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Supplement For My Daughter's Eighth Grade Math Book's First
Chapter
by Rick Garlikov
The lack of some explanations and the bad explanations of some of the
key concepts in the first chapter are disappointing. I want to try to fill
in the gaps in order to make understanding and doing algebra easier.^{(1)}
First, the book makes the knowing the names of some the principles and some of the properties discussed seem more important than understanding the concepts, whereas it is understanding the concepts that is important. In some cases, the names the author gives the principles or the properties are arbitrary and non-descriptive; and in some cases they are very misleading. The eighth grade chapter 1 pre-test had at least one question in which knowing the names of some principle was necessary in order to get the answer right, even though knowing the name is not especially important for being able to do algebra. Knowing the name of the principles is neither necessary nor sufficient for understanding the principles; and many kids will know the names and even be able to work some problems, but only paradigmatic ones, without knowing what they are doing or why it works. I think this is not good. It is much more important that students understand the concepts, properties, relationships, and principles involved than that they know names of these things, particularly when knowing the names does not show or mean understanding. If grades are to be given, it should be on understanding the items, not naming them, particularly when the names used in the text are arbitrary.
The reason understanding the concepts is important in algebra is that math is generally about the logical relationships of certain things, such as numbers, and the logical properties these things have. In algebra, you usually have to use just (relevant) logical properties and (relevant) logical relationships, along with a few given facts, in order to derive some things you need to know but are not given or told. Basically all you have to work with is your understanding of number properties and relationships. And it is your understanding properties and relationships, not your naming them, that is important for doing algebra. And the better your understanding of logical properties and logical relationships, the better chance you have of figuring out which ones are relevant to a given problem, though knowing what is relevant often takes luck or a flash of insight^{(2)}. Naming only allows an abbreviation for discussing properties and relationships with someone else --provided they know the same names you do.
With regard to what the book calls the distributive principle (of multiplication):
(1) that principle involves two principles actually: a) what my teachers called "multiplying (an expression) out" an expression of the form a(b+c), and (b) "factoring" or finding factors for an expression of the form ab+ac. For example, not only is it important to know
a(b+c) equals ab+ac, [equation 1]
but it is just as important to know this in the "opposite direction", that
ab+ac equals a(b+c). [equation 2]
This is because in some algebra problems or steps it will be important to "multiply out" the expression a(b+c) in order to get ab+ac; but in other problems or steps it will be important to change expressions into their "factors", so that, in this case, you would want to change ab+ac into a(b+c). The reason it will be important to go one way rather than the other is that one form of the equation will be more useful than the other in some cases. For example, if you have to solve: 5(3x - 11) = 7x - 3(x+10), multiplying it out is important in order to find out how many total x's you are dealing with, and what total purely numerical quantity you are dealing with. However: in a problem like (9x^{2} - 9y^{2}) ÷ 3(x-y), it will be easier to solve the problem by factoring the first expression into 9(x^{2}-y^{2}) and then factoring (x^{2}-y^{2}) into the two factors which are (x+y)(x-y), as you will learn later in algebra.^{(3)} This gives 9(x+y)(x-y)÷3(x-y), which is easier to see gives 3(x+y). ^{(4)}
(2) The book does not explain why distributive multiplication works; and most students don't understand how it works. Enclosed is a diagram that shows how it works. It is a diagram of rows and columns showing why (a+b)(c+d)= ac+ad+bc+bd. You could make up a similar diagram for yourself of a(b+c). Or suppose a=8, b=3, and c=4. Then, according to the principle of distribution (or of "multiplying out"), 8(4+3) should be the same as (8x4)+(8x3). Since 32+24 = 56, you can see it works in that case. But also think of it as laying out 8 rows and 7 columns of something, such as chocolate chips. No matter how you divide up the rows and columns with partitions [for example, 8(3+4) or 8(6+1) or 8(2+3+2) or (5+3)(2+3+1+1)], the total number of chips will not change; and the subgroups will all be little rectangles of (sub-rows x sub-columns), which, when added up will be the same number of chips as the total number of rows times the total number of columns. You may understand the principle in some other way, but you should figure out for yourself some way to see why or how the principle works. It should make sense to you. Then you don't have to try to remember it or worry about getting it confused with some other equation that is not universal.
(3) The distributive principle is what is behind the way we multiply "by hand", to get the product of numbers such as 35x82.
35
x 82
70
280
2870
If you were taught to multiply in the above way, multiplying 35 by 2 and then 35 by 80 (or 8, but starting the answer in the "tens column") you are essentially multiplying 35(80+2) and getting (35)(80)+(35)(2). That is why you end up adding together the two numbers 70 and 2800 after you multiply the parts separately. [Actually, if you want to get precise, you are multiplying (30+5)(80+2) to get (30x80)+(30x2)+(5x80)+(5x2), only in a slightly different order, but it all comes out the same.]
Reflexive, Symmetrical, and Transitive Properties
This is not explained well at all in the book. "Equality" or "being equal" is one of many relationships which are logically reflexive, symmetrical, and transitive. There are many such properties in the universe. Being "the same height as" is such a property, for (1) everyone is the same height as him/herself (reflexive), (2) everyone is the same height as anyone the same height as him (symmetrical), and (3) everyone is the same height as a third person who is the same height as someone else who is the same height as the original person (transitive). The last one of these is a little hard to follow, but it essentially says something like, if you are the same height as Frank, and Frank is the same height as Sally, then you are the same height as Sally.
A property or relationship is reflexive if and only if everything has that property in relationship to itself. Being a grandfather is not a reflexive property because not everyone is their own grandfather (actually no one is, of course). Being "as smart as" is reflexive, since everyone is as smart as him/herself.
A property or relationship is symmetrical if and only if in every case where one thing has that relationship to a second thing, the second thing logically has that relationship to the first thing. Being "a sister to" is not reflexive, since Sue can be a sister to George, but that does not mean George is a sister to Sue. Being "a sister to a female" IS reflexive, however, isn't it?
A property or relationship is transitive if and only if in every case, whenever one thing has that relationship to a second thing, and the second thing has that relationship to a third thing, the first thing logically has to have that relationship to the third thing also. Being "a cousin of" is not transitive because not all the cousins of your cousins are cousins to you. For example, you are the cousin of both your father's brother's kids and your mother's sister's kids, but they are not cousins to each other. Being "taller than" is transitive because if you are taller than Paul, and Paul is taller than John, you must be taller than John also.
Being "equal to" is reflexive, symmetrical, and transitive since everything is equal to itself (reflexive), equal to whatever is equal to it (symmetrical), and equal to anything equal to anything equal to it (transitive). The last one is, of course, hard to understand put into those words, but it essentially says that if A is equal to B, and B is equal to C, then A must also be equal to C.
The reason any of this is even said about "equality" is that when you are trying to solve problems in algebra, often you will use these properties of equality to make substitutions for equivalent formulas, quantities, or expressions in order to do calculations more easily. It is just that you will do it without thinking of the names "reflexive", "symmetrical", and "transitive". For example, if you have
(3a+3b)÷(b+a),
you can substitute 3(a+b) for (3a+3b), since they are equal, and then substitute (a+b) for (b+a) because they are equal, and that makes it easy to see how to divide, and get the answer, 3, since the problem is the same as
3(a+b)÷(a+b).
And substituting (a+b) for (b+a) also shows why it is important to keep
in mind that addition and multiplication can each be done in any order
separately ("the commutative property") though not mixed
up together, since by arbitrary, but binding, convention (unless it
is changed) 3+6x8 [51] is not the same as 6+3x8 [30]. Similarly, there
may be times of regrouping or re-associating addends or multiplicands
in order to see relationships more clearly. I cannot think of such a case,
but perhaps you will come across one some day.
What the book calls "the additive and multiplicative identity properties" are, of course, just the facts that anything zero is added to does not change, and anything multiplied by 1 does not change. It is certainly easier to remember these ideas than the names the author gives them. They should be able to be expressed clearly in terms other than the names the author has mentioned.
Order of Operations
Now the trick in understanding "order of operations" is realizing that there is nothing to "understand", since "order of operations" is not a matter of logic, but a matter of convention. That is, it is simply an edict that IF numbers are written without groupings by parentheses, THEN the order one is supposed to deal with them is by the arbitrary, but agreed upon, rules of "order of operations". This is a convention simply because it is the way that was chosen even though other ways could have been chosen that would have been different and might have worked just as well. They would have worked just as well as long as everyone knew what the rules were. Driving on the right side of the road in the United States, and on the left side of the road in England are examples of conventions. It doesn't matter which side of the road everyone drives on, as long as a side is chosen and everyone THEN knows what it is and honors it. Similarly it is a convention that English is written and read from left to right and from top to bottom rather than from bottom to top or from right to left. The reason for written language conventions, which is what "order of operations" is, is so that people can read what others (or themselves later) have written. In math, order of operations tells you how to read what I have written if I happen to leave out the parentheses and write:
This work is available here free, so that those who cannot afford it can still have access to it, and so that no one has to pay before they read something that might not be what they really are seeking. But if you find it meaningful and helpful and would like to contribute whatever easily affordable amount you feel it is worth, please do do. I will appreciate it. The button to the right will take you to PayPal where you can make any size donation (of 25 cents or more) you wish, using either your PayPal account or a credit card without a PayPal account. |
1. Some of the explanations and mathematical proofs I give will be clear,
I think, if you patiently work your way through them. Unfortunately,
these things are hard to say in a way that makes them clear without your
having to think about them and work through them yourself. I try to show
a way to work through them, and though this way makes sense to me,
you have to make it-- or your own way of thinking about these things--
make sense to you. Someone else cannot automatically make such complex
things make sense to you no matter how they say them; only you can make
sense out of complex explanations --by thinking about them. (Return to text.)
2. When Richard Feynman, a gifted mathematician and physicist, was on his high school math team, one of the problems he solved in a contest long before everyone else was this one: Suppose you and a crew of friends are rowing upstream at the rate of 7 miles per hour relative to the shore and that the current is flowing at the rate of 2 miles per hour, relative to the shore. Your hat falls into the water and starts floating downstream fifteen minutes before you notice it. How long would it take you to catch up to the hat if you instantaneously reversed direction? Feynman immediately saw that the answer was 15 minutes because he saw that the rates given in relation to the shore were irrelevant, just like when you leave something at home by accident and drive 15 minutes before you realize it. It will take you 15 minutes at the same speed to get back to it, even though the earth has moved relative to the sun and has also been rotating at the rate of 1000 miles per hour in perhaps the opposite direction you were driving. I still have difficulty seeing this with regard to the water case, but Feynman saw it right away.
Often it is easy to focus on the "wrong" or least helpful
relationships. There is a famous trick problem that is not difficult to
solve if you don't know calculus, but is difficult to solve if you do know
calculus. The reason is that the easier solution does not use calculus,
but the problem is set up to look like the sort that it takes calculus
to solve. That problem is: Two trains start out 1500 miles apart on the
same track, going straight toward each other. One train is moving 40 miles
per hour and the other is going 35 miles per hour. At the exact moment
they start, a bee that can fly 100 miles per hour starts from one and flies
to the other, instantaneously reversing direction when it gets to the other
train. The bee does this over and over --flying shorter and shorter distances
each time between the trains as they travel toward each other-- until it
is crushed between the two trains when they crash into each other. How
far in total distance will the bee have flown. To do this problem using
calculus requires what is called "summing an infinite series",
which is a complex calculation. But in arithmetic, it is easy: the trains
are headed toward each other at the combined rate of 75 miles per hour.
Thus it will take them 20 hours to cover the 1500 miles. Since the bee
is flying 100 miles per hour for that 20 hours, the bee flies 2000 miles.
It is not always easy to know which logical relationships are the most
productive to use when solving a math problem. (Return to text.)
3. (x+y)(x-y) is, by (the relationship involved in) the "distributive property",
(x+y)(x)-(x+y)(y)
and these then, by the distributive property, multiply out to be
(x^{2}+xy)-(xy+y^{2})
which then is the same as
(x^{2}+xy-xy-y^{2})
and which is x^{2}-y^{2}. In the given problem, however, reversing this procedure to end up with the factors (x+y)(x-y) is the more important direction.
By the way, the formula or equation (x+y)(x-y)=(x^{2}-y^{2)}
comes in handy in multiplying some numbers in your head. For example, 48
x 52 is the same as (50-2)(50+2), and since this is equal to 50^{2}
- 2^{2}, the answer is fairly easily computed to be 2496, since
that is 2500 - 4. You can do this in multiplying together any two
numbers
equidistant from some easily squared number. Appearing to do such
multiplications"in your head" will impress some people; don't tell them
the
trick. (Return to text.)
4. There is a famous fake "proof" that 2=1, using some of these principles:
Let x=y
Then multiplying both sides by x, we get
x^{2 }= xy
Then subtract y^{2} from both sides, giving
x^{2} -y^{2 }=xy-y^{2}
Factoring both sides, we get
(x+y)(x-y)=y(x-y)
Dividing both sides by (x-y) gives
x+y = y
Since x=y and therefore y=x, we can substitute y for x, giving
y+y = y
which is 2y = y
and dividing both sides by y, gives
2 = 1
The "bad" step is dividing both sides by (x-y), since in
this case, where x and y are equal, that is dividing by 0.
A more interesting, real proof using some of these principles gives a surprising, but actual result. Consider this problem:
Suppose you had a smooth ball the size of the earth, and tied a 24,000 mile long ribbon around its middle, so that it was tight against the ball. Then add a piece of extra ribbon to it -- a piece that is just 1 yard (i.e., 36 inches) long; and smooth the ribbon out all the way around, so that the loop disappears and the ribbon now forms a circle again. Will the ribbon be very far off the surface of the ball? How far? It seems like it would hardly raise the ribbon at all; but actually it will raise the ribbon about six inches off the ball all the way around. In fact, it will turn out that not only is the circumference of every ball roughly six times its radius, but increasing any circumference of any circle by some amount will increase its radius by nearly one-sixth that amount also.
The circumference, C, of any circle is 2piR, where R is the radius of the circle.
Dividing both sides of C=2piR by 2pi, we get the formula for the radius in terms of the circumference: R=C/2pi
In the problem with the ball and the ribbon, what we are looking for is the difference in the radius of the ribbon when it is 24,000 miles long and when it is 24,000 miles and 36 inches long. That will give us the distance the ribbon is away from the ball.
If we use subscripts to designate the first and second radius and the first and second circumference, the new radius will be R_{2} = C_{2}/2pi
and since the new circumference is 36 inches longer than the first
circumference, we can substitute (C_{1}+36") for C_{2} to get: R_{2 }= (C_{1}+36")/2pi
Since a/x + b/x = (a+b)/x, (a+b)/x = a/x + b/x,
And by this principle we can rewrite our result to get: R_{2} = (C_{1}/2pi)+(36"/2pi)
Since (C_{1}/2pi) is the old radius, this means R_{2 }= R1+(36"/2pi)
And since 2pi is 6 when rounded off to a whole number, this means that the new radius is the old radius plus approximately 6 inches. And, in general, the new radius of any circle will be the old radius plus one-sixth of the amount added to the circumference. So whether you add 36 inches to the circumference of a dime or to the circumference of the earth, you increase the radius of each of them by the same amount, roughly six inches. Hard to believe, but true. (Return to text.)