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#7

Taking the Fearout of Math

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The definition of subtraction as “unadding” remains the same.

© Math As A Second Language All Rights Reserved

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That is when we say that…“f – s means the number that we must

add to s to obtain f as the sum”…it does not matter whether f and s are

whole numbers or rational numbers.

Suppose we want to compute the difference 67/9 – 42/9 and express the

answer as a mixed number.


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There are a few ways of finding the answer.

The “Direct” Method


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Since subtraction is “unadding” and since we add mixed numbers by adding the

whole numbers and adding the fractions, we want to solve the following two

problems…

2/9 + _____ = 7/9

and…

4 + _____ = 6

The “Direct” Method


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Since 4 + 2 = 6 and 2/9 + 5/9 = 7/9, we see that the answer is 25/9.

42/9 + 25/9 =

As a check , we see that…

(4 + 2/9) + (2 + 5/9) =

(4 + 2) + (2/9+ 5/9) = 67/9

Converting Mixed Numbers to Improper Fractions


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A common strategy for solving an unfamiliar problem is to rephrase it in

the equivalent form of one or more familiar problems.

In this case, since we already know the arithmetic of common fractions, we can convert the mixed numbers to improper

fractions.


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We may rewrite the problem as follows …

67/9 – 42/9 =

61/9 – 38/9 =

23/9 = 25/9

Converting Mixed Numbers to Improper Fractions

► Doing a mixed number problem by converting the mixed numbers to

improper fractions is not as transparent as it is when we work directly

with the mixed numbers.


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► However, it does avoid a pitfall that students often fall into when they subtract

one mixed number from another.

► Specifically, students often make a mistake when they borrow orregroup using mixed numbers.


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► As we mentioned in the previous presentation the idea of exchanging ten of

one denomination for one of the next higher denomination is valid only when the

denominator (if there is one) is 10. However, students often forget this when they are

borrowing or regrouping by rote.


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To illustrate the above note, suppose that we want to express the difference

62/3 – 33/4 as a mixed number and that we don’t want to rewrite the mixed numbers

as equivalent improper fractions.

Just as in our previous illustration, in terms of “unadding” the problem is

asking us to fill in the blank…

33/4 + ____ = 62/3


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We begin as we did in the previous problem by filling in the blanks in

the two statements...

and…

3 + _____ = 6

3/4 + _____ = 2/3


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We can only add or subtract fractions if they have the same denominator (noun).

Therefore, we rewrite 3/4 _____ = 2/3 in the form…

9/12 + _____ = 8/12

The fact that 9/12 is greater than 8/12 means that we will have to “borrow” in order

to solve the problem.

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nextnext This in turn means that we will begin

by rewriting 62/3 in the form 68/12 and then use the following steps…

68/12 = (5 + 1) + 8/12

= 5 + (1 + 8/12)

= 5 + (12/12 + 8/12)

= 5 + 20/12

We use this information to rewrite the problem in the form…

39/12 + ____ = 520/12

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We now proceed just as we did before by filling in the blanks

in the two statements..

We fill in the blank in the top equation with 2 and the blank in the bottom equation with 11/12.

3 + ____ = 5

9/12 + ____ = 20/12

2

11/12

So altogether, we have to add 211/12 to 33/4 to obtain 62/3 as our sum.

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► The most common mistake that students are prone to make is that

when they want to “borrow” 1 from 6 8/12, they will write 5 18/12 rather than 5 20/12 .


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When borrowing using mixed numbers, the new numerator is the sum of the

original numerator and the denominator.

► This is analogous to the fact that if you have 8 doughnuts and you buy another dozen, you have 20 doughnuts, not 18!


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► Or if you have 3 quarters and you exchange a dollar for quarters,

you now have 7 quarters, not 13!

► While converting mixed numbers to improper fractions is usually a more

tedious way to subtract mixed numbers, it does avoid the

“borrowing” problem.


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► For example, we may rewrite 62/3 and 33/4 as…


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62/3 = (6 + 2/3)

= (6/1 + 2/3)

= 20/3

= (18/3 + 2/3)

= 80/12

33/4 = (3 + 3/4)

= (3/1 + 3/4)

= 15/4

= (12/4 + 3/4)

= 45/12

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► Hence, 62/3 – 33/4


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= 80/12 – 45/12

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= 35/12

= 211/12

However, as you can see, this method, while giving us the correct answer, gives

us no hint as to what is actually happening.

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As usual, the abstract approach becomes more transparent when we supply

nouns (units) for the numbers to modify.

Let’s return to the problem we were discussing in our previous presentation.

Recall that we had assumed that you rented a particular machine and used it for

6 hours and 40 minutes one day and for 3 hours and 45 minutes the second day; and the question involved having to find

the total time you used the machine.

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This time, however, let’s suppose that we knew that we had used

the machine for a total of 10 hours and 25 minutes during the two days and that we also

knew that we used it for 3 hours and 45 minutes the second day, and now we want to know how long we used it on the first day.

A Subtraction Problem

10 hours + 25 minutes =


nextnext Probably, we would subtract 3 hours 45 minutes from 10 hours 25 minutes

and in order to subtract in an easier way we would have written…

(9 hours + 1 hour) + 25 minutes =

9 hours + (1 hour + 25 minutes) =

9 hours + 85 minutes =9 hours 85 minutes

9 hours + (60 minutes + 25 minutes)1 =

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1 Notice that when we exchanged an hour for the equivalent number of minutes, we replaced the hour by 60 minutes, not 10 minutes.

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The above subtraction would now look like…

10 hours 25 minutes – 3 hours 45 minutes

9 hours 85 minutes – 3 hours 45 minutes

6 hours 40 minutes

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6 hours =


nextnext If we wanted to use only one noun, we

could have either,-- used the fact that 60 minutes = 1 hour

and written…

6 × 60 minutes = 360 minutes

6 hours 40 minutes =

400 minutes

360 minutes + 40 minutes =

40 minutes =


nextnext or,

-- again using the fact that 1 hour = 60 minutes and write…

40/60 of an hour = 2/3 of an hour

6 hours 40 minutes =

6 hours + 40 minutes =

6 hours + 2/3 of an hour =

62/3 hours


However, notice how quickly the transparency of what we did above

disappears as soon as we elect to omit the noun “hours” (and/or “minutes”).

Suppose rather than writing 10 hours and 25 minutes, we simply wrote 105/12, and

instead of writing 3 hours and 45 minutes, we wrote 33/4.

We would then be faced with the equivalent problem of computing the

difference 105/12 – 33/4 and expressing the answer as a mixed number.

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Most likely we would have expressed the two fractions in an equivalent form in which they have the same denominator

and written…

The plausible error is that we might have rewritten 105/12 as 915/12 forgetting that

1 = 12 twelfths, not 10 twelfths.

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105/12 – 33/4 =

105/12 – 39/12

Or we might have rewritten the problem in the form of equivalentimproper fractions and obtained…


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= 125/12 – 45/12

= (125 – 45)/12

= 80/12

105/12 – 33/4

= 125/12 – 15/4

= 68/12= 62/3


Notice that if we now reinsert the noun “hours”, we see that 62/3 hours is the same as 6 hours and 40 minutes,

which agrees with the answer we obtained previously.

However, notice how much more transparent the problem becomes if

we insert units for the mixed numbers to modify.

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Notice also that any common denominator can be used. In fact, if

we had been used to supplying a unit based solely on the above steps, the chances are we would have chosen

“dozens” or “feet” instead of “hours” because there are 12 in a dozen and

12 inches in 1 foot.

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In our next section we will discuss the process of multiplying one mixed

number by another.


62/3 × 33/4 = ?

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