Numbers: Roots
Roots of integers
The root of a real number
If #a# is a non-negative real number, then there is exactly one non-negative real number #b#, in such a way that #b^2 = a#. This number is indicated by #\sqrt{a}# and called the root of #a#.
We show that there is only one non-negative root of #a#. Suppose #b# and #c# are both roots. Then we have #b^2=a=c^2#, hence #b^2-c^2=0#.
From this follows that #(b-c)\cdot(b+c)=0#, from which again follows that #bc=0# or #b+c=0#. Hence #b=c# or #b=-c#. But if they are both non-negative, #b=c# should be true. This proves that there is only one root of #a# is.
If #a# is a negative real number, then there is no real number #b# with #b^2 = a#. After all, if #b# would exist, then #b^2# would have to be a non-negative number (because squares are never negative). But, since #a=b^2#, this is in contradiction with the assumption that #a# is negative.
Some roots are well known: #\sqrt{1} = 1#, #\sqrt{4} = 2# and #\sqrt{9} = 3#. But not all roots are integers, or even rational numbers.
Rules of calculation for roots Let #a# and #b# be non-negative numbers. The following rules apply.
- #\left(\sqrt{a}\right)^2 =a= \sqrt{a^2} #
- #\sqrt{a}\cdot\sqrt{b} =\sqrt{a\cdot b}#
The first equality of the first line follows directly from the definition: if #c=\sqrt{a}#, then the definition says #c# is non-negative and #c^2=a#. If we fill in #\sqrt{a}# for #c# once again, we find #\left(\sqrt{a}\right)^2=a#.
The second equality of the first line follows from the definition of #d= \sqrt{a^2}#. After all the definition says #d# is non-negative and satisfies #d^2=a^2#. But from this follows #d=a# or #d=-a#; because #d# and #a# neither are negative, it must mean that #d=a#, which is proven by #a=\sqrt{a^2}#.
Finally, if #e=\sqrt{a}\cdot\sqrt{b}#, then #e^2 = \left(\sqrt{a}\cdot\sqrt{b}\right)^2 = \left(\sqrt{a}\right)^2\cdot \left(\sqrt{b}\right)^2= a\cdot b#. Hence we see that #e^2=a\cdot b#. Because #e# is non-negative, #e=\sqrt{a\cdot b}#. Accordingly #\sqrt{a}\cdot\sqrt{b}=\sqrt{a\cdot b}#. This proves the equality in the second line.
With these rules, you can write a product with roots as a product with no more than one root. This for example helpts to write down each product of integers and roots of integers in a unique way, the standard form.
Standard form for roots of integers Every positive integer #a# can be written as #b^2\cdot c#, in which #b# and #c# are integers and #c# a product of prime numbers that each appear exactly once. The standard form for #\sqrt{a}# is #b\sqrt{c}#.
Proof: use the notation of #a# as a product of primes. If #p# is prime that divides #a#, then the question is what the highest power of #p# is that divides #a#. If it #p# itself, the prime #p# corresponds with #c#. If it is an even exponent, for example, #p^6#, then the root thereof is #p^3#; we add that factor to #b#. If it is an odd exponent, eg #p^5#, we add #p^2# to #b#, and #p# to #c#. So we get integers #b# and #c#, in such a way that no factor of #c# is a square. From #a=b^2c# and the rules of calculation rules stated above it follows that #\sqrt{a} = \sqrt{b^2}\cdot\sqrt{c} = b\sqrt{c}#.
This can be seen as follows:
\[\begin{array}{rcl}-2\sqrt{30}\cdot -3\sqrt{2}&=&-2 \cdot -3 \sqrt{30\cdot 2}\\ &=&6\sqrt{{2}^2\cdot 15 }\\ &=& 6\cdot 2\sqrt{15}\\ &=&12 \sqrt{15}\end{array}\]
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