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22M:012 Theory of Arithmetic Final Exam Solutions

1. Suppose a shipment of milk to a local grocery consists of individual gallons of either skim milk or whole milk. Suppose the shipment contained 40% more skim milk than whole milk. If the total shipment consisted of 48 gallons altogether, how many gallons of each type were contained in the shipment?

SOLUTION:

Let x = # gallons of whole milk. Then the problem language translates into the equation

x + 1.40 x = 48, or

 2.40*x = 48

Solving gives x = 48/2.40 = 20

 So there are 20 gallons of whole milk and 20*1.4 = 28 gallons of skim milk.

2. Give a description, based on any of the manipulative or geometric models (colored disks, number line arrows, mail time, etc.) we discussed, of the how you might explain the fact that

- 6   -  - 2  = - 4

to children who are first learning about the integers and the arithmetic operations.

If you use the number line model be sure to indicate whether you are using the missing addend model or an alternative model (e.g. the take-away model).

Any diagrams you use must be accompanied by a correct written identification of what the model is showing (e.g. the number line model of the missing addend approach, the set model with red/black chips, etc.).

SOLUTION:
A solution with red and black chips can be given as follows. Red chips have value -1 and black chips have value +1.
Start with a set with net value -6 

●             ●         ● ●             ●         ●

(set with net value -6)

 

Now to model -6 - -2 we remove (i.e. subtract) from this set a subset with net value -2 . So we just delete two of the six red dots from this set. The resulting set is below

●             ●         ● ●

This set has net value -4. And this is a set model of -6 - -2 = -4.

 

 

 

 

A solution involving the number line model of subtraction can be given as follows.

If a and b are numbers, here is how to determine the difference a - b
as a point on the number line:

**Start with the arrow corresponding to a (its starting point is 0 and its ending point (arrow) points to a).

Now take the arrow which has its initial point (tail)  at the number b and ends at the ending point of the a arrow. Translate or move this new arrow left or right so its new starting point is 0. The ending point or arrow head of this newly moved arrow points to the number a – b. **

Informally we might say that a - b is obtained by drawing the arrow starting 'at' b and pointing to a, then translating the tail of that arrow to 0. The arrow head of the translated arrow is the number a - b.

This is the missing addend model of subtraction.

We use exactly the same definition as the one above between the double asterisks **, we just allow a and/or b to be negative integers.

For example here is the number line model for  - 6 - -2 :
                 
                                                                               
                                                                       
                 <--------------------------------------------------------------        
                                                the -6 arrow

                                                <------------------
                                                                the -2 arrow
----------+---------+---------+---------+---------+---------+---------+---------+-------+------+----------->
            -6           -5             -4           -3          -2          -1            0            1          2                    3                                                                                       
              <----------------------------------------
                This arrow points from -2 to -6.        It points to the left 4 units, so this arrow represents -4, and the result of the computation to -6 - -2.           If you translate this arrow so its initial point is at 0 (as shown in the picture below), it will point to -4, the correct result from computing -6 - - 2.              

----------+---------+---------+---------+---------+---------+---------+---------+-------+------+----------->
            -6           -5             -4           -3          -2          -1            0            1          2                    3                                                                                       
                                           <----------------------------------------
                This arrow points from -2 to -6.       
The model above of integer subtraction, CALLED THE MISSING ADDEND MODEL,  is based on this idea of subtraction:

 

a – b means a number c such that a =  b + c.

Call this the MISSING ADDEND  definition of  subtraction.

3. Give an induction proof for the following sum of the first n odd natural numbers:

 

1 + 3 + 5 + 7 + …+ ( 2n - 1 ) = n²  

where n is any natural number 1, 2, 3, 4,  …

 

SOLUTION:

Label this equation, which we want to verify,  1 + 3 + 5 + 7 + …+ ( 2n - 1 ) = n 2   with an asterisk (*).

 1 + 3 + 5 + 7 + …+ ( 2n - 1 ) = n²   (*).

 

We can verify by hand calculation that (*) is correct when n is 1:

When n = 1, the left side of (*) is just 1.

And the right side of (*) is 1²  , which is also 1. So the left and right side of (*) are indeed equal when n  = 1.

 

Now assume for the induction hypothesis that (*) is correct, we now have to check whether or not (*) is correct when we replace n by n+1 on each side of (*).
So here is what we get to assume this is true:

 1 + 3 + 5 + 7 + …+ ( 2n - 1 ) = n²    (the "induction hypothesis")

And here is what we have to try to prove:

1 + 3 + 5 + 7 + …+ ( 2n - 1 ) + ( 2 [ n + 1 ] - 1 ) =?? (n+1)²    call this (**).

We know from the induction hypothesis that 1 + 3 + 5 + 7 + …+ ( 2n - 1 ) = n²    

Using this induction hypothesis to simplify the expression 1 + 3 + 5 + 7 + …+ ( 2n - 1 )  by just replacing this whole sum with n² , we can write the left side of (**) in a much simpler-looking way:

1 + 3 + 5 + 7 + …+ ( 2n - 1 ) + ( 2 [ n + 1 ] - 1 ) = n²   +  ( 2 [ n + 1 ] - 1 )

because we know from the induction hypothesis that 1 + 3 + 5 + 7 + …+ ( 2n - 1 ) = n²    

Therefore we now know

1 + 3 + 5 + 7 + …+ ( 2n - 1 ) + ( 2 [ n + 1 ] - 1 ) = n²   +  ( 2 [ n + 1 ] - 1 )

And now notice that

 n²   +  ( 2 [ n + 1 ] - 1 ) = n²  +  2  n + 2 - 1  = n²  +  2n + 1  = (n+1)²     

We have now completed the proof of (**) because we showed that

1 + 3 + 5 + 7 + …+ ( 2n - 1 ) + ( 2 [ n + 1 ] - 1 ) does in fact =  (n+1)²  .

This completes the induction step and as a result we can now assert that (*) is true for every natural number n.

 

 

4. How can you use the number line model of whole numbers and whole number arithmetic to illustrate the idea of the greatest common divisor of two natural numbers a and b, GCD(a, b)? For example, we have GCD(108, 144) = 36. What might be a number line interpretation of this fact?

Your explanation should be simple enough that you might use it in an elementary school class studying math and should apply to GCD(a,b) no matter what natural numbers a and b are. You are welcomed to draw on real-life examples as long as they accurately support the mathematical ideas used in defining the GCD.
 

 

                                                                       
SOLUTION:

Explanation for university math class discussion:
One way you might answer is as follows. GCD(108, 144) = 36 means that 36 is the largest number which evenly divides both 108 and 144. In a number line model, this means you can take an arrow corresponding to 36 (i.e. the arrow starting at 0 and ending at 36) and place repeated copies of this arrow end to end you will arrive at the numbers 108, and also 144, after a whole number of such arrows are placed end to end.

In fact, because 108 = 36 * 3 it takes 3 such arrows to reach exactly 108, and b/c 144 = 4 * 36, it takes 4 such arrows to reach exactly 144.

Now many other sized arrows might be copied and placed end to end, and also reach both 108 and 144 after a whole number or repeats (for example, arrows of length 1, 2, 4 , 8, 12, ..).
What distinguishes the length 36 arrow from any of these other arrows is that because 36 is the GREATEST common divisor of 108 and 144, the 36 arrow is the longest arrow for which this is possible.

Explanation for elementary school math class discussion:
If you think of the arrow as representing the length, maybe in millimeters, of a jump which (say) a frog can make, then 36 mm is the longest jump a frog could make, and still reach Lilly pads 108 and 144 millimeters away from the start with a whole # of jumps.

5. Write two different explanations of how one can describe the quotient 45 / 9. In the first use the measurement, also known as the repeated subtraction model, in the second the partition, also known as the sharing model. You do not need to include real life applications illustrating these models (unless you feel including those helps to illustrate the mathematics). Your answers should only be around two to four sentences each (or less!).

 

i.   Measurement or repeated subtraction model:

 

SOLUTION:
Repeatedly remove or delete 9 elements from a set with 45 elements. These removals can occur 45/9 = 5 times and then there are no elements left in the set.

 

 

 

 

ii.   Partition or sharing model:

 

SOLUTION:
45 objects are to be separated into 9 equal-sized piles, each pile containing the same number of objects. The number of objects which will end up in each pile is 45/9 = 5.

 

6. This problem invovles the numbers 35 and 100 (among others). The numbers 35 and 100 were essentially chosen at random, there is no special significance to those choices other than we chose to use them.

 

Consider the part of the number line consisting of all the numbers starting at 0 and ending at 35.

|___|___|___|_______________________________|  

0    1     2    3…                                                         35

(image that there are marks “|” for each of the whole numbers from 0 to 35).

If you divide this part of the number line into one hundred equal-sized pieces, you would expect that each piece has length 35/100.

We use the number line defintion of fractions in this class, and that definition gives the number 35/100 as a dot on the line between 0 and 1 defined in a precise way.

Using the number line definition of 35/100, explain why or not the following statement is correct or is not correct.

"If you divide the above part of the number line (consisting of all the numbers starting at 0 and ending at 35)  into one hundred equal-sized pieces, then each piece has length 35/100."

SOLUTION

For the convenience of the reader, we review the number line definition of fractions below.

 

|______________|______________|______________|______________|__________...

0                            1                              2                           3                          4

THE NUMBER LINE DEFINITION OF FRACTIONS

We take as our starting object the number line pictured above with the whole numbers marked off as dots or dashes on it.

We now take the following statement as a given axiom or assumption about an operation that is always allowable for us to perform with the number line.

AXIOM OF THE NUMBER LINE: Given any whole number l with l ≠ 0 and given any segment of the number line, it is possible to split that entire segment up into exactly l pieces, each one of which has the same length as each and every one of all the other pieces.

Below in the box is a number line definition of fractions, which is only based on the idea of whole numbers and on the number line we have just given. This appears in a number of texts.

 

Now to answer question 6, we say that the quoted statement is correct. To verify this we start with the number line definition of 35/100.

It is the dot on the number line obtained by dividing each segment [ 0, 1], [ 1, 2], [2, 3], [3, 4], . . . into 100 segments of equal length, then starting at 0 we count to the right until we hit the thrity-fifth one. This dot is labeled 35/100.

Notice that the segment [ 0 , 35/100 ] is 35 times longer than the segment [ 0 , 1/100 ], because the segment [ 0 , 35/100 ] is obtained by putting 35 of these [ 0 , 1/100 ]-length segments one right after the other, starting at 0. 

From the number line definition of 1/100, we know that if we put 100 of the [ 0 , 1/100 ]-length segments one after the other starting at 0, these will form a combined segment of length 1 (i.e. the last [ 0 , 1/100 ]-length segment (furthest to the right) will end at 1, filling up the segment [ 0 , 1 ] ).

 

Since the segment [ 0 , 35/100 ] is 35 times longer than the segment [ 0 , 1/100 ], it seems reasonable to conclude that putting together 100 of these [ 0 , 35/100 ]-length segments one after the other starting at 0, the result will form a combined segment of length 35 times longer than the segment [ 0 , 1 ] .

Of course this means that the last such
 [ 0 , 35/100 ]-length segment (furthest to the right) will end at 35, filling up the segment [ 0 , 35 ] ), because the segment [ 0 , 35 ] is 35 times as long as the segment
[ 0 , 1 ].

 

This argument shows that 100 of these [ 0 , 35/100 ]-length segments fill up the line segment from 0 to 35, in equal-sized pieces. And then each piece must be the same size as the segment [ 0 , 35/100 ], that size being 35/100.

7.

i.  Give an example of a fraction which does not have a terminating decimal expression, and give its nonterminating decimal expression. Explain briefly, using ideas or results we have discussed in the class or that appear in the course text, why it is true that it has no terminating decimal expression.

 

SOLUTION:
One example is 1/3 = .333… (repeating 3s). This fraction has no terminating decimal expression because it is written in lowest form (no common factors in the numerator and denominator) and in the prime factorization of the denominator (that prime factorization is just 3) it is not true that only products of 2s and 5s occur.

 

 

 

ii.  Give an example of a fraction which has a terminating decimal expression, and give its terminating decimal expression. Then express this same fraction as a nonterminating repeating decimal expression (with non zero repeating part).

SOLUTION:
One example is 1/2 = .5 = .4999… (repeating 9s).

 

 

 

iii. Describe the difference or differences between a rational and an irrational number. What are the key features which would allow you to determine if a given number was rational or irrational? To earn credit, your answer must be more informative than something like, "An irrational number is a number which is not a rational number".

SOLUTION:

An irrational number is any number that has a nonterminating, nonrepeating decimal expression. Every rational number has a repeating decimal expression, and some rational numbers also have terminating decimal expressions. But only irrational numbers have nonterminating, nonrepeating decimal expressions.

8. Give an example of a real-world application problem, suitable for discussion with elementary school students, which uses the idea of proportion or proportional quantities in an essential way for its solution. And solve the resulting problem. Your answer must not be anything of the form, "One person asks another person to solve the proportion
2/4 = 6 / n for n, and the other person figures it out." Your problem must instead involve a realistic application which might occur outside of the classroom setting, but still be understandable to an elementary school student.

 

SOLUTION:

One possible example is the following. For every pair consisting of two people who come to a picnic, you will need to bring 4 snacks. If 3 pairs come (6 people) how many snacks will you need?

9. Give an example of a real-valued function f [x] of a real variable x for which the graph is a line in the x-y plane which contains the points (1 , 4) and (-3 , 8).

 

SOLUTION:

There are many ways to solve this. Here is one. The line must have slope (4-8)/ (1--3) = -4/4 = -1. The function must therefore have the form

f [ x ] = -1*x + b or  - x + b .

 Now we know from the given points that f [ 1 ]  = 4. Hence

                4 = f [ 1 ] = -1 + b and solving gives b = 5.

Therefore f [ x ] = -x + 5

10. 
i. Show how to solve the gardener problem below using a problem solving strategy such as one of the Polya strategies mentioned in the text. If you use guess and check your answer must include at least three different guesses and checks, and that strategy should show how your guesses led to additional guesses.

Your solutions should state clearly which problem-solving strategy is being used and it must be clear that the solution actually used that strategy.

 

SOLUTION:
Solution strategy: use a variable. Let w stand for the width of the flower patch in feet. The length must be 2*w and the perimeter of the rectangle has to be 120 ft. This means
w + w + 2*w +2 *w (the perimeter of this rectangular patch) = 120
or
6*w = 120.
Therefore w = 20 and l = 2*w = 40 (the units are feet).

 

 

ii. Show how to solve the sequence of letters problem below using a problem solving strategy such as one of the Polya strategies which is DIFFERENT from the one you used above. If you use guess and check (which you should not do if part i. was solved using guess and check) your answer must include at least three different guesses and checks, and that strategy should show how your guesses led to additional guesses.

Your solution should state clearly which problem solving strategy is being used and it must be clear that the solution actually used that strategy.

 

 

SOLUTION:
Solution strategy: look for a pattern

The correct answer is c.

If you look at the sequence you see it consists of 4 letters a,b,c,d repeated over and over. The fourth letter of every one of these a,b,c,d groups is the letter d. So the d's all occur as multiples of 4: the #4, #8, #12, #16,…numbered letters are all d's.

If we divide 951 by 4 we find out how many groups of 4 there are. The division 951/4 yields 237 R3, that is, 951 = 4*237 + 3. This means there are 237 groups of 4 letters, (a, b, c, d) then three more letters after that to get to 951. Since we know the letter number 237*4 = 948 must be d, then letter number 237*4 + 3 must be c, because after d comes a then b then c.

 

 

 

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