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Difference between revisions of "Exponentiation"
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== Introduction ==
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'''Exponentiation''' is an arithmetic operation, just like [[addition]], [[multiplication]], etc.  It is often written in the form <math>a^b</math>, where <math>b</math> is the exponent (or power) and <math>a</math> is the [[base]].
  
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To understand '''exponents''' and the '''exponential function''', let's first review how [[multiplication]] came about.  Let's say we wanted to add 3 ten times.  We could write this out as <math> 3 + 3 +3 + 3 +3 + 3 +3 + 3 +3 + 3</math> but this gets burdensome if we wanted to add, for example, 3 three hundred times.  Thus, we define the multiplication function (usually denoted <math>\times</math> or <math>\cdot</math>) such that <math> 3\times 300=3+3+\cdots+3</math> where there are 300 threes in the sum.  For integers, it is straightforward how multiplaction works.  We can then extend the notion of multiplication to non-integers.
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In the [[order of operations]], it is the second operation performed if a equation has [[parentheses]] or the first one performed when there is no parentheses.   
  
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Similarly, the exponential function is defined as the repitition of multiplication.  For example, writing out <math>3\cdot 3\cdot 3\cdot 3\cdot 3</math> can get boring fast.  So we define the exponential function to express this in a much more compact form so that the preceeding example can be written as <math>3^5</math> (read 3 to the 5th or 3 to the 5 power).  What this means is that we are multiplying 3 by itself 5 times.
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== Introduction ==
  
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Before we proceed, we define 3 terms:
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To understand how exponents arise, let's first review how we can build multiplication from addition.  Let's say we wanted to capture the notion of "the amount equal to 3, ten times."  We could write this out as <math> 3 + 3 +3 + 3 +3 + 3 +3 + 3 +3 + 3</math>, but this gets burdensome quickly: if we wanted to capture the idea of "the amount equal to two hundred 3s."  Thus, we define the multiplication function, usually denoted <math>\times</math> or <math>\cdot</math>, such that <math> 3\times 200=3+3+\ldots+3</math> where there are 200 threes in the sum.  This process (actually an [[induction | inductive]] definition) defines the operation of "multiplication by [[positive integer]]s."  We can then extend the notion of multiplication to non-integers.
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* exponent or power - in <math>4^6</math> the exponent is 6; this tells us how many times we multiply the 4
 
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* [[base]] - in <math>10^9</math> the base is 10; this tells us what we will be multiplying 9 times
 
  
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Our definition of exponentiation makes sense if the exponent is a positive integer.  How about negative integers such as <math>2^{-4}</math>? How do we multiply 2 by itself -4 times!?  Let's think about what a negative sign means a little more.  When we append a negative sign to a number (say 4, for example) we are basically saying go four units in the ''opposite'' direction.  So we want to do the opposite of multiplication four times.  In other words, we want to divide by 2 four times. Therefore <math> 2^{-4}=\frac 1{2^4}.</math>
+
Similarly, the exponentiation is defined as the repetition of multiplication.  For example, writing out <math>3\cdot 3\cdot 3\cdot 3\cdot 3</math> can get boring fast,  so we define the exponential function to express this in a much more compact form so that the preceeding example can be written as <math>3^5</math> (read 3 to the 5th or 3 to the 5 power). What this means is that we are multiplying 3 by itself 5 times.  The base is 3 (what is repeatedly multiplied) and the exponent (or power) is 5 (the number of times to repeat multiplication).
  
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It is also possible to extend the exponential function to all non-integers.
+
Our definition of exponentiation makes sense if the exponent is a positive integer.  How about negative integers such as <math>2^{-4}</math>?  How do we multiply 2 by itself -4 times!?  Let's think about what a negative sign means a little more.  When we append a negative sign to a number (say 4, for example), we are basically saying go four units in the ''opposite'' direction. We want to do the opposite of multiplication four times.  In other words, we want to divide by 2 four times.  Therefore, <math> 2^{-4}=\frac 1{2^4}.</math>
  
 
== Basic Properties ==
 
== Basic Properties ==
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Listed below are some important properties of exponents:
 
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# <math> b^x\cdot b^y = b^{x+y}</math>
 
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# <math> b^{-x}=\frac 1{b^x} </math>
 
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# <math> \frac{b^x}{b^y}=b^{x-y} </math>
 
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# <math> (b^x)^y = b^{xy} </math>
 
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# <math> (ab)^x = a^x b^x </math>
 
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# <math> b^0 = 1 </math> (if <math>b \neq 0</math>.  <math>0^0</math> is undefined.)
 
  
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Here are explanations of the properties listed above:
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Listed below are some important properties of exponents (with explanations and notes):
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# On both sides we are multiplying '''b''' together '''x+y''' times. Thus, they are equivalent.  
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* <math> b^x\cdot b^y = b^{x+y}</math>
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# This is described in the previous section.
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** We are multiplying <math>b</math> together <math>x</math> times and then <math>y</math> times, so in total, we are multiplying <math>b</math> together <math>x+y</math> times.
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# This results from using the previous two properties.
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* <math>b^0 = 1</math> if <math>b \neq 0</math>
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# We are multiplying <math>b^x</math> by itself '''y''' times which is the same as multiplying '''b''' by itself '''xy''' times.
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** From the property of same base multiplication, <math>b^x \cdot b^0 = b^{x+0} = b^x</math>, so dividing both sides by <math>b^x</math> results in <math>b^0 = 1</math>.
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# After multiplying '''ab''' by itself '''x''' times we can collect '''a''' and '''b''' terms thus establishing the property.
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** <math>0^0</math> is undefined.
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#Hoping that property #1 will be true when <math>y=0</math>, we see that <math>b^x\cdot b^0</math> should (hopefully) be equal to <math>b^x</math>.  Thus, we ''define'' <math>b^0</math> to be equal to <math>1</math> in order to make this be true.
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* <math> b^{-x} = \frac{1}{b^x}</math> if <math>b \neq 0</math>
 +
** From the property of same base multiplication and zero exponent, <math>b^x \cdot b^{-x} = b^{x + -x} = b^0 = 1</math>, so dividing both sides by <math>b^x</math> results in <math>b^{-x} = \frac{1}{b^x}</math>.
 +
* <math>\frac{b^x}{b^y} = b^{x-y}</math>
 +
** This results from using the property of same base multiplication and negative exponent property.
 +
* <math> (b^x)^y = b^{xy} </math>
 +
** We are multiplying <math>b^x</math> by itself <math>y</math> times, and because of the first property, we are multiplying <math>b</math> by itself <math>x \cdot y</math> times.
 +
* <math> (ab)^x = a^x b^x </math>
 +
** We are multiplying <math>ab</math> together <math>x</math> times, which is the same as multiplying <math>a</math> together <math>x</math> times and <math>b</math> together <math>x</math> times.
  
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== Fractional Exponents ==
  
 +
So far we discussed expressions with integer exponents.  However, it is also possible to extend the exponential function to all non-integers.
  
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== Fractional exponents ==
 
  
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If <math>b</math> is a number and each of <math>x</math> and <math>y</math> is a positive integer, then as explained above (property 1) <math>b^x b^y = b^{x+y}</math>. For example, <math>b^2 b^3 = (b\cdot b)(b\cdot b \cdot b) = b^5</math>.
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How could we make sense of an expression like <math>b^{\frac{1}{2}}</math>? If you don't already know the answer, this is a good exercise; I recommend puzzling over it for awhile.
  
  
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How could we make sense of an expression like "<math>b^0</math>"?  Well, hoping that property 1 will remain true when <math>y=0</math>, we see that <math>b^x b^0</math> should (hopefully) be equal to <math>b^{x+0}=b^x</math>.  For that reason, we ''define'' <math>b^0 = 1</math>, in order to make that be true.  (And we only make this definition in the case where <math>b \neq 0</math>.  We choose to leave <math>0^0</math> undefined.)
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Answer: Hoping that property 1 will remain true when <math>x</math> or <math>y</math> is a fraction, we see that <math>b^{\frac{1}{2}} b^{\frac{1}{2}}</math> should (hopefully) be equal to <math>b^{\frac{1}{2} + \frac{1}{2}} = b^1 = b</math>.  Thus, we ''define'' <math>b^{\frac{1}{2}}</math> to be <math>\sqrt{b}</math>, in order to make this be true.
  
  
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We can make sense of an expression like "<math>b^{-5}</math>" in a similar way.  Hoping that property 1 will remain true even if <math>x</math> or <math>y</math> is negative, we see that <math>b^5 b^{-5}</math> should (hopefully) be equal to <math>b^{5 + -5} = b^0 = 1</math>.  Thus, we ''define'' <math>b^{-5}</math> to be <math>\frac{1}{b^5}</math>, in order to make this be true.  Similarly, if <math>x</math> is a positive integer, we define <math>b^{-x}</math> to be <math>\frac{1}{b^x}</math>.  (This depends on having <math>b \neq 0</math>.  Otherwise we'd be dividing by <math>0</math>.)
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For the time being, how to deal with other fractions in the exponent can be an exercise for the reader.
  
 +
Hint: What would <math>b^{\frac{1}{3}}</math> be? What about <math>b^{\frac{1}{4}}</math>? Do you notice anything?
  
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How could we make sense of an expression like <math>b^{\frac{1}{2}}</math>? If you don't already know the answer, this is a good exercise; I recommend puzzling over it for awhile.
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Try to figure out <math>b^{\frac{2}{3}}</math> -- how does it relate to <math>b^{\frac{1}{3}}</math>?
  
 +
==Problems==
  
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Answer: Hoping that property 1 will remain true when <math>x</math> or <math>y</math> is a fraction, we see that <math>b^{\frac{1}{2}} b^{\frac{1}{2}}</math> should (hopefully) be equal to <math>b^{\frac{1}{2} + \frac{1}{2}} = b^1 = b</math>. Thus, we ''define'' <math>b^{\frac{1}{2}}</math> to be <math>\sqrt{b}</math>, in order to make this be true.
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* Practice Problems on [https://artofproblemsolving.com/alcumus/ Alcumus]
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** General Positive Exponents (Prealgebra)
−
 
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** Negative and Zero Exponents (Prealgebra)
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For the time being, how to deal with other fractions in the exponent can be an exercise for the reader.
 
−
 
 
  
 
== See also ==
 
== See also ==
 
* [[Logarithms]]
 
* [[Logarithms]]
 
* [[Algebra]]
 
* [[Algebra]]
 +
* [[Hyperexponentiation]]
 +
* [[Radical]]

Latest revision as of 15:25, 10 August 2020

Exponentiation is an arithmetic operation, just like addition, multiplication, etc. It is often written in the form $a^b$, where $b$ is the exponent (or power) and $a$ is the base.

In the order of operations, it is the second operation performed if a equation has parentheses or the first one performed when there is no parentheses.

Introduction

To understand how exponents arise, let's first review how we can build multiplication from addition. Let's say we wanted to capture the notion of "the amount equal to 3, ten times." We could write this out as $3 + 3 +3 + 3 +3 + 3 +3 + 3 +3 + 3$, but this gets burdensome quickly: if we wanted to capture the idea of "the amount equal to two hundred 3s." Thus, we define the multiplication function, usually denoted $\times$ or $\cdot$, such that $3\times 200=3+3+\ldots+3$ where there are 200 threes in the sum. This process (actually an inductive definition) defines the operation of "multiplication by positive integers." We can then extend the notion of multiplication to non-integers.

Similarly, the exponentiation is defined as the repetition of multiplication. For example, writing out $3\cdot 3\cdot 3\cdot 3\cdot 3$ can get boring fast, so we define the exponential function to express this in a much more compact form so that the preceeding example can be written as $3^5$ (read 3 to the 5th or 3 to the 5 power). What this means is that we are multiplying 3 by itself 5 times. The base is 3 (what is repeatedly multiplied) and the exponent (or power) is 5 (the number of times to repeat multiplication).

Our definition of exponentiation makes sense if the exponent is a positive integer. How about negative integers such as $2^{-4}$? How do we multiply 2 by itself -4 times!? Let's think about what a negative sign means a little more. When we append a negative sign to a number (say 4, for example), we are basically saying go four units in the opposite direction. We want to do the opposite of multiplication four times. In other words, we want to divide by 2 four times. Therefore, $2^{-4}=\frac 1{2^4}.$

Basic Properties

Listed below are some important properties of exponents (with explanations and notes):

  • $b^x\cdot b^y = b^{x+y}$
    • We are multiplying $b$ together $x$ times and then $y$ times, so in total, we are multiplying $b$ together $x+y$ times.
  • $b^0 = 1$ if $b \neq 0$
    • From the property of same base multiplication, $b^x \cdot b^0 = b^{x+0} = b^x$, so dividing both sides by $b^x$ results in $b^0 = 1$.
    • $0^0$ is undefined.
  • $b^{-x} = \frac{1}{b^x}$ if $b \neq 0$
    • From the property of same base multiplication and zero exponent, $b^x \cdot b^{-x} = b^{x + -x} = b^0 = 1$, so dividing both sides by $b^x$ results in $b^{-x} = \frac{1}{b^x}$.
  • $\frac{b^x}{b^y} = b^{x-y}$
    • This results from using the property of same base multiplication and negative exponent property.
  • $(b^x)^y = b^{xy}$
    • We are multiplying $b^x$ by itself $y$ times, and because of the first property, we are multiplying $b$ by itself $x \cdot y$ times.
  • $(ab)^x = a^x b^x$
    • We are multiplying $ab$ together $x$ times, which is the same as multiplying $a$ together $x$ times and $b$ together $x$ times.

Fractional Exponents

So far we discussed expressions with integer exponents. However, it is also possible to extend the exponential function to all non-integers.


How could we make sense of an expression like $b^{\frac{1}{2}}$? If you don't already know the answer, this is a good exercise; I recommend puzzling over it for awhile.


Answer: Hoping that property 1 will remain true when $x$ or $y$ is a fraction, we see that $b^{\frac{1}{2}} b^{\frac{1}{2}}$ should (hopefully) be equal to $b^{\frac{1}{2} + \frac{1}{2}} = b^1 = b$. Thus, we define $b^{\frac{1}{2}}$ to be $\sqrt{b}$, in order to make this be true.


For the time being, how to deal with other fractions in the exponent can be an exercise for the reader.

Hint: What would $b^{\frac{1}{3}}$ be? What about $b^{\frac{1}{4}}$? Do you notice anything?

Try to figure out $b^{\frac{2}{3}}$ -- how does it relate to $b^{\frac{1}{3}}$?

Problems

  • Practice Problems on Alcumus
    • General Positive Exponents (Prealgebra)
    • Negative and Zero Exponents (Prealgebra)

See also

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