keystone illustrations keystone illustrations next set 10 © 2007 herbert i. gross

Keystone IllustrationsKeystone

Illustrations

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Set 10© 2007 Herbert I. Gross

You will soon be assigned problems to test whether you have internalized the material

in Lesson 10 of our algebra course.

Instructions for the Keystone Illustrations

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© 2007 Herbert I. Gross

The Keystone Illustrations below are prototypes of the problems you'll be doing.

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Most likely the concept of mathematics as a game is new to you, and consequently it

may take time for you to become comfortable with how this game is played. Therefore, rather than a keystone problem,

we will give several different types of illustrations in this presentation to show

what we mean by a proof.

Prefacenext


In many ways, it is beyond the scope of an introductory algebra course to spend too much time on proofs. Yet it is important

to understand the nature of a proof and how it uses “facts" to derive other “facts”. We will limit our illustrations to showing how our rules justify some of the things that we ordinarily take for granted in an

algebra course.

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Be sure that you understand how every statement that's made in the proof is an

accepted property of the number system.

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Once we feel that the idea has been adequately presented; we will become more informal and resort to proofs only when we feel that a statement is not “self-evident”.

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Keystone Illustrations for Lesson 10

#1 Let's say you were called upon to find the value of x for which x + 3 = 7. Quite likely

you realized that by subtracting 3 from both sides of the equation you would obtain the

result that x = 4.

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You knew that since 3 – 3 = 0, x + 3 – 3 = x, and in vertical form your

solution might then look like…

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x + 3 = 7

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– 3 – 3 x = 4

Probably no one would have disagreed with your proof.

However suppose someone who accepted our rules of the game had

observed that by the closure property of addition x – 3 (i.e., x + -3) is one number. Hence, when you subtract 3 from it, the

correct way to indicate this is as (x – 3) + 3 rather than as

x + (3 – 3).

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So no matter how obvious the proof may seem to you, the obligation is to show the skeptic that your approach was justified

by the accepted rules of the game.

To this end, suppose someone who accepted our rules of the game wanted us

to prove that if x + 3 = 7, then x = 4.

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-- Starting with x + 3 = 7, we might begin by adding -3 to x + 3. Then by replacing x + 3 by 7 (substitution), we could write that…

(x + 3) + -3 = 7 + -3 (= 4).

-- Then by the associative property for addition we are allowed to rewrite…

(x + 3) + -3 as x + (3 + -3).

-- Hence, we may rewrite (x + 3) + -3 = 4 as… x + (3 + -3) = 4.

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-- by the additive inverse we know that3 + -3 = 0. Hence, by substitution we may

replace x + (3 + -3) by x + 0.

-- Knowing that 0 is the additive identitytells us that x + 0 = x. Hence, we may replace the equation x + 0 = 4 by the

equation x = 4.

-- Therefore, we may rewrite x + (3 + -3) = 4in the form x + 0 = 4.

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To prove that if x + 3 = 7 then x = 4.

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ProofStatement Reason

(1) (x + -3) + 3 = 7 + -3 (=4) (1) Substituting 7 for x + 3

(2) (x+ 3) + 3 = x + (3 + -3) (2) Associative Property (+)

(3) x + (3 + -3) = 4 (3) Substituting (3) into (1)

(4) 3 + -3 = 0 (4) Additive Inverse Property

(5) x + 0 = 4 (5) Substituting (4) into (3)

(6) x + 0 = x (6) 0 is the Additive Identity

(7) x = 4 (7) Substituting (6) into (5)

So what we've shown is that replacing (x + 3) – 3 by x is an inescapable

consequence of the rules of the game. Therefore, we may now use it as a “fact”

whenever we wish without having to demonstrate the validity of this again.

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If we restrict our assumed knowledge to the accepted rules of the

game, only the numbers 0 and 1 exist. In particular, the numbers 7, 3, and 4 do not

yet exist. This leads us to our next illustration…

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Caution

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#2 We have all been taught that 3 + 2 = 5

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However, suppose a skeptic were to ask, “What rule tells us that 3 + 2 = 5?” To answer this, we

could start with the fact that since 1 is a number, the closure property for addition tells

us that 1 + 1 is also a number. We call that number 2. Since 2 and 1 are numbers,

so is 2 + 1; which we will call 3.

Aside: By the additive inverse property once 1, 2, and 3 are numbers, so also are -1, -2, and -3.

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Continuing in this way we have, by definition…

4 + 1 = 5; 5 + 1 = 6; 6 +1 = 7; 7 + 1 = 8; etc.

Hence, if we start with the expression…

3 + 2

we may replace 2 above by 1 + 1 to obtain…

3 + 2 = 3 + (1 + 1)

By the associative property for addition we know that…

3 + (1 + 1) = (3 + 1) + 1

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1 + 1 = 2; 2 + 1 = 3; 3 + 1 = 4;

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Therefore by substitution we may replace 3 + (1 + 1) by its value (3 + 1) + 1 to obtain…

3 + 2 = (3 + 1) + 1

By definition… 3 + 1 = 4

So we may replace 3 + 1 by its value 4 to obtain…

3 + 2 = 4 + 1

And since by definition 4 + 1 = 5, we may rewrite 3 + 2 = 4 + 1 as…

3 + 2 = 5

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To prove that 3 + 2 = 5next

More formally


(1) 1 is a number (1) Multiplicative Identity

(2) 1 + 1 is a number (2) Closure for Addition

(3) 1 + 1 = 2 (3) Definition of 2

(4) 3 + 2 = 3 + (1 + 1) (4) Substituting 1+1 for 2

(5) 3 + (1 + 1) = (3 + 1) + 1 (5) Associative Property (+)

(6) 3 + 2 = (3 + 1) + 1 (6) Substituting (5) into (4)

(7) 3 + 1 = 4 (7) Definition of 4

(8) 3 + 2 = 4 + 1 (8) Substituting (7) into (6)

(9) 4 + 1 = 5 (9) Definition of 5

(10) 3 + 2 = 5 (10) Substituting (9) into (8)

While our discussion in the above situation might seem a bit abstract, it captures the

way young children tend to use their fingers to do addition. More specifically, if we use tally marks to represent fingers, we

may view 3 + 2 in the form…

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Notenext

This array can be regrouped by moving one of the two tally marks on the right closer to the three tally marks on the left to obtain…

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And finally we may move the remaining tally mark on the right closer to the left to obtain…

3 + 24 15

Consider the equality…

3 dimes + 2 nickels = 40 cents.

Preface to the Next Illustration

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In this case 3, 2, and 40 are adjectives modifying the nouns dimes, nickels, and

cents respectively.

If we were to omit the nouns, the above equality would become…

3 + 2 = 40; which would seem to be nonsense!

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Based on the ambiguity as to whether 3 + 2 = 5 or 3 + 2 = 40, it seems that the equality 3 + 2 = 5 should be emended to read…

3 + 2 = 5 provided that 3, 2, and 5 are adjectives modifying the same noun.

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If we now use x as a generic name for the noun, the above equality becomes…

3x + 2x = 5x

Therefore…

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#3 In terms of our rules of the game, we now know that 3 + 2 = 5

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We will now show that as a consequence3x + 2x = 5x, where x is any number.

In the spirit of deductive reasoning, one form of our proof would be…

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To prove that if 3 + 2 = 5 then 3x + 2x = 5x.

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(1) 3x = x3 and 2x = x2 (1) Commutative Property (×)

(2) 3x + 2x = x3 + x2 (2) Substitution

(3) x(3 + 2) = x3 + x2 (3) Distributive Property

(4) 3x + 2x = x(3 + 2) (4) Substituting (3) into (2)

(5) 3 + 2 = 5 (5) Previously Proved

(6) 3x + 2x = x5 (6) Substituting (5) into (4)

(7) x5 = 5x (7) Commutative Property (×)

(8) 3x + 2x = 5x (8) Substituting (7) into (6)

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Enrichment Illustration for Lesson 10

#4 In terms of working with 0 and 1, all we know from the listed properties is that

a + 0 = a and a × 1 = a.

Nowhere do we have a rule or a definition that tells us that for any number a, a × 0 =0.

Of course, this “fact” might seem obvious to us from what we have learned in arithmetic.

For example, we defined 3 × 0 to mean0 + 0 + 0, which is clearly 0.

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However, there are many students in arithmetic who feel that a × 0 should be a.

The reason lies in the “excuse" that is often given as to why a + 0 = a. Namely…

“Since we didn't add anything to a the sum is still a. So why shouldn't the answer still

be a if we multiply it by nothing?”

However, the spirit of our game requires that we can only use results that are contained in

the definitions and the rules we accepted; and a × 0 = 0 is not one of the rules we have

accepted.

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If we are able to show that a × 0 = 0 follows inescapably from the rules of our game, it means that every student who has accepted the “rules”, as they

were presented in this lesson, must accept as a fact that a × 0 = 0.

We will prove below that a × 0 = 0, but for the sake of brevity we will write

a × 0 in the form a0

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Since 0 is the additive identity , we know that…

A Proof That a0 = 0

Hence by substitution, we may interchange 0 + 0 and 0 in any mathematical expression.

In particular, we may multiply both sidesof the above equation by a to obtain…

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0 + 0 = 0

a(0 + 0) = a0

By the distributive property we know that…

a(0 + 0) = a0 + a0

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Hence, again by substitution, we may replace a(0 + 0 ) by a0 + a0 to obtain…

Since a and 0 are numbers, and since the numbers are closed with respect to

multiplication, we know that a0 is a number. So, as an abbreviation, let b stand for the

number a0. That is…

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a0 + a0 = a0

a0 = b

a(0 + 0) = a0

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Substituting b for a0 we obtain…

If we now subtract b from both sides of the equation, we see that…

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b + b = b

b + b = b

a0 + a0 = a0

–b = –b b = 0

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And since b = a0, and b = 0 we see that…

Thus, a0 = 0 is an inescapable conclusion based on the accepted properties of our

number system. Therefore, it may be treated as a fact in our “game”.Summarized more formally…

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a0 = 0b = a0

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To prove that a0 = 0next


(1) 0 + 0 = 0 (1) Additive Identity Property

(2) a(0 + 0) = a(0) (2) Substitution

(3) a(0 + 0) = a0 + a0 (3) Distributive Property

(4) a0 + a0 = a0 (4) Substituting (3) into (2)

(5) b + b = b (5) Definition of b (b = a0)

(6) (b + b) + -b = b + -b (6) Substituting (5) into (b+b) + -b

(7) b + (b + -b) = b + -b (7) Associative Property (+)

(8) b + 0 = 0 (8) Additive Inverse Property

(9) b = 0 (9) Additive identity Property

(10) a0 = 0 (10) Substituting a0 for b

In our informal discussion for proving that a0 = 0; when we arrived at the equation b + b = b, we nonchalantly subtracted

b from both sides to obtain b = 0. However, none of our rules justified this

assertion.

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A Possible Oversight

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In our more formal proof, steps (6) through (9) rectified this oversight.

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Hopefully, the above situations help you internalize what we mean by a proof and why it's necessary that we have proofs. However,having done this, our strategy in this course will be to prove only those facts that might not seem to be obvious to us from our past

experiences in mathematics. In other words, we shall continue to state as facts such things as: we may “subtract equals from

equals” or “when adding numbers, the sum does not depend on how the terms are

rearranged and/or regrouped” etc. More practice is left for the Exercise Set.

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