3.1 Roots - Revision history

Phil: Further minor changes to notation and language.

2009-01-02T16:08:03Z

Further minor changes to notation and language.

Phil at 15:59, 2 January 2009

2009-01-02T15:59:16Z

←Older revision		Revision as of 15:59, 2 January 2009
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	The well-known symbol <math>\sqrt{a}</math>, the square root of <math>a</math>, is used to describe the number that when multiplied by itself gives <math>a</math>. However, one has to be a little more precise in defining this symbol.		The well-known symbol <math>\sqrt{a}</math>, the square root of <math>a</math>, is used to describe the number that when multiplied by itself gives <math>a</math>. However, one has to be a little more precise in defining this symbol.

-	The equation <math>x^2 = 4</math> has two solutions <math>x = 2</math> and <math>x = -2</math>, since both <math>2\~~cdot~~ 2 = 4</math> and <math>(-2)\~~cdot~~(-2) = 4</math>. It would then be logical to suppose that <math>\sqrt{4}</math> can be either <math>-2</math> or <math>2</math>, i.e. <math>\sqrt{4}= \pm 2</math>, but by convention, <math>\sqrt{4}</math> only denotes the positive number <math>2</math>.	+	The equation <math>x^2 = 4</math> has two solutions <math>x = 2</math> and <math>x = -2</math>, since both <math>2\times 2 = 4</math> and <math>(-2)\times(-2) = 4</math>. It would then be logical to suppose that <math>\sqrt{4}</math> can be either <math>-2</math> or <math>2</math>, i.e. <math>\sqrt{4}= \pm 2</math>, but by convention, <math>\sqrt{4}</math> only denotes the positive number <math>2</math>.


	<div class="regel">		<div class="regel">
	The square root <math>\sqrt{a}</math> means '''the non-negative number''' that, when multiplied by		The square root <math>\sqrt{a}</math> means '''the non-negative number''' that, when multiplied by
-	itself, gives <math>a,</math>; that is, the non-negative solution of the equation <math>x^2 = a</math>.	+	itself, gives <math>a</math>; that is, the non-negative solution of the equation <math>x^2 = a</math>.

	The square root of <math>a</math> can also be written as <math>a^{1/2}</math>.		The square root of <math>a</math> can also be written as <math>a^{1/2}</math>.

Phil: Clarifying my own explanation of rationalising the denominator

2009-01-02T13:05:55Z

Clarifying my own explanation of rationalising the denominator

←Older revision		Revision as of 13:05, 2 January 2009
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	which is usually preferable. This is called rationalising the denominator.		which is usually preferable. This is called rationalising the denominator.

-	In slightly more complicated cases, we can use the difference of two squares, <math>(a+b)(a-b) = a^2 – b^2</math>, to eliminate the root from the denominator. The trick is this: if the denominator is <math>p+~~q\sqrt r~~</math>, then we multiply the numerator and denominator by <math>p-~~q\sqrt r~~</math>. For example,	+	In slightly more complicated cases, we can use the difference of two squares, <math>(a+b)(a-b) = a^2 – b^2</math>, to eliminate the root from the denominator. The trick is this: if the denominator is of the form <math>a+b</math>, where either <math>a</math> or <math>b</math> (or both) contains a square root, then we multiply the numerator and denominator by <math>a-b</math>; if the denominator is of the form <math>a-b</math>, we multiply by <math>a+b</math>. For example,

	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}

Phil at 12:53, 2 January 2009

2009-01-02T12:53:25Z

←Older revision		Revision as of 12:53, 2 January 2009
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	which is usually preferable. This is called rationalising the denominator.		which is usually preferable. This is called rationalising the denominator.

-	In slightly more complicated cases, we can use the difference of two squares, <math>(a+b)(a-b) = a^2 – b^2</math>, to eliminate the root from the denominator. The trick is this: if the denominator is <math>a+b\sqrt c</math>, then we multiply the numerator and denominator by <math>a-b\sqrt c</math>. For example,	+	In slightly more complicated cases, we can use the difference of two squares, <math>(a+b)(a-b) = a^2 – b^2</math>, to eliminate the root from the denominator. The trick is this: if the denominator is <math>p+q\sqrt r</math>, then we multiply the numerator and denominator by <math>p-q\sqrt r</math>. For example,

	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}

Phil at 12:46, 2 January 2009

2009-01-02T12:46:42Z

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	Of course, numbers like <math>\sqrt2</math> can be expressed ''approximately'' as decimals, but in fact it is often preferable to leave expressions like <math>\sqrt2</math> unevaluated, at least until the end of a calculation; that way, there is no loss of accuracy along the way.		Of course, numbers like <math>\sqrt2</math> can be expressed ''approximately'' as decimals, but in fact it is often preferable to leave expressions like <math>\sqrt2</math> unevaluated, at least until the end of a calculation; that way, there is no loss of accuracy along the way.

-	Expressions containing unevaluated irrational square roots, or more generally <math>n</math> roots, are often called surds.	+	Expressions containing unevaluated irrational square roots, or more generally <math>n</math>th roots, are often called surds.

Phil: Edited for UK idiom, notation and terminology: section on surds added.

2009-01-02T12:40:40Z

Edited for UK idiom, notation and terminology: section on surds added.

(Difference between revisions)

Jdinglis at 19:32, 26 October 2008

2008-10-26T19:32:47Z

←Older revision		Revision as of 19:32, 26 October 2008
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	For those of you who want to deepen your understanding or need more detailed explanations consider the following references		For those of you who want to deepen your understanding or need more detailed explanations consider the following references

-	[http://en.wikipedia.org/wiki/Root_(mathematics) Learn more about square roots ~~in the English~~ Wikipedia ]	+	[http://en.wikipedia.org/wiki/Root_(mathematics) Learn more about square roots from Wikipedia ]

	[http://www.mathacademy.com/pr/prime/articles/irr2/ How do we know that the square root of 2 is not a fraction?]		[http://www.mathacademy.com/pr/prime/articles/irr2/ How do we know that the square root of 2 is not a fraction?]

Jdinglis at 14:01, 26 October 2008

2008-10-26T14:01:39Z

←Older revision		Revision as of 14:01, 26 October 2008
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	'''Study advice'''		'''Study advice'''

-	'''Basic and final tests:'''	+	'''Basic and final tests'''

	After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.		After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.

-	'''Keep in mind that: '''	+	'''Keep in mind that... '''

	The square root of a number is always non-negative (that is, positive or zero)!		The square root of a number is always non-negative (that is, positive or zero)!
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-	'''Reviews:'''	+	'''Reviews'''

	For those of you who want to deepen your understanding or need more detailed explanations consider the following references		For those of you who want to deepen your understanding or need more detailed explanations consider the following references
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-	'''Useful web sites:'''	+	'''Useful web sites'''

	[http://mathforum.org/dr.math/faq/faq.sqrt.by.hand.html How to find the root of a number, without the help of calculators?]		[http://mathforum.org/dr.math/faq/faq.sqrt.by.hand.html How to find the root of a number, without the help of calculators?]
	</div>		</div>

Jdinglis at 17:04, 24 October 2008

2008-10-24T17:04:24Z

←Older revision		Revision as of 17:04, 24 October 2008
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	[http://en.wikipedia.org/wiki/Root_(mathematics) Learn more about square roots in the English Wikipedia ]		[http://en.wikipedia.org/wiki/Root_(mathematics) Learn more about square roots in the English Wikipedia ]

-	[http://www.mathacademy.com/pr/prime/articles/irr2/ How do we know that the ~~<math>\sqrt{~~2~~}</math>~~ is not a fraction?]	+	[http://www.mathacademy.com/pr/prime/articles/irr2/ How do we know that the square root of 2 is not a fraction?]

Jdinglis at 17:03, 24 October 2008

2008-10-24T17:03:03Z

←Older revision		Revision as of 17:03, 24 October 2008
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	The square root of a number is always non-negative (that is, positive or zero)!		The square root of a number is always non-negative (that is, positive or zero)!

-	Rules for roots are actually a special case of laws of exponents, since, for example, <math>\sqrt{x} = x^{1/2}<\math>.	+	Rules for roots are actually a special case of laws of exponents, since, for example, <math>\sqrt{x} = x^{1/2}</math>.

←Older revision		Revision as of 16:08, 2 January 2009
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	<ol type="a">		<ol type="a">
-	<li><math>\sqrt{0}=0 \quad</math> because <math>0^2 = 0 \~~cdot~~ 0	+	<li><math>\sqrt{0}=0 \quad</math> because <math>0^2 = 0 \times 0
	= 0</math> and <math>0</math> is not negative. </li>		= 0</math> and <math>0</math> is not negative. </li>
-	<li><math>\sqrt{100}=10 \quad</math> since <math> 10^2 = 10 \~~cdot~~ 10	+	<li><math>\sqrt{100}=10 \quad</math> since <math> 10^2 = 10 \times 10
	= 100 </math> and <math>10</math> is a positive number. </li>		= 100 </math> and <math>10</math> is a positive number. </li>
	<li> <math>\sqrt{0_.25}=0_.5 \quad</math> since <math>0_.5^2		<li> <math>\sqrt{0_.25}=0_.5 \quad</math> since <math>0_.5^2
-	= 0_.5 \~~cdot~~ 0_.5 = 0_.25 </math> and <math>0{_.}5</math> is positive. </li>	+	= 0_.5 \times 0_.5 = 0_.25 </math> and <math>0{_.}5</math> is positive. </li>
	<li><math>\sqrt{2} \approx 1{_.}4142 \quad</math> since <math>1{_.}4142		<li><math>\sqrt{2} \approx 1{_.}4142 \quad</math> since <math>1{_.}4142
-	\~~cdot~~ 1{_.}4142 \approx 2</math> and <math>1{_.}4142</math> is positive. </li>	+	\times 1{_.}4142 \approx 2</math> and <math>1{_.}4142</math> is positive. </li>
	<li> The equation <math>x^2=2</math> has the solutions <math>x=\sqrt{2}		<li> The equation <math>x^2=2</math> has the solutions <math>x=\sqrt{2}
	\approx 1{_.}4142</math> and <math>x = -\sqrt{2} \approx -1{_.}4142</math>.</li>		\approx 1{_.}4142</math> and <math>x = -\sqrt{2} \approx -1{_.}4142</math>.</li>
	<li><math>\sqrt{-4}\quad</math> is not defined, since there is no real number <math>x</math> that satisfies <math>x^2=-4</math>.</li>		<li><math>\sqrt{-4}\quad</math> is not defined, since there is no real number <math>x</math> that satisfies <math>x^2=-4</math>.</li>
	<li><math>\sqrt{(-7)^2} = 7 \quad</math> because <math> \sqrt{(-7)^2}		<li><math>\sqrt{(-7)^2} = 7 \quad</math> because <math> \sqrt{(-7)^2}
-	= \sqrt{(-7) \~~cdot~~ (-7)} = \sqrt{49} = \sqrt{ 7 \~~cdot~~ 7} = 7</math>.</li>	+	= \sqrt{(-7) \times (-7)} = \sqrt{49} = \sqrt{ 7 \times 7} = 7</math>.</li>
	</ol>		</ol>
	</div>		</div>
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	<li><math>\sqrt[\scriptstyle 5]{-243} = -3\quad</math> because <math>(-3)		<li><math>\sqrt[\scriptstyle 5]{-243} = -3\quad</math> because <math>(-3)
	\times (-3) \times (-3) \times (-3) \times (-3) = -243</math>.</li>		\times (-3) \times (-3) \times (-3) \times (-3) = -243</math>.</li>
-	<li><math>\sqrt[\scriptstyle 6]{-17}\quad</math> is not defined as <math>6</math> is even and <math>-17</math> is a negative number. </li>	+	<li><math>\sqrt[\scriptstyle 6]{-17}\quad</math> is not defined, since <math>6</math> is even and <math>-17</math> is a negative number. </li>
	</ol>		</ol>
	</div>		</div>
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	{{Displayed math\|\|<math>\begin{align*}		{{Displayed math\|\|<math>\begin{align*}
	\sqrt[\scriptstyle n]{ab}		\sqrt[\scriptstyle n]{ab}
-	&= \sqrt[\scriptstyle n]{\vphantom{b}a}\~~cdot~~	+	&= \sqrt[\scriptstyle n]{\vphantom{b}a}\times
	\sqrt[\scriptstyle n]{b}\\[4pt]		\sqrt[\scriptstyle n]{b}\\[4pt]
	\sqrt[\scriptstyle n]{\frac{a}{b}}		\sqrt[\scriptstyle n]{\frac{a}{b}}