Transcript Mathematics as a Second Language

Taking the Fear
out of Math
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#10
2-8
Extending
the
Definition
of
Exponents
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In illustrating the use of positive
integer exponents, we have chosen
several different applications including
how to compute the number of outcomes
when a coin is flipped a certain number
of times.
This was a good application to use
because a coin can only be flipped a
whole number of times. It makes no
sense to talk about what happens, for
example, when a coin is flipped
“negative 2” times.
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However, there are other applications in
which we have occasion to use
exponents that include other integers
(that is, 0 and the negative integers).
One such application is when we want to
extend the use of exponential notation
to represent the powers of 10 when we
deal with decimals numbers.
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When all we had were the positive
integers, the denominations were
represented as shown below…
Place Value
Notation
Exponential
Notation
10,000
1,000
100
10
104
103
102
101
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As we have mentioned in previous
lessons, definitions and rules are often
based on things we would like to be true
or patterns we would like to continue.
Notice that in the
table, each number
is divided by 10 to
get to the number
below it.”
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Place Value
Notation
10,000
1,000
100
Exponential
Notation
104
103
102
10
101
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Therefore, if we want this pattern to
continue, the next entries would have to be
based on the facts that 10 ÷ 10 = 1,
1 ÷ 10 = 1/10; 1/10 ÷ 10 = 1/100 = 1/102 etc.
Thus, the
extension of
the first
column
would have
to look like.
Place Value
Notation
10,000
1,000
100
10
1
1/
10
1/
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1/
100
1000
Exponential
Notation
104
103
102
101
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Again, assuming that we want the
same pattern to continue, we notice that
as we read down the second column, the
exponent decreases by 1 from its value
in the row above.
In other words, the sequence of
exponents would look like…
104, 103, 102, 101, 100, 10-1, 10-2, 10-3, etc…
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Our chart would then look like…
Place Value
Notation
10,000
1,000
100
Exponential
Notation
104
103
102
10
101
1
1/
10
100
10-1
1/
100
10-2
1000
10-3
1/
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So if we want the chart to continue in the
same way, what must be true is that…
100 = 1
10-1 = 1/101
10-2 = 1/100 = 1/10 2
10-3 = 1/1000 = 1/103
In other words, 10n and 10-n are reciprocals
of one another. That is, if n is any
positive integer 10-n = 1 ÷ 10n
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Notes
Notice that defining 100 to be 1 is
reasonable in the sense that if n is any
positive integer, 10n is a 1 followed by n
zeroes. So it seems natural that if the
exponent was 0, the number would be a 1
followed by no 0’s (that is, 1).
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Notes
We know that to multiply a decimal number
by 1,000, we move the decimal point
3 places to the right, and to divide a
decimal number by 1,000 we move the
decimal point 3 places to the left. In that
context, it seems natural that if 10+3 tells us
to move the decimal point 3 places to the
right, that 10-3 should tell us to move the
decimal point 3 places to the left.
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Notes
However, mathematicians have a better
reason for defining integer exponents
the way we do. Without going into the
reasons behind the decision, it turns out
that there are good reasons to make sure
that the rules that govern the arithmetic of
non-zero whole number exponents should
also apply to any other exponents as well.
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With this in mind, let’s see how we would
have to define b0. We already know that for
positive integer exponents
bm × bn = bm+n.
So if we were to let n = 0, the rule would
become…
bm × b0 = bm+0
Since m + 0 = m, this would mean that…
bm × b0 = bm
If we now divide both sides of the above
equation by bm, we see that b0 = 1.1
next
note
1 If b = 0 then bm is also equal to 0. However, we are not allowed to divide by 0.
Therefore, we have to add the restriction that b ≠ 0.
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Note
Another way of saying this is to
observe that since bm × b0 = bm, this
equation tells us that b0 is that number
which when multiplied by bm yields bm as
the product, and this is precisely what it
means to multiply a number by 1.
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Note
By way of review, the reason we had to add
the restriction that b ≠ 0 when defining b0 is
based on the fact that any number multiplied
by 0 is 0. Notice that if we replace b by 0 in the
equation bm × b0 = bm, we obtain the result
that 03 × 00 = 03.
Since we know that 03 = 0, this says that
0 × 00 = 0. But, any number times 0 is 0!
Therefore 00 can equal any number. When
this happens, we say that the value of 00 is
indeterminate; meaning that it can be any
number.
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For example, suppose that for some
“strange” reason we wanted to define
00 to be 7. If we replace 00 by 7, 0 × 00 = 0
becomes 0 × 7 = 0, which is
a true statement.
For b ≠ 0, we have defined b0 to be 1. It
does not have to be defined to equal 1, but
if we don’t define b0 to equal 1, then we
cannot use the rule bm × bn = bm+n if either
m or n is equal to 0. In other words, by
electing to let b0 = 1, we are still allowed to
use the rule for multiplying like bases.
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Key Point
Students often feel that b0 should equal 0
because there are no factors of b. What
we can tell students when this happens is
that there is nothing wrong with thinking
that it should be 0, but if we let it equal 0,
we lose the use of the rules that make
computing so convenient.
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Perhaps it will make it easier for students to
feel comfortable with our defining b0 to be
equal 1 if we point out that since bn × 1 = bn
to the exponent, n, tells us the number of
times we multiply 1 by b. If we don’t
multiply 1 by b, then we still have 1. As we
have already discussed, this is especially
easy to visualize when b = 10. Namely, in
this case 10n is a 1 followed by n zeroes.
Thus, 100 would mean a 1 followed by no 0’s,
which is simply 1.
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In the case where n = 0, 20 would
be the number of outcomes that can
occur if we don’t flip the coin at all.
There is ONE outcome - the current
state of the coin.
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One reason that mathematicians prefer
to work abstractly is that physical
models do not always exist, and even
when they exist, they might not make
sense in some instances.
For example, we cannot use
the “flipping a coin” model to explain the
use of negative exponents. However, as we
have explained earlier, negative exponents
make sense when we are dealing with
powers of 10.
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The “powers of 10” model gives us a
clue as to why b-n = 1/bn might still be a
correct rule to use even if b ≠ 10. More
specifically, we have already given a
plausible explanation as to why this is true
in the case where b = 10. So based on the
fact that 10-n = 1/10n, we might want to
conjecture that for any base b, b-n = 1/bn.
A more consistent reason might be that we
still want bm × bn = bm+n to be true even
when m and/or n is a negative integer.
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To this end we already know that when
we multiply like bases we add the
exponents, and we also know that for any
number, n, n + -n = 0.
Therefore…
bn × b-n = bn + -n = b0
And since we have already accepted the
definition that b0 = 1, it follows that…
bn × b-n = bn + -n = b0 = 1
So if now we divide both sides of the
equality bn × b-n = 1 by bn, it follows that
b-n = 1 ÷ bn.
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More generally, if m and n are any
integers and b and c are any numbers,
then it is still true that…
(1)
bm × bn = bm+n
(2) b0 = 1 provide that b ≠ 0; 00 is indeterminate2
(3)
bn = 1 ÷ b-n
(4)
bm ÷ bn = bm-n
(5)
(bm)n = bm×n
(6)
(b × c)n = bn × cn
note
2 Whenever an exponent is 0 or negative the base b cannot equal 0.
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To help you internalize the rules let’s
suppose that you wanted to rewrite the
expression 102 × 10-5 as a decimal number.
By Rule (1) (with b = 10, m = 2 and n = -5)
102 × 10-5 = 10-3 = 0.001. As a check, notice
that…
102 × 10-5 = 102 × 1/10-5
= 100 × 1/100,000
= 100/100,000
= 1/1000 = 0.001
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An Enrichment Note
on
Integer Exponents
In the same way that something may
increase exponentially, another thing might
decrease exponentially. One such example
is in terms of radioactive decay where we
talk about the half-life of a radioactive
substance. The half life is the amount of
time it takes for the substance to “shrink”
to half its present weight (mass).
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Rather than talk about radioactivity,
let’s suppose instead that you have
received a gift of $128 and you don’t want
to spend it all at once. Instead you
decide to spend half of it now and then
each of the following weeks, you will
spend half of what is left.
So the first week you spend half of the
$128, leaving you with $64; the next week
you spend half of $64, leaving you with
$32; the next week you would spend half
of the $32, leaving you with $16, etc.
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Generally, if P denotes the present
amount, a week later the amount left is
1/ × P, after the second week the amount
2
left is 1/2 × (1/2 × P) or (1/2)2× P, and after
the third week the amount left is
(1/2)3 × P., etc.
However, in order not to have to use
fractional notation we may rewrite an
expression such as (1/2)3 × P in the form
2-3 × P. After n weeks the amount left
could then be expressed as 2-n × P.
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An Enrichment Note
on
Fractional Exponents
Until now there seems to be no pressing
reason to define fractional exponents.
However, let’s suppose that we knew that
the cost of living was increasing at a rate of
4% per year. That means what costs $1.00
this year will cost $1.04 next year. So if C
represents the cost of living this year, the
cost of living next year will be 1.04 × C
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Thus, the next year it will be 1.04 × C,
not C, that is increasing by 4% a year. So at
the end of the second year the cost of living
is 1.04 × (1.04 × C), or 1.042 × C. And in a
similar way at the end of the third year the
cost of living is 1.043 × C. More generally, at
this rate at the end of n years the cost of
living would be 1.04n × C.
However, if we wanted to know the cost of
living 6 months (i.e., 1/2 year) from now,
using the above formula, it would be given
by 1.041/2 × C .
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This motivates us to try to define 1.041/2
More specifically, if we still want it to
be true that bm × bn = bm+n, we may
replace m and n by 1/2 to obtain…
b1/2 × b1/2 = b1/2 + 1/2 = b1 = b.
In other words, b1/2 is that number which
when multiplied by itself is equal to b.
By definition of the square root (that is, the
square root of a given number is the
positive number which when multiplied by
itself is equal to the given number), it
means that b1/2 = √b .
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As a check that this definition is
plausible, we can use our calculator and
representing 1/2 as 0.5, we see, for example,
that 9.0.5 = 3, and as a check, we know that
3 x 3 = 9. Hence, 9.0.5 is the square root of 9.
In a similar way, we see that
1.041/2 = 1.01980…, and this, in turn tells
us that if something costs $100 today,
6 months from now it will cost
1.01980… × $100 or, to the nearest cent,
$101.98.
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Algebra is next.
This discussion
completes our arithmetic
course.
We hope you will join us
for our algebra course.
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