tables or slide rules.

There are numerous methods to compute square roots. See Methods of computing square roots.

Square roots of complex numbers

To every non-zero complex number z there exist precisely two numbers w such that w² = z. The usual definition of √z is as follows: if z = r exp(iφ) is represented in polar coordinates with -π AND1lt; φ ≤ π, then we set √z = √r exp(iφ/2). Thus defined, the square root function is holomorphic everywhere except on the non-positive real numbers (where it isn't even continuous). The above Taylor series for √(1+x) remains valid for complex numbers x with |x| AND1lt; 1.

When the number is in rectangular form the following formula can be used:

$\sqrt{x+iy} = \sqrt{\frac{\left|x+iy\right| + x}{2}} \pm i \sqrt{\frac{\left|x+iy\right| - x}{2}}$

where the sign of the imaginary part of the root is the same as the sign of the imaginary part of the original number.

Note that because of the discontinuous nature of the square root function in the complex plane, the law √(zw) = √(z)√(w) is in general not true. Wrongly assuming this law underlies several faulty "proofs", for instance the following one showing that -1 = 1:

$-1 = i \times i = \sqrt{-1} \times \sqrt{-1} = \sqrt{-1 \times -1} = \sqrt{1} = 1$

The third equality cannot be justified. (See invalid proof.)

However the law can only be wrong by a factor -1 (it is right up to a factor -1), √(zw) = ±√(z)√(w), is true for either ± as + or as - (but not both at the same time). Note that √(c²) = ±c, therefore √(a²b²) = ±ab and therefore √(zw) = ±√(z)√(w), using a = √(z) and b = √(w).

Square roots of matrices and operators

If A is a positive definite matrix or operator, then there exists precisely one positive definite matrix or operator B with B² = A; we then define √A = B.

More generally, to every normal matrix or operator A there exist normal operators B such that B² = A. In general, there are several such operators B for every A and the square root function cannot be defined for normal operators in a satisfactory manner. Positive definite operators are akin to positive real numbers, and normal operators are akin to complex numbers.

Infinitely nested square roots

Under certain condition infinitely nested radicals such as

$x = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}$

represent rational numbers. This rational number can be found by realizing that x also appears under the radical sign, which gives the equation

$x = \sqrt{2+x}.$

If we solve this equation, we find that x = 2. More generally, we find that

$\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}} = \frac{1 + \sqrt {1+4n}}{2}.$

The same procedure also works to get

$\sqrt{n-\sqrt{n-\sqrt{n-\sqrt{n-\cdots}}}} = \frac{-1 + \sqrt {1+4n}}{2}.$

This method will give a rational $x$ value for all values of $n$ such that

n = x 2 + x .

Square roots of the first 20 positive integers

√ 1 = 1
√ 2 ≈ 1.4142135623 7309504880 1688724209 6980785696 7187537694 8073176679 7379907324 78462
√ 3 ≈ 1.7320508075 6887729352 7446341505 8723669428 0525381038 0628055806 9794519330 16909
√ 4 = 2
√ 5 ≈ 2.2360679774 9978969640 9173668731 2762354406 1835961152 5724270897 2454105209 25638
√ 6 ≈ 2.4494897427 8317809819 7284074705 8913919659 4748065667 0128432692 5672509603 77457
√ 7 ≈ 2.6457513110 6459059050 1615753639 2604257102 5918308245 0180368334 4592010688 23230
√ 8 ≈ 2.8284271247 4619009760 3377448419 3961571393 4375075389 6146353359 4759814649 56924
√ 9 = 3
√10 ≈ 3.1622776601 6837933199 8893544432 7185337195 5513932521 6826857504 8527925944 38639
√11 ≈ 3.3166247903 5539984911 4932736670 6866839270 8854558935 3597058682 1461164846 42609
√12 ≈ 3.4641016151 3775458705 4892683011 7447338856 1050762076 1256111613 9589038660 33818
√13 ≈ 3.6055512754 6398929311 9221267470 4959462512 9657384524 6212710453 0562271669 48293
√14 ≈ 3.7416573867 7394138558 3748732316 5493017560 1980777872 6946303745 4673200351 56307
√15 ≈ 3.8729833462 0741688517 9265399782 3996108329 2170529159 0826587573 7661134830 91937
√16 = 4
√17 ≈ 4.1231056256 1766054982 1409855974 0770251471 9922537362 0434398633 5730949543 46338
√18 ≈ 4.2426406871 1928514640 5066172629 0942357090 1562613084 4219530039 2139721974 35386
√19 ≈ 4.3588989435 4067355223 6981983859 6156591370 0392523244 4936890344 1381595573 28203
√20 ≈ 4.4721359549 9957939281 8347337462 5524708812 3671922305 1448541794 4908210418 51276

External links

Japanese soroban techniques - Professor Fukutaro Kato's method
Japanese soroban techniques - Takashi Kojima's method
Algorithms, implementations, and more - Paul Hsieh's square roots webpage

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Square root?

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