Here's a problem that can easily be solved incorrectly if the order of operations is not used properly.
6 divided by 2(1+2)
The
answer is 9, some might think it's 1, but really it's 9. Do the work
in parenthesis first, so you have 6 divided by 2(3).. Now in the order
of operations multiplication comes before division.. IF you do it that
way you would get 1 because you'd multiply 2 and 3 to get 6, then 6
divided by 6 is 1. But the order of operations also states
multiplication and division is done from left to right, so the division
is first... IF the problem was 6/[2(1+2)] THEN it would be 6/[2(3)] =
6/6 = 1. But the problem doesn't state this.
Sunday, December 30, 2012
Thursday, December 27, 2012
Tuesday, December 25, 2012
We learned how to find a zero of a function using synthetic division. Now we will learn how to determine the number and nature of zeroes, (also known as roots) that a polynomial function has. To determine the number of positive and negative real roots, we use a technique founded by Rene Descartes, thus named Decartes' Rule of Signs.
To find the number of positive real roots, start with the sign of the coefficient of the term with the highest power of the variable. Count the sign changes as you proceed through the polynomial. The number of sign changes is the number of positive real roots or less than it by a multiple of two. For example, if there are 2 sign changes there are either 2 or (2 – 2) =
0 positive real roots.
Example: Find the number of positive real roots of f(x) = -4x^5 – 11x^4 + 2x^3 + 9x^2 - x + 3.
Starting with -4x^5 , there is a sign change at 2x^3, another at -x and a third at 3. Therefore there are 3 sign changes and either 3 or (3 – 2) = 1 positive real roots.
To find the number of negative real roots of f(x), substitute -x into f(x) to get f(-x). Then proceed in the same manner as you would when finding the number of positive real roots. In the above example,
f(-x) = -4(-x)^5 – 11(-x)^4 + 2(-x)^3 + 9(-x)^2 - (-x) + 3
f(-x) = 4x^5 - 11x^4 – 2x^3 + 9x^2 + x + 3
Starting with 4x^5 , there is a sign change at 11x^4 and another at 9x^2 . Therefore, there are 2 or (2 – 2) = 0 negative real roots.
• Note that the number of roots in a polynomial function equals the highest power. In the above example, there are 5 roots. The sum of the positive and negative roots will not always equal the highest power. In such cases, those polynomial functions have imaginary roots, which we will deal with in a later chapter.
Example: Find the number of positive and negative real roots of f(x) = 6x^4 – 9x^3 – 2x^2 + 5x – 5.
For positive real roots f(x) = 6x^4 – 9x^3 – 2x^2 + 5x – 5. Start with 6x4 and count the sign changes. Notice a change at -9x^3, another at 5x and a third at -5. Therefore there are 3 or (3 – 2) = 1 positive real roots.
For negative real roots f(-x) = 6(-x)^4 – 9(-x)^3 – 2(-x)^2 + 5(-x) – 5.
f(-x) = 6x^4 + 9x^3 – 2x^2 – 5x – 5. Start with 6x4 and count the sign changes. Notice a change at -2x^2 and no other changes. Therefore there is 1 negative real root.
To find the number of positive real roots, start with the sign of the coefficient of the term with the highest power of the variable. Count the sign changes as you proceed through the polynomial. The number of sign changes is the number of positive real roots or less than it by a multiple of two. For example, if there are 2 sign changes there are either 2 or (2 – 2) =
0 positive real roots.
Example: Find the number of positive real roots of f(x) = -4x^5 – 11x^4 + 2x^3 + 9x^2 - x + 3.
Starting with -4x^5 , there is a sign change at 2x^3, another at -x and a third at 3. Therefore there are 3 sign changes and either 3 or (3 – 2) = 1 positive real roots.
To find the number of negative real roots of f(x), substitute -x into f(x) to get f(-x). Then proceed in the same manner as you would when finding the number of positive real roots. In the above example,
f(-x) = -4(-x)^5 – 11(-x)^4 + 2(-x)^3 + 9(-x)^2 - (-x) + 3
f(-x) = 4x^5 - 11x^4 – 2x^3 + 9x^2 + x + 3
Starting with 4x^5 , there is a sign change at 11x^4 and another at 9x^2 . Therefore, there are 2 or (2 – 2) = 0 negative real roots.
• Note that the number of roots in a polynomial function equals the highest power. In the above example, there are 5 roots. The sum of the positive and negative roots will not always equal the highest power. In such cases, those polynomial functions have imaginary roots, which we will deal with in a later chapter.
Example: Find the number of positive and negative real roots of f(x) = 6x^4 – 9x^3 – 2x^2 + 5x – 5.
For positive real roots f(x) = 6x^4 – 9x^3 – 2x^2 + 5x – 5. Start with 6x4 and count the sign changes. Notice a change at -9x^3, another at 5x and a third at -5. Therefore there are 3 or (3 – 2) = 1 positive real roots.
For negative real roots f(-x) = 6(-x)^4 – 9(-x)^3 – 2(-x)^2 + 5(-x) – 5.
f(-x) = 6x^4 + 9x^3 – 2x^2 – 5x – 5. Start with 6x4 and count the sign changes. Notice a change at -2x^2 and no other changes. Therefore there is 1 negative real root.
Saturday, December 22, 2012
Many problems in algebra are in the form of words. In order to solve
them, we must know how to translate the words into an equation. The key
steps in solving a problem are analyzing the problem, forming the
equation, solving the equation and checking the result.
Words and phrases can represent different mathematical operations. It's important to be able to translate words and phrases into equations when solving word problems. Here are some common words and phrases and their associated mathematical operation.
John and Steve have been collecting baseball cards since 2003. They now have accumulated a total of 40,000 cards. If John has 7,500 more cards than Steve, how many cards do each of them have?
First, assign a variable for the number of cards that either John or Steve has.
Let x = number of cards that Steve has. Since John has 7,500 more cards than Steve, he has 7,500 + x cards. Therefore, the number of cards that Steve has + Number of cards that John has = total number of cards. The equation that must be solved then is x + 7,500 + x = 40,000.
Here's another example which can be applied to real life. Suppose you wish to decide between two long distance phone plans. The first plan is $0.20 for the first minute, $0.07 for each additional minute. The second plan is $0.15 for the first minute and $0.10 for each additional minute. How many minutes must you talk for the first plan to be cheaper than the second plan?
For the first plan, let m = total minutes. Therefore, the cost for first minute plus cost for each additional minute is 0.20 + 0.07(m - 1).
For the second plan, the cost for first minute plus cost for each additional minute is 0.15 + 0.10(m - 1). We set equations equal to each other and solve for m to see how many minutes will the cost be equal.
Suppose a school wants to have an enclosed rectangular playground and has enough money in the budget to afford 550 feet of fencing. If the length is to be 125 feet longer than the width, what are the dimensions of the fence that encloses the playground?
Let W = width of the fence. The length is W + 125, therefore 2 times the length + 2 times the width = 550. The equation we must solve is 2(W + 125) + 2W = 550. Once we get a value for W, substitute back into the equation to solve for the length.
The important thing to remember is to analyze the problem, assign variable, set up equations which make sense in the context of the problem and check solutions. Be sure to try and write the equation as you are reading the problem.
Words and phrases can represent different mathematical operations. It's important to be able to translate words and phrases into equations when solving word problems. Here are some common words and phrases and their associated mathematical operation.
- sum of, added to, increased, plus: addition
- minus, decreased, less than, reduced by: subtraction
- twice, product, multiplied by, times, of: multiplication
- quotient, divided by, ratio, into: division
John and Steve have been collecting baseball cards since 2003. They now have accumulated a total of 40,000 cards. If John has 7,500 more cards than Steve, how many cards do each of them have?
First, assign a variable for the number of cards that either John or Steve has.
Let x = number of cards that Steve has. Since John has 7,500 more cards than Steve, he has 7,500 + x cards. Therefore, the number of cards that Steve has + Number of cards that John has = total number of cards. The equation that must be solved then is x + 7,500 + x = 40,000.
Here's another example which can be applied to real life. Suppose you wish to decide between two long distance phone plans. The first plan is $0.20 for the first minute, $0.07 for each additional minute. The second plan is $0.15 for the first minute and $0.10 for each additional minute. How many minutes must you talk for the first plan to be cheaper than the second plan?
For the first plan, let m = total minutes. Therefore, the cost for first minute plus cost for each additional minute is 0.20 + 0.07(m - 1).
For the second plan, the cost for first minute plus cost for each additional minute is 0.15 + 0.10(m - 1). We set equations equal to each other and solve for m to see how many minutes will the cost be equal.
Suppose a school wants to have an enclosed rectangular playground and has enough money in the budget to afford 550 feet of fencing. If the length is to be 125 feet longer than the width, what are the dimensions of the fence that encloses the playground?
Let W = width of the fence. The length is W + 125, therefore 2 times the length + 2 times the width = 550. The equation we must solve is 2(W + 125) + 2W = 550. Once we get a value for W, substitute back into the equation to solve for the length.
The important thing to remember is to analyze the problem, assign variable, set up equations which make sense in the context of the problem and check solutions. Be sure to try and write the equation as you are reading the problem.
Wednesday, December 19, 2012
Here's a few examples of some word problems.
Michael has a budget of $1200 for the golf season. The course he plays costs $40 for a round of golf. He also wants to takes lessons which costs $70 each. He wants to play twice as many rounds as lessons taken.
How many rounds of golf can Michael play for the season? How many lessons can he have?
Solution:
Let x = Number of rounds played
x/2 = Number of lessons
Total cost of rounds played plus total number of lessons must be less than or equal to $1,200.
40x = Total cost of rounds played
70(x/2) = 35x = Total cost of lessons taken
40x + 35x ≤ 1,200
75x ≤ 1,200
x ≤ 16. Therefore, Michael can play 16 times and take 8 lessons for the season
The temperatures on a day in January satisfy the inequality |t - 30| ≤ 12. The temperatures on a day in August satisfy the equation |t - 70| ≤ 16. What are the temperature ranges for January and August and which month has the widest range of temperatures?
Solution:
Solve each inequality.
|t - 30| ≤ 12
Recall for absolute value there is a positive case and a negative case.
The positive case is t - 30 ≤ 12 and the negative case is -12 ≤ t – 30.
t - 30 ≤ 12, -12 ≤ t - 30
t ≤ 42 18 ≤ t
Therefore, the temperature ranges from 18 to 42 in January.
|t - 70| ≤ 16
t - 70 ≤ 16, -16 ≤ t - 70
t ≤ 86 54 ≤ t
Therefore, the temperature ranges from 54 to 86 in August.
The range for January temperatures is 24. (42 - 18 = 24)
The range for August temperatures is 32. (86 - 54 = 32)
Michael has a budget of $1200 for the golf season. The course he plays costs $40 for a round of golf. He also wants to takes lessons which costs $70 each. He wants to play twice as many rounds as lessons taken.
How many rounds of golf can Michael play for the season? How many lessons can he have?
Solution:
Let x = Number of rounds played
x/2 = Number of lessons
Total cost of rounds played plus total number of lessons must be less than or equal to $1,200.
40x = Total cost of rounds played
70(x/2) = 35x = Total cost of lessons taken
40x + 35x ≤ 1,200
75x ≤ 1,200
x ≤ 16. Therefore, Michael can play 16 times and take 8 lessons for the season
The temperatures on a day in January satisfy the inequality |t - 30| ≤ 12. The temperatures on a day in August satisfy the equation |t - 70| ≤ 16. What are the temperature ranges for January and August and which month has the widest range of temperatures?
Solution:
Solve each inequality.
|t - 30| ≤ 12
Recall for absolute value there is a positive case and a negative case.
The positive case is t - 30 ≤ 12 and the negative case is -12 ≤ t – 30.
t - 30 ≤ 12, -12 ≤ t - 30
t ≤ 42 18 ≤ t
Therefore, the temperature ranges from 18 to 42 in January.
|t - 70| ≤ 16
t - 70 ≤ 16, -16 ≤ t - 70
t ≤ 86 54 ≤ t
Therefore, the temperature ranges from 54 to 86 in August.
The range for January temperatures is 24. (42 - 18 = 24)
The range for August temperatures is 32. (86 - 54 = 32)
Sunday, December 16, 2012
Here's some tips for kids dealing with multiplication tables.
There is an important rule shown
in the multiplication tables. Any number multiplied by 0 is equal to
0, and is not included in the multiplication table. Any number
multiplied by 1 is itself.
Here
are the multiplication tables for 1 through 12
1
x 1 = 1 2 x 1 = 2 3 x 1 = 3 4 x 1 = 4
5 x 1 = 5 6 x 1 = 6 7 x 1 = 7
1
x 2 = 2 2 x 2 = 4 3 x 2 = 6 4 x 2 = 8
5 x 2 = 10 6 x 2 = 12 7 x 2 = 14
1
x 3 = 3 2 x 3 = 6 3 x 3 = 9 4 x 3 = 12
5 x 3 = 15 6 x 3 = 18 7 x 3 = 21
1
x 4 = 4 2 x 4 = 8 3 x 4 = 12 4 x 4 = 16
5 x 4 = 20 6 x 4 = 24 7 x 4 = 28
1
x 5 = 5 2 x 5 = 10 3 x 5 = 15 4 x 5 = 20
5 x 5 = 25 6 x 5 = 30 7 x 5 = 35
1
x 6 = 6 2 x 6 = 12 3 x 6 = 18 4 x 6 = 24
5 x 6 = 30 6 x 6 = 36 7 x 6 = 42
1
x 7 = 7 2 x 7 = 14 3 x 7 = 21 4 x 7 = 28
5 x 7 = 35 6 x 7 = 42 7 x 7 = 49
1
x 8 = 8 2 x 8 = 16 3 x 8 = 24 4 x 8 = 32
5 x 8 = 40 6 x 8 = 48 7 x 8 = 56
1
x 9 = 9 2 x 9 = 18 3 x 9 = 27 4 x 9 = 36
5 x 9 = 45 6 x 9 = 54 7 x 9 = 63
1
x 10 = 10 2 x 10 = 20 3 x 10 = 30 4 x 10 = 40 5 x
10 = 50 6 x 10 = 60 7 x 10 = 70
1
x 11 = 11 2 x 11 = 22 3 x 11 = 33 4 x 11 = 44 5
x 11 = 55 6 x 11 = 66 7 x 11 = 77
1
x 12 = 12 2 x 12 = 24 3 x 12 = 36 4 x 12 = 48 5 x
12 = 60 6 x 12 = 72 7 x 12 = 84
8
x 1 = 8 9 x 1 = 9 10 x 1 = 10 11 x 1 = 11
12 x 1 = 12
8
x 2 = 16 9 x 2 = 18 10 x 2 = 20 11 x 2 = 22
12 x 2 = 24
8
x 3 = 24 9 x 3 = 27 10 x 3 = 30 11 x 3 = 33
12 x 3 = 36
8
x 4 = 32 9 x 4 = 36 10 x 4 = 40 11 x 4 = 44
12 x 4 = 48
8
x 5 = 40 9 x 5 = 45 10 x 5 = 50 11 x 5 = 55
12 x 5 = 60
8
x 6 = 48 9 x 6 = 54 10 x 6 = 60 11 x 6 = 66
12 x 6 = 72
8
x 7 = 56 9 x 7 = 63 10 x 7 = 70 11 x 7 = 77
12 x 7 = 84
8
x 8 = 64 9 x 8 = 72 10 x 8 = 80 11 x 8 = 88
12 x 8 = 96
8
x 9 = 72 9 x 9 = 81 10 x 9 = 90 11 x 9 = 99
12 x 9 = 108
8
x 10 = 80 9 x 10 = 90 10 x 10 = 100 11 x 10 = 110
12 x 10 = 120
8
x 11 = 88 9 x 11 = 99 10 x 11 = 110 11 x 11 = 121
12 x 11 = 132
8
x 12 = 96 9 x 12 = 108 10 x 12 = 120 11 x 12 = 132
12 x 12 = 144
There are a few things to notice
when looking at the multiplication tables. Staring with the 1's,
each answer goes up by 1. We say that the multiples of 1 are
1,2,3,4,5,6, and so on.
For the 2's tables, each answer
goes up by 2. The multiples of 2 are 2,4,6,8,10,12, and so on.
For the 3's tables, each answer
goes up by 3. The multiples of 3 are 3,6,9,12,15,18, and so on.
The same pattern is true for all
the multiplication tables.
The answer to a multiplication
problem is also called the product.
Notice that any number multiplied
by 10 ends in 0.
It may seem hard to learn the
whole table, but notice that 1 x 2 is the same as 2 x 1, 3 x 2 is
the same as 2 x 3, and so on
Friday, December 14, 2012
When dealing with confidence intervals for proportions, the formula is
p^ +/- Zcritical(standard deviation)
Where p^ = r/n, r is the number of observations and n is the sample size
For a 95% confidence interval, Zcritical = 1.96
The standard deviation is square root[(p^)(1 - p^)/n]
The margin of error, denoted at E = Zcritical(standard deviation)
p^ +/- Zcritical(standard deviation)
Where p^ = r/n, r is the number of observations and n is the sample size
For a 95% confidence interval, Zcritical = 1.96
The standard deviation is square root[(p^)(1 - p^)/n]
The margin of error, denoted at E = Zcritical(standard deviation)
Wednesday, December 12, 2012
To find inflection points and concavity, take the second derivative and set equal to 0. Solve for x, then test a value on the left of the value for x and one on the right. If the second derivative of this value is less than zero, it's concave down on that interval, if the second derivative of this value is greater than zero, then it's concave up.
For example:
f(x) = 3x^3 + 2x^2 + 5x + 6
first derivative : f ' (x) = 9x^2 + 4x + 5
second derivative : f " (x) = 18x + 4
set the second derivative equal to 0 and solve for x
18x + 4 = 0
18x = -4
x = -4/18 = -2/9
Test a value to the left of -2/9, I choose -1
f " (-1) = -14
Test value to the right of -2/9, I choose 0
f " (0) = 4
Since f " (-1) is negative, the concavity is downward from negative infinity to -2/9
Since f " (0) is positive, the concavity is upward from -2/9 to infinity.
For example:
f(x) = 3x^3 + 2x^2 + 5x + 6
first derivative : f ' (x) = 9x^2 + 4x + 5
second derivative : f " (x) = 18x + 4
set the second derivative equal to 0 and solve for x
18x + 4 = 0
18x = -4
x = -4/18 = -2/9
Test a value to the left of -2/9, I choose -1
f " (-1) = -14
Test value to the right of -2/9, I choose 0
f " (0) = 4
Since f " (-1) is negative, the concavity is downward from negative infinity to -2/9
Since f " (0) is positive, the concavity is upward from -2/9 to infinity.
Monday, December 10, 2012
During a course of algebra, teachers discuss functions and composition of
functions. The topics can be confusing to many students, who also don't see any
practical uses beyond the classroom. The next few paragraphs will clear any
confusion you have on these topics.
Suppose we wish to represent the function g ( x ) by a token machine. The token machine yields one token for every quarter that is deposited into the machine. The tokens can be used to purchase prizes. We think of the quarters as the input x and the number of tokens as the output g ( x ). Suppose further that there is another machine that requires the use of tokens to obtain prizes. A certain number of tokens are needed to purchase each prize. We'll define the prize machine as f ( x ). The input is the number of tokens, which we defined as g ( x ). Notice that purchasing a prize out of the second machine is dependent on the number of tokens from the first machine. Such dependence can be interpreted in mathematical terms as composition of functions.
In the previous example, the domain x yields g ( x ), the number of tokens. Then g ( x ) becomes the input into f ( x ) to produce the output, which is the prize purchased. The end result is a composition function f º g , also noted as f ( g ( x )).
Next, notice the composition function f º g , also noted as f ( g ( x )), joining the two machines together as one machine which automatically deposits tokens into the prize generator, which ejects the appropriate prize corresponding to the number of tokens generated.
Here's a practical example using the composition of functions. A meteorologist predicts a low pressure area to move across the region over the next 36 hours. The current temperature of 60 degrees Fahrenheit is forecast to drop 1 degree every 3 hours. What is the composition function that expresses the Celsius temperature as a function of the number of hours from now? Note that expresses the Celsius temperature as a function of the number of hours from now? Note that C = (5/9)( F - 32).
Suppose we wish to represent the function g ( x ) by a token machine. The token machine yields one token for every quarter that is deposited into the machine. The tokens can be used to purchase prizes. We think of the quarters as the input x and the number of tokens as the output g ( x ). Suppose further that there is another machine that requires the use of tokens to obtain prizes. A certain number of tokens are needed to purchase each prize. We'll define the prize machine as f ( x ). The input is the number of tokens, which we defined as g ( x ). Notice that purchasing a prize out of the second machine is dependent on the number of tokens from the first machine. Such dependence can be interpreted in mathematical terms as composition of functions.
In the previous example, the domain x yields g ( x ), the number of tokens. Then g ( x ) becomes the input into f ( x ) to produce the output, which is the prize purchased. The end result is a composition function f º g , also noted as f ( g ( x )).
Next, notice the composition function f º g , also noted as f ( g ( x )), joining the two machines together as one machine which automatically deposits tokens into the prize generator, which ejects the appropriate prize corresponding to the number of tokens generated.
Here's a practical example using the composition of functions. A meteorologist predicts a low pressure area to move across the region over the next 36 hours. The current temperature of 60 degrees Fahrenheit is forecast to drop 1 degree every 3 hours. What is the composition function that expresses the Celsius temperature as a function of the number of hours from now? Note that expresses the Celsius temperature as a function of the number of hours from now? Note that C = (5/9)( F - 32).
To solve this we need to know the current
temperature and the rate of change of the temperature. We know the current
temperature is 60 and there is a 1/3 degree drop expected every hour. We will
represent time in hours since the temperature is 60 degrees at t. The
temperature in Fahrenheit at time t will be expressed by the function
F (t ). The temperature at time t expressed in Celsius will
be given by the composite function C (F ( t )).
The above example is just one application of composite functions in real life situations. The goal of the articles was to explain composition of functions as far as their structure is concerned and to show a real life application. I believe my explanation will clear questions one might have on these topics.
The above example is just one application of composite functions in real life situations. The goal of the articles was to explain composition of functions as far as their structure is concerned and to show a real life application. I believe my explanation will clear questions one might have on these topics.
Friday, December 7, 2012
On the long running hit game show, "Let's Make a Deal," Monty Hall would ask the
contestant which of three doors he or she wants. Behind each door is a prize,
only one of which is valuable. After the contestant chooses a door, Hall opened
one of the other doors and revealed a worthless prize. Then he would ask the
contestant if he or she wanted to switch doors. To maximize chances of winning,
one should always switch.
To analyze the problem, suppose I were to pick door number 1 and then was revealed door number 3 to be the worthless prize. Now I know the prize is either behind the door I chose or door number 2. It appears that the probability of winning is 50% and switching doesn't increase or decrease my chances.
Think of the problem this way. The probability that the prize is behind door number 1 given that the prize is not behind door number 3. The probability behind door number 1 equals 1/3 since there are 3 doors to choose from. The probability the prize is behind door number 1 or 2 knowing it is not behind door number 3 is 1/2.
The problem involves the use of conditional probability. Suppose you have events A and B, then the probability of A given B equals the probability of A and B divided by the probability of B. This is denoted as P(A/B) = P(A and B)/P(B).
Using the formula for conditional probability, you will find the probability of winning if switching is (1/3)/(1/2) = 2/3. Another way to think of this is that if you switch, the only way that you lose is if the prize was behind the door you initially picked, which has probability of winning 1/3. So the probability that you win is 1 - 1/3 = 2/3.
Although switching doesn't seem to give the best chance of winning, you can conduct an experiment. Play the game with someone assuming the role of Monty Hall and you'll find out as you increase the number of games you play, you'll win approximately 2 out 3 times.
To analyze the problem, suppose I were to pick door number 1 and then was revealed door number 3 to be the worthless prize. Now I know the prize is either behind the door I chose or door number 2. It appears that the probability of winning is 50% and switching doesn't increase or decrease my chances.
Think of the problem this way. The probability that the prize is behind door number 1 given that the prize is not behind door number 3. The probability behind door number 1 equals 1/3 since there are 3 doors to choose from. The probability the prize is behind door number 1 or 2 knowing it is not behind door number 3 is 1/2.
The problem involves the use of conditional probability. Suppose you have events A and B, then the probability of A given B equals the probability of A and B divided by the probability of B. This is denoted as P(A/B) = P(A and B)/P(B).
Using the formula for conditional probability, you will find the probability of winning if switching is (1/3)/(1/2) = 2/3. Another way to think of this is that if you switch, the only way that you lose is if the prize was behind the door you initially picked, which has probability of winning 1/3. So the probability that you win is 1 - 1/3 = 2/3.
Although switching doesn't seem to give the best chance of winning, you can conduct an experiment. Play the game with someone assuming the role of Monty Hall and you'll find out as you increase the number of games you play, you'll win approximately 2 out 3 times.
Tuesday, December 4, 2012
Sunday, December 2, 2012
Over the course of mathematics, we learn how to raise a quantity to an exponent.
We know, for example, that xn means we multiply x by itself n times.
But suppose n is zero. How do we multiply a number by itself zero times? We are
taught that any number raised to the zero power equals one, but rarely does
anyone explain why this is the case.
One way to explain the zero exponent phenomenon is to use rules for exponents. When multiplying like bases, exponents are added. For example, 35 x 30 = 3(5 + 0). Therefore 35 x 30 = 35, so 30 must equal 1.
When dividing like bases, exponents are subtracted. For example, (46)/(40) = 4(6 - 0). Therefore (46)/(40) = 46, so 40 must equal 1.
Another way to show that any base raised to the zero power is one is to examine some patterns. Notice the pattern in the following:
Finally, I can use the concept of limits to show that any number raised to the zero power is one. Suppose we take 5n, and we start with n = 1. We know that 51 = 1. Take the square root of 5, which is equivalent of 5(1/2), which is approximately 2.24. Now take a smaller value for n, such as 1/3, which gives us 5(1/3) = 1.71. Continue to take values for n smaller and smaller but not less than or equal to zero. You'll start to notice what is happening, 5(1/10) = 1.17 , 5(1/1000) = 1.002, 5(1/100000) = 1.00002. Notice how the result is getting closer and closer to 1. We say that the limit as n approaches 0 of 5n = 1.
I just explained three methods that clarify why any number raised to the zero power equals one. Next time someone is puzzled by this fact, you can explain the reason behind this somewhat vague and often explained topic in mathematics.
One way to explain the zero exponent phenomenon is to use rules for exponents. When multiplying like bases, exponents are added. For example, 35 x 30 = 3(5 + 0). Therefore 35 x 30 = 35, so 30 must equal 1.
When dividing like bases, exponents are subtracted. For example, (46)/(40) = 4(6 - 0). Therefore (46)/(40) = 46, so 40 must equal 1.
Another way to show that any base raised to the zero power is one is to examine some patterns. Notice the pattern in the following:
- 24 = 16
- 23 = 8
- 22 = 4
- 21 = 2
- 20 = ?
Finally, I can use the concept of limits to show that any number raised to the zero power is one. Suppose we take 5n, and we start with n = 1. We know that 51 = 1. Take the square root of 5, which is equivalent of 5(1/2), which is approximately 2.24. Now take a smaller value for n, such as 1/3, which gives us 5(1/3) = 1.71. Continue to take values for n smaller and smaller but not less than or equal to zero. You'll start to notice what is happening, 5(1/10) = 1.17 , 5(1/1000) = 1.002, 5(1/100000) = 1.00002. Notice how the result is getting closer and closer to 1. We say that the limit as n approaches 0 of 5n = 1.
I just explained three methods that clarify why any number raised to the zero power equals one. Next time someone is puzzled by this fact, you can explain the reason behind this somewhat vague and often explained topic in mathematics.
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