square roots (surds) - Wurzeln
Definition
Given any non-negative number a (a is greater or equal to zero). The square root of a is a number that multiplied with itself is equal to the number a.
notation
We use following notation
$${ \sqrt[2]{a} }$$
or we can also write
$${ \sqrt{a} }$$
This is called “the square root of a” (die Quadratwurzel von a)
Where $${ \sqrt{\space\space} }$$ is the square root symbol (das Wurzelzeichen). The number under the root symbol is called radicand (Radikand - Ger.)
examples
Evaluate $${ \sqrt{961} }$$ means find a non-negative number that multiplied with itself is 961.
So, $${ \sqrt{961} = 31 \space\space \text{as}\space\space 31 \cdot 31 = 961 }$$ Similarly, $${ \sqrt{\frac{4}{25}} = \frac{2}{5} \space\space \text{as}\space\space \frac{2}{5} \cdot \frac{2}{5} = \frac{4}{25} }$$ $${ \sqrt{0.09} = 0.3 \space\space \text{as}\space\space 0.3 \cdot 0.3 = 0.09}$$ $${ \sqrt{0} = 0 \space\space \text{as}\space\space 0 \cdot 0 = 0}$$
note
square roots are always non-negative (positive or equal to zero)
cubic root
Definition
Given any non-negative number a (a is greater or equal to zero). The cubic root of a is a number taken to the power of 3 is equal to the number a. It is written as $${ \sqrt[3]{a} }$$ It is called the “the third root of a”
examples
Evaluate $${ \sqrt[3]{512} }$$ means find a non-negative number that taken to the power of 3 is equal to 512.
So, $${ \sqrt[3]{512} = 8 \space\space \text{as}\space\space 8 \cdot 8 \cdot 8 = 8^3 = 512 }$$ Similarly, $${ \sqrt[3]{\frac{27}{64}} = \frac{3}{4} \space\space \text{as}\space\space \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} = \left(\frac{3}{4}\right)^3 = \frac{27}{64}}$$ $${ \sqrt[3]{0.512} = 0.8 \space\space \text{as}\space\space 0.8 \cdot 0.8 \cdot 0.8 = 0.8^3 =0.512}$$
note
We have defined cubic roots as non-negative numbers although $${ \sqrt[3]{-8} = -2 }$$ would make sense. This is to avoid looking at the different cases with any kind of root.
exercises
side length of a square with area 2
The blue square has side lengths equal to 1 and therefore an area of 1. What area is the larger green square?
We can see the length of the sides of that square. Move the red point (glider) towards the x-axis to see how long it actually is. You can zoom in more and more if you like. The x-value of the point on the x-axis is an approximated value for the length of the side.
Actually it is not possible to determine all the decimal places and it is possible to prove that the square root of 2 can not be represented by a decimal number. It can only be approximated.
What does this length now actually represent? Is it a square root? And if so, what square root is it?
approximate measurement of drawing
to do
irrational numbers
to do
real numbers
to do
nested intervals
Intervallschachtelung und Halbierungsverfahren
Babylonian method (Heron-Verfahren)
Also known as Hero’s method or Heron-Verfahren zur näherungsweisen Bestimmung von Quadratwurzeln
long division
The process of long division is an algorithm of gradually approximating the root by finding the sides of a perfect square.
squaring and taking the square root
to do
rules Rechenregeln
Theorem (Satz)
For all $${a \in \mathbb{R} \space\space \text{following holds:}\space\space \sqrt{a^2} = | a |}$$
this means that solving the equation in the form $${ x^2 = 16}$$ leads to two solutions $${\sqrt{x^2} = \{ \begin{matrix} \sqrt{16} = 4 & \newline -\sqrt{16} = -4 & \end{matrix}}$$
multiplication
$${ \sqrt{a} \cdot \sqrt{b} = \sqrt{ab} }$$
division
$${ \frac{ \sqrt{a} } {\sqrt{b} } = \sqrt{\frac{a}{b}} }$$
simplifying roots - teilweise radizieren
When a prime factor appears twice under the square root it can be taken out as one factor.
$${ \sqrt{ac^2} = \sqrt{a{\cdot \color{orange}{c \cdot c}}} = \color{orange}{c}\cdot \sqrt{a} = \color{orange}{c}\sqrt{a}}$$
example
$${ \sqrt{12} = \sqrt{4\cdot 3} = \sqrt{2\cdot 2 \cdot 3} = 2 \sqrt{3}}$$
estimating roots after simplifying
Following table of roots of prime numbers will help find an estimate of a root.
| surd | estimate |
|---|---|
| $${\sqrt{2}}$$ | $${\approx 1.4}$$ |
| $${\sqrt{3}}$$ | $${\approx 1.7}$$ |
| $${\sqrt{5}}$$ | $${\approx 2.3}$$ |
| $${\sqrt{7}}$$ | $${\approx 2.6}$$ |
| $${\sqrt{11}}$$ | $${\approx 3.3}$$ |
| $${\sqrt{13}}$$ | $${\approx 3.6}$$ |
| $${\sqrt{17}}$$ | $${\approx 4.1}$$ |
Using the previous example.
$${ \sqrt{12} = \sqrt{4\cdot 3} = \sqrt{2\cdot 2 \cdot 3} = 2 \sqrt{3} \approx 2 \cdot 1.7 = 3.4}$$
rationalising the denominator - den Nenner rational machen
In the past, there were no tables with division by roots. So, in order to find the answer of such a problem the denominator had to be “rationalised”. And, with the root in the numerator it was possible to multiply the root by any other number using given tables.
$${ \frac{ \sqrt{a} } {\sqrt{b} } = \frac{ \sqrt{a} } {\sqrt{b}}\cdot \frac{ \sqrt{b} } {\sqrt{b} } = \frac{ \sqrt{a} \cdot \sqrt{b}} {b} = \frac{1} {b}\sqrt{ab}}$$
$${ \frac{ \sqrt{5} } {\sqrt{7} } = \frac{ \sqrt{5} } {\sqrt{7}}\cdot \frac{ \sqrt{7} } {\sqrt{7} } = \frac{ \sqrt{5} \cdot \sqrt{7}} {7} = \frac{1} {7}\sqrt{35}}$$
$${ \frac{ \sqrt{a} + \sqrt{b} } { \sqrt{a} - \sqrt{b} } = \frac{ \sqrt{a} + \sqrt{b} } { \sqrt{a} - \sqrt{b}} \cdot \frac{ \sqrt{a} + \sqrt{b} } { \sqrt{a} + \sqrt{b} } = \frac{ (\sqrt{a} + \sqrt{b}) \cdot (\sqrt{a} + \sqrt{b}) } { (\sqrt{a})^2 - (\sqrt{b})^2} = \frac{ (\sqrt{a} + \sqrt{b})^2 } { a - b} }$$
Here, you expand the fraction with the conjugate of the denominator so that you are forming the 3. binomische Formel (difference of squares) in the denominator. This eliminates the surd in the denominator after simplifying the expression.
$${\begin{align} \frac{ \sqrt{2} + \sqrt{3} } { \sqrt{2} - \sqrt{3} } & = \frac{ \sqrt{2} + \sqrt{3} } { \sqrt{2} - \sqrt{3}} \cdot \frac{ \sqrt{2} + \sqrt{3} } { \sqrt{2} + \sqrt{3} } \newline & = \frac{ (\sqrt{2} + \sqrt{3}) \cdot (\sqrt{2} + \sqrt{3}) } { (\sqrt{2})^2 - (\sqrt{3})^2} \newline & = \frac{ (\sqrt{2} + \sqrt{3})^2 } { 2 - 3} \newline & = \frac{ (\sqrt{2})^2+ 2 \cdot \sqrt{2} \cdot \sqrt{3} + (\sqrt{3})^2 } { 2 - 3} \newline & = \frac{ 2+ 2 \sqrt{6} + 3 } {-1} \newline & = - (5 + 2 \sqrt{6}) \newline & = - 5 - 2 \sqrt{6} \end{align}}$$
using the geometric mean (Höhensatz) to construct a root
Here are the instructions
using Pythagoras’ theorem to construct roots
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