Higher roots

Numbers: Roots

Higher roots

Let #n# be a natural number.

Higher root

If #a# is a non-negative real number, then there is exactly one non-negative real number #b#, in such a way that #b^n = a#. This number is written as #\sqrt[n]{a}#; it is called the #n#-th root of #a#.

Solutions #b# of #b^n = a#	#a\gt 0#	#a=0#	#a\lt0#
#n# even	#b=\pm\sqrt[n]{a}#	0	-
#n# odd	#b=\sqrt[n]{a}#	0	#b=-\sqrt[n]{-a}#

We show that if #n=3#, there is no more than one non-negative real number #b# in such a way that #b^n = a#. Suppose #c# also is such a number: #c^3=a# and #c# is non-negative. Then #b^3=c^3# holds, and therefore also #b^3-c^3=0#. However, the left-hand member can be written as a product of two factors: #b^3-c^3=(b-c)\cdot (b^2 + b\cdot c + c^2)#. Since this expression is equal to #0#, at least one of both sides has to be equal to zero. But #b^2 + b\cdot c + c^2# is zero only if both #b# and #c# are zero (because both #b# are #c# are non-negative). Hence follows that #b-c# is equal to #0#. But that would mean that #b=c#.

For #n=2# we find the normal square root #\sqrt[2]{a}=\sqrt{a}#.

Often #\sqrt[3]{-7}#, the cube root from #-7#, a negative number, is also defined as the number of which the third power is equal to #-7#. However, the quantity below the root sign is always assumed non-negative in this course. The number meant with #\sqrt[3]{-7}# can be expressed as #-\sqrt[3]{7}#.

Rules of calculation for higher roots

Let #a# and #b# be positive numbers, and #m# and #n# positive integers.

#\left(\sqrt[n]{a}\right)^m=\sqrt[n]{a^m}#
#\left(\sqrt[n]{a}\right)^n=\sqrt[n]{a^n} = a#
#\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}#
#\frac{\sqrt[n]{a}}{ \sqrt[n]{b}}=\sqrt[n]{\frac{a}{ b}}=\frac{1}{b}\sqrt[n]{{a}\cdot { b}^{n-1}}#
#\sqrt[m\cdot n]{a^{m}} = \sqrt[n]{a}#

Standard notation for higher roots of fractions

The expression #\frac{a}{b}\sqrt[n]{c}#, in which #a#, #b#, and #c# are positive integers, is called a standard notation for a higher root of a positive rational number if

#\frac{a}{b}# is a simiplified fraction,
#c# is free of factors that are a #n#-th power, and
#c# is unequal to a #d#-th power for each divisor #d# of #n#.

The last condition is new with respect to the case #n=2#, and is linked to the last of the rules of calculation above.

The standard notation of #\sqrt[8]{324}# for example is # \sqrt[4]{18}#, because #324 = 2^2\cdot 3^4=(2\cdot 3^2)^2 = 18^2#, in such a way that #\sqrt[8]{324}=\sqrt[8]{18^2}=\sqrt[4]{18}#.

Write the expression \(\frac{\sqrt[4]{7}}{\sqrt[4]{72}} \) in standard notation.

You can use the fact that #7 \cdot { 72 }^{ 4 -1}=2^9\cdot 3^6\cdot 7#.

\(\frac{\sqrt[4]{7}}{\sqrt[4]{72}}\,=\,\) \( { {{1}\over{6}} } \sqrt[4]{ 126 } \)

We rewrite the expression by first making sure numerator and denominator are in standard notation, next eliminating the root from the denominator, next removing all #4#th exponents from below the root sign, and finally simplifying the fraction in front of the root sign: \[ \frac{\sqrt[4]{7}}{\sqrt[4]{72}} =\frac{ \sqrt[ 4 ]{ 7 }}{ \sqrt[ 4 ]{ 72 }} = \frac{ 1 }{ 72 }\cdot \sqrt[ 4 ]{ 7 \cdot { 72 }^{ 4 -1}} = {{1}\over{72}}\cdot \sqrt[4]{2^9\cdot 3^6\cdot 7} = { {{1}\over{6}} } \sqrt[4]{ 126 } \tiny.\]

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