Download Area of A Trapezoid

Mathematical Proofs
Table of Contents
Proof
Area of a Trapezoid
Area of a Circle
Volume of a Right Circular Cylinder
Volume of a Sphere
Algebraic Identities
Irrationality of 2
Division by 0
Dividing by a Fraction
Slope of a Line is Constant
Product Property of Radicals
Square Roots
Pythagorean Theorem and Converse
Definition of pi
Circumference of a Circle
Zero Product Rule
N is irrational if N is a positive integer that
is not a perfect square
Density – A Property of Rational Numbers
Square and Triangular Numbers
Multiplication of Negative Numbers
Zero Property of Multiplication
Commutative, Associative, and Distributive
Properties
Terminating Decimals
1
Standard
7MG2.1
Geometry 8.0
6MG1.1
6MG1.2
7MG2.1
Geometry 8.0
6MG1.3
7MG2.1
Geometry 8.0
Geometry 8.0
Algebra I 10.0
7NS1.4
Algebra I 1.0
Algebra I 25.0
3NS2.6
6NS2.2
7NS1.2
7AF3.3
Algebra I 6.0
Algebra I 10.0
Algebra I 1.0
7MG3.3
Geometry 14.0
6MG1.1
6MG1.2
6MG1.1
6MG1.2
7AF1.3
Algebra I 11.0
7NS1.4
Algebra I 1.1
Algebra I 1.1
7NS1.0
Algebra I 1.1
6NS2.3
7NS1.2
7 NS1.0
Algebra I 1.1
7NS1.0
Algebra I 25.0
7NS1.5
Page(s)
2
3, 19
4
5
6, 7, 8, 27
9, 10, 21
11
12
13
14
15
16, 17
18
18
19
20
22
24
25
26
28
30
Area of a Trapezoid
Proof of A Trapezoid 
1
b2  b1 h 
2
b1
Region 1
h
h
Region 2
Region 3
x
b2  b1  x
b1
b2
The proof of the area formula falls into place as follows:

The area of triangle Region 1:

The area of rectangle Region 2:
1
Aleft traiangle   x h 
2
Arectangle  b1 h 
1
Aright triangle   b2  b1  x h 
2
 The combined area of the three Regions:
1
1
A Trapezoid  x h   b1 h   b2  b1  x h  … Adding the three regions
2
2
1
1
1
1
A Trapezoid  x h   b1 h   b2 h   b1 h   x h 
2
2
2
2
1
1
A Trapezoid  b1 h   b2 h   b1 h 
… Addition Property
2
2
1
1
A Trapezoid  b2 h   b1 h 
… Simplify
2
2
1
A Trapezoid  b2  b1 h 
Q.E.D.
2

The area of triangle Region 3:
Area of a Circle
2
Using the concept of the Area of a Triangle to show that Area of a Circle =  r 2 .
r
2 r
If we cut the circles along the radius of the disk and let them fan out to become straight lines, we get
a triangle (because the ratio of the circumference of the circles to their diameters is constant). The
base length of the resulting triangle is equal to the circumference of the original circle, and its
height is equal to the radius of this circle.
Thus, the area of a circle is equal to half of the product of the radius and the circumference.
ATriangle 
1
area of base height   which equals to the area of the DISK or CIRCLE.
2
1
area of base height
2
1
 2  r r    r 2
2
Since, ATriangle 
ATriangle

Volume of a Right Circular Cylinder
3
Volume = (Area of Base)(Height)
V   r2 h
If you bend a rectangular sheet of paper, bringing two opposite sides together, you will get an open
tube.
2 r
h
h
This is called an open cylinder.
The surface area of this open cylinder is the area of its curved surface given by the product of the
distance round the rim and the height.  2  r  h  2  r h
If the top and bottom of the cylinder are covered, we will have a closed cylinder.
Thus, the formula for volume of prism or cylinder is given by
V = Volume = (Area of Base)(Height)
V   r2 h
4
Volumes
Cavalieri’s Principle: Solids with the same height and
with cross-sections of equal area have the same volume; in
particular, prisms or cylinders with equal bases and heights
have the same volume.
Volume of a Sphere
An Experimental Approach:
If we build a cylinder with height 2r and the diameter 2r and fill the cylinder with water and then
1
remove the sphere from the cylinder, the water will only take up of the cylinder. Thus, the
3
2
volume of the sphere is
the volume of the cylinder having the same diameter as the sphere and
3
height equal to the diameter.
2r
2r
Proven earlier
Given
Substitution
Equation 1
So,
….
V   r2 h
….
h  2r
2
V   r  2r  ….

V  2 r 3
Simplify
Equation 2
Volume of the Cylinder:
….
We know from the experiment that:

5
Volume of the Sphere 
2
the Volume of Cylinder
3
Volume of the Sphere 
2
3
Volume of the Sphere =
4
 r3
3
 2 r 
3
Q.E.D.
Algebraic Identities
Algebraic manipulation very often can be explained with the aid of geometrical figures.
Objective 1: To show that  a  b   a 2  2ab  b 2
a
2
I
II
III
IV
a
The figure on the right is a square made up of
four Regions: I, II, III & IV.
Each side of the square has length
 a  b  units.

b
b
Area of the square is:
2
 a  b  a  b    a  b  square units.
Area Region I:
A   a  a   a2
Area Region II:
A   a b   ab square units
Area Region III:
A  b  a   ba
square units
Area Region IV:
A  b b   b2
square units
square units
Area of the Square = Area Region I + Area Region II + Area Region III + Area Region IV
6
a 2  ab  ba  b 2 , which is  a 2  2ab  b 2
…
Sum of the areas in the large square
a2  2ab  b2   a  b  (a  b)
…
Equate the two expressions for the area
Objective 2: To show that  a  b   a 2  2ab  b 2 (see an alternate proof on page 27)
2
b
a-b
I
a
b
IV
b
The figure on the left is made up of
four Regions: I, II, III, and IV.
Regions: I, II, III make up a
SQUARE, which has an AREA of
a 2 square units.
II
III
Region IV has an area of b 2 square
units.
a
Area Region I:
A  b  a  b
square units
Area Region II:
A   a b   ab
square units
Area Region III:
A  a  b
square units
Area Region IV:
A  b b   b2
square units
Area Regions I & IV:
A   a  b  b b   ab
square units
We know
7
2
Region I + Region II + Region III = a 2
b(a  b)  ab  (a  b)2  a 2
(a  b)2  a 2  b(a  b)  ab
(a  b)2  a 2  ab  b2  ab
(a  b)2  a 2  2ab  b2
Sum of areas in the square
Isolating (a  b)2
Distributive Property
Associative and Commutative
Properties
Objective 3: To show that a  ba  b  a 2  b 2
a
a
II
II
I
a b
I
b
Figure 1
b
a
a b
Figure 2
A small square (in red), area b2 square units is removed from a bigger square of area a2 square units
as shown in the Figure 1 above. Thus, the area of regions I and II is a2 – b2 square units.
In Figure 2, the two remaining regions, I and II, are rearranged to form the rectangle shown.
The area of the newly formed rectangle is equal to a  ba  b square units
Hence, a  ba  b  a 2  b 2
Further justification:
Total Area of Figure 1:
Total Area of Figure 2:
8
a2  b2
a  ba  b  a 2  b 2
Proof 2 is irrational
The Greeks discovered that the diagonal of a square whose side is 1 unit long has a diagonal whose
length cannot be rational. By the Pythagorean Theorem, the length of the diagonal equals the
square root of 2. So the square root of 2 is irrational!
The following proof is a classic example of a proof by contradiction: We want to show that A is
true, so we assume it's not, and come to contradiction. Thus A must be true since there are no
contradictions in mathematics!
Proof:
Suppose 2 is rational. That means
p and q , where q ≠ 0.
2
2 can be written as the ratio of two integers
p
q
……… Equation #1
Where we may assume that p and q have no common factors. (If there are any common
factors we cancel them in the numerator and denominator.) Then,
p2
…
Squaring both sides of Eq. #1
2 2
q
This implies
…
Isolating p 2
p 2  2q 2
Thus, p 2 is even. This means that p itself must be even.
Why? Because if p was odd, then  p p would also be odd. (An odd number times an odd
number is always odd.)
So p is an even number. Then p is 2 times some other whole number, or p  2k , where k
is this other number. We don't need to know exactly what k is; it doesn't matter. Soon we
will get our contradiction:
Then p 2 = 4k2. So 4k2 = 2q2, hence q2 = 2k2.
Hence q 2 is even and therefore q itself must be even.
So p and q are BOTH even. They both have a common factor of 2. This contradicts our
assumption that p and q have no common factors.
 The square root of 2 is NOT rational.
9
The Irrationality of
Prove that
is an irrational number.
Solution:
The number,
, is irrational, ie., it cannot be expressed as a ratio of integers a and b. To prove that
this statement is true, let us assume that
is rational so that we may write
= a/b
1.
for a and b = any two integers. To show that
is irrational, we must show that no two such
integers can be found. We begin by squaring both sides of eq. 1:
2 = a2/b2
2.
2b2 = a2
or
2a.
From eq. 2a, we must conclude that a2 (and, therefore, a) is even; b2 (and, therefore, b) may be even
or odd. If b is even, the ratio a2/b2 may be immediately reduced by canceling a common factor of 2.
If b is odd, it is possible that the ratio a2/b2 is already reduced to smallest possible terms. We assume
that b2 (and, therefore, b) is odd.
Now, we set a = 2m, and b = 2n + 1, and require that m and n be integers (to ensure integer values
of a and b). Then
a2 = 4m2
2
and
3.
2
b = 4n + 4n + 1
4.
Substituting these expressions into eq. 2a, we obtain
2(4n2 + 4n + 1) = 4m2
2
or 4n + 4n + 1 = 2m
2
5.
6.
The L.H.S. of eq. 6 is an odd integer. The R.H.S., on the other hand, is an even integer. There are no
solutions for eq. 6. Therefore, integer values of a and b which satisfy the relationship
= a/b cannot be found. We are forced to conclude that
is irrational.
10
Division by 0 is Meaningless
Since division is defined in terms of multiplication, the rules for the sign of a product lead directly
to the corresponding rules for the sign of a quotient, which are given below:
Symbols
a  0; b  0
a
a b 
b
a a
 a   b  

b b
a
a
a b 

b
b
a
a
a   b  

b
b
0
0b   0
b
0
0   b  
0
b
Divisor and dividend
Quotient
Both Positive
Positive
Both Negative
Positive
One Positive
Negative
One Negative
Dividend 0
0
Divisor not 0
Note: For b  0 , in the quotient
0
Q
b
You must have
Q0
Since the product
bQ
Can be 0 only if one of the factors is 0. Therefore,
0
0
b
Also, for b  0 , if you had
b
Q,
0
Then it would follow that
b  0Q
But this is impossible, since
b  0 and 0Q  0
For example, suppose you wished to find the quotient Q in:
Q  20
You would have to find a number Q such that Q0  2 . But that is impossible because you have
already seen that the product of 0 and any number must be 0. Thus, we can conclude that division
by 0 is meaningless.
11
Dividing by a Fraction
If
Let
a
c
a c a d
and
are rational numbers with b , c , and d  0 , then    .
b
b d b c
d
a c
 Q,
b d
where b  0 , c  0 , and d  0 (you should ordinarily substitute
particular numbers for a, b, c, and d )
Finding the quotient Q means finding the missing factor Q in
c
a
Q 
d
b
Since,
a
c
 Q and
name the same number
b
d
The product of this number and
d
can be written either as
c
d c

   Q  or as
c d

d a

c b
That is,
d c
 d a
 Q  
c d
 c b
Therefore,
d a
d c 
  Q  
c b
c d
12
1 Q 
d a

c b
Q
d a

c b
…
Associatively
Q.E.D.
Slope of a Line is Constant
Prove that the slope of a nonvertical line does not depend on which two points are used to
calculate it.
Concept:
Slope of a line is CONSTANT and it does NOT depend on which two
points are used to calculate it.
Proof:
Let A, B, C , D be any four points on the line. Prove that the slope obtained using A and B equals
the slope obtained using C and D.
D
y2  rise2
C
x2  run 2
A
y1  rise1
x1  run1
Study Tip:
If two figures are similar, then the ratios of the lengths of corresponding sides
are equal.
y1 x1

y2 x2
y1  x2  x1  y2
y1  x2 x1  y2

x1  x2 x1  x2
y1 y2

x1 x2
Slope using
points A and B
13
Slope using
= points C and D
…
Definition of similar triangles
…
Cross product property of equality
…
Divide both sides by x1  x2
…
Simplify
Q.E.D.
B
Product Property of Radicals
3
Prove:
a  3 b  3 ab
Proof:
 a   a and  b   b
 a  b   ab
 ab   ab
  a  b    ab 
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
a  3 b  3 ab
…
Definition of cube root
…
Substitution, Commutativity, Associativity
…
Definition of cube root
…
Substitution or transitive property of equality
…
Uniqueness of the real cube root of a number
…
an
a
Definition. If b  0 , then    n
b
b
…
Definition of cube root
…
Substitution
…
Definition of cube root
…
Substitution or transitive property of equality
…
Uniqueness of the real cube root of a number
Q.E.D.
3
Prove:
3
a 3a

c
c
, c0
Proof:
3
3a
 3  
 c
 a
3
3
 a
 c
3
3
3
n
3
 a and
 c
3
3
c
3
3a a
 3  
c
 c
3
 a a
 3  
c
 c
3
3a  a
  3  3 
 c  c

 

3
a 3a

3
c
c
3
Q.E.D.
14
Square Roots
Prove: Any positive real number has a POSITIVE real number as a square root
The area of a SQUARE is obtained by squaring the length of a side. The side of a square with an
area of 2 square units must have a length whose square is 2. That is, the side has length 2 .
Figure 1 on the right shows
squares with AREAS 1, 2, 3, and
4 and sides measuring 1, 2 , 3 ,
and 2, respectively.
2
3
2
1
Figure 1
These squares show that any positive real number k has a POSITIVE real number as one of its
square roots. Specifically, the square whose area is equal to k square units has a side whose length
is equal to the positive square root of k.
Since every real number has an additive inverse, and since a 2   a  for each real number a,
every positive real number clearly has a negative square root as well.
2
Thus the statement
a2  a
k , k  0 , shall name only a nonnegative number. We also
is a consequence of our definition that
have the statement
 a2   a
This statement simply reasserts the definition of  k . This idea is of importance in later
mathematics courses when, for example, students are asked to solve equations such as:
 x  3
15
2
 4.
Pythagorean Theorem
Consider Figure 1, a square whose side has length a + b. As indicated, there are four identical right
triangles and an interior quadrilateral. The legs of each triangle have lengths a and b. The
hypotenuse of each triangle is c. The measures of the two acute angles of each triangle have a sum
of 90°. Now look at the point P. Since the two acute angles at that point have a sum of 90°, the
interior angle of the quadrilateral must have a measure of 90°. The same is true for each of the
other vertices of the interior quadrilateral. And the four sides of the interior quadrilateral all have
length c. Thus the interior quadrilateral is a square.
Hence, the area of the large square = the sum of the areas of the 4 triangles plus the area of the
1
1
interior square  4   ab  a 2  b 2  4   ab  c 2 .
2
2
But the large square can also split into two squares and 4 triangles, as shown in Figure 2. The area
1
of Figure 2 is  4   ab  a 2  b 2 . These two expressions for the area of the large square must be
2
equal. Hence,
1
1
4   ab  a 2  b 2  4   ab  c 2
2
2
a2 + b2 = c2
a
a
b
a
c
b
b
b
c
b
c
c
a
c
c
a
a
b
Figure 1
16
b
a
b
Figure 2
a
Converse of Pythagorean Theorem
Consider triangle ABC with sides of length a, b, and c respectively. In the figure, angle C is
unknown. We assume that a2 + b2 = c2, and we want to prove that angle C is a right angle. Now
consider triangle A′B′C′. As indicated, this is a right triangle with the right angle at C and legs of
length a and b. Since we know that a2 + b2 = c2 and triangle A′B′C′ is a right triangle, it follows that
the length of its hypotenuse must be c. Thus in the two triangles, the three pairs of corresponding
sides are equal in length. Hence these triangles must be congruent by the Side-Side-Side theorem of
congruent triangles. Hence their corresponding angles are equal. Since angle C is a right angle, so
is angle C′.
B
B
a
c
a
?
C
A
b
17
c
C
A
b
Definition of π
The number π is defined to be the length of the perimeter (usually called the circumference) of a
circle whose diameter is 1 unit. Its value is irrational but is approximately equal to 3.14159.
Circumference of a Circle
Notice that any two circles are similar to each other. In similar geometric figures, the lengths of
corresponding measures are proportional. Now consider the two circles below, with radii of ½ and
r. Call the diameter and circumference of the second circle d and C respectively. Since these
circles are similar, the ratio of the circumference of the first circle to its diameter is equal to the
ratio of the circumference of the second circle to its diameter. This means that π/1 = C/d. When
we simplify this equation, we get C = πd, or equivalently, since d = 2r, we have C = 2πr.
C
1/2
d
r
18
Area of a Circle
Consider the following figure. The circle contains an inscribed regular hexagon. As indicated, the
length of each side of the hexagon is s and the apothem (the distance of the perpendicular segment
from the center of the circle to the side of the hexagon) is a. Notice that the hexagon can be split
into six congruent triangles. Then the area of one of these triangles is (1/2)sa. Hence the area of
the entire hexagon is 6(1/2)sa = (1/2)(6s)a = (1/2) times the perimeter of the hexagon times the
apothem.
Now suppose instead of an inscribed regular hexagon, we have an inscribed regular polygon with
12 sides. Using the same symbols as above, the area of this polygon is (1/2) times the perimeter of
the polygon times the apothem. We can also see that the area of the polygon is closer to the area of
the circle than the area of the hexagon. Clearly, as the number of sides of the polygon increases, the
area of the polygon gets closer and closer to the area of the circle. The formula for the area of this
polygon is again (1/2) times its perimeter times the apothem. Finally, notice that as the number of
sides increases, the perimeter of the polygon approaches the circumference of the circle, 2πr, and
the apothem approaches the radius of the circle r. Thus the area of the circle is this limiting value,
(1/2) times 2πr times r. When simplified, this is just πr2.
S
a
a
Zero Product Rule: If ab = 0, then either a = 0 or b = 0.
Suppose that the product ab is equal to zero. Suppose also that a is not zero. Then the
multiplicative inverse of a, a-1, exists. So we have:
ab = 0
(a-1)ab = (a-1)0
(a-1a)b = 0
1b = 0
b=0
19
Multiply both sides by a-1
Associativity of multiplication
Definition of multiplicative inverse
Definition of multiplicative identity
A proof that
N is irrational if N is a positive integer that is not a perfect square
The proof that we are about to render is dependent on the well ordering principle for positive
integers. This principle states that any nonempty set of positive integers contains a smallest positive
integer. As in the usual proof that the 2 is irrational, we will proceed by contradiction. Thus
assume that the N is rational. Hence there are positive integers k and l with l  0 such that
k
N  . Consider the set
l
k
S  l : N  for some positive integers k and l} .
l
The set S is a nonempty set of positive integers; therefore, it contains a smallest positive integer by
the well ordering principle. Denote the smallest positive integer in S by n. Thus there is a positive
m
integer m such that N  . Now divide the integer m by the integer n to obtain a quotient q and
n
a remainder r satisfying m  nq  r , where 0  r  n ; note that the remainder r is greater than zero
since the number N is not a perfect square. Now from the equation
m nq  r
N  
n
n
we obtain the equation
(nq  r )2 n2 q 2  2nqr  r 2
.
N

n2
n2
Hence
Nn2  n2 q 2  2nqr  r 2 .
Now rewrite this equation to obtain the equations
( N  q 2 )n2  2nqr  r 2
( N  q 2 )n2  nqr  nqr  r 2
n[( N  q 2 )n  qr ]  r (nq  r )
Now we obtain the fractions
( N  q 2 )n  qr nq  r

 N.
r
n
Since r and N are positive integers, we conclude that the numerator ( N  q 2 )n  qr is a positive
integer. Therefore the positive integer r is in the set S. But r is strictly less than n; thus we have
obtained a contradiction to the fact that the integer n is the smallest positive integer in the set S.
{
20
A second proof that
2 is irrational
The proof that we are about to render is dependent on the well ordering principle for positive
integers. This principle states that any nonempty set of positive integers contains a smallest positive
integer. As in the usual proof that the 2 is irrational, we will proceed by contradiction. Thus
assume that the 2 is rational. Hence there are positive integers k and l with l  0 such that
k
2  . Consider the set
l
k
S  l : 2  for some positive integers k and l}.
l
The set S is a nonempty set of positive integers; therefore, it contains a smallest positive integer by
the well ordering principle. Denote the smallest positive integer in S by n. Thus there is a positive
m
integer m such that 2  . Since 2  1 , the integer m satisfies m  n . Thus there is a positive
n
integer k such that m  n  k . Furthermore, since
nk m
  22
n
n
(n  k )2
we conclude that the positive integer k satisfies the inequality 1  k  n . Now 2 
. Hence
n2
n 2  2kn  k 2  2n 2 .
Rewrite the previous equation to obtain the equation
n 2  kn  kn  k 2 .
Therefore,
n( n  k )  k ( n  k ) .
Finally we obtain the equation
nk nk

 2.
k
n
The integer n  k is a positive integer since 1  k  n . This inequality together with the previous
equation shows that the integer k is in the set S. But the positive integer k is less than the integer n.
We have established a contradiction.
{
21
Density – A Property of Rational Numbers
The set of rational numbers has the property that between any two different rational numbers there
are infinitely more rational numbers. This property is referred to as density.
One way
"between any two real numbers a and b there exists a rational number r such that a  r  b " then
the proof is as follows:
We first need the Archimedean Property:
If x > 0 and y > 0 are real numbers, then there exists a positive integer n such that n  x  y . This
tells us that even if a is quite small and b is quite large, some integer multiple of a will exceed b .
Another way to think of this is, "given enough time, one can empty a swimming pool with a tea
spoon."
We need to show that given real numbers a and b we can construct a 
m
 b for some integers
n
m and n.
We can assume that n > 0 without loss of generality. If our fraction
m
is to be negative, just let m
n
be the negative number.
Since n > 0 we can multiply everything by n and we don't have to worry about flipping the
inequality. Thus we need an  m  bn .
Since b  a we have b  a  0 and so, by the Archimedean property, there exists a positive integer n
such that n(b  a )  1 (pick x  b  a and y  1 in the statement of the Archimedean property above;
we can do this since b  a and 1 are both greater than zero).
Since bn  na  1it is fairly evident that there is an integer m between an and bn . Does this make
sense? If the difference between two numbers is greater than one unit (i.e., if two numbers are more
than one unit apart), there must be an integer between them, since the integers are spaced exactly
one unit apart. If this isn't clear, try to pick two numbers that are more than one unit apart such that
there isn't an integer between them. I think you'll find yourself unable to do so. Thus there exist
integers m and n such that an  m  bn , which implies
a
m
 b which was to be shown.
n
Notice that this proof doesn't tell us how to find
are very common in higher mathematics.
22
m
m
. All it does is show that
exists. Such proofs
n
n
Note: One way of identifying a number between two given numbers is to find their average. That
the average of two numbers lies exactly halfway between them can be proven briefly as follows: if
ab
, does not lie exactly
a and b are rational numbers such that a  b , assume that their average,
2
halfway between them. Then
ab
ab
a  b
2
2
 a  b  2a  2b   a  b
2  a  b   2a  2b
2  a  b  2  a  b
Since the last statement is false, the assumption is also false, and
and b , where a  b .
23
ab
does lie midway between a
2
Square and Triangular Numbers
To compute a square of a given number 22 = 4, 32 = 9, 42 = 16, and so on. Why is the operation of
multiplying a number by itself called squaring? The reason is best grasped from the following
picture.
A square with a side of length n can be visualized as comprising a grid of n  n smaller squares of
size 1 1 .
Triangular numbers owe their name to a similar construction, now of triangles. Actually we already
used triangular numbers in the proof of Lemma 3 above. Triangular numbers are in the form 1, 3, 6,
n(n  1)
10, 15, ... The general formula is that for every n the number N 
is triangular and
2
describes the number of points arranged in a triangular shape with n points on a side as shown
below.
Of course it needs to be proven that the total number of crosses arranged as in the diagram is given
n(n  1)
by the formula N 
. However, we can observe that the number of crosses in every triangle
2
equals the sum
1 + 2 + ... + n,
where n is the number of crosses on the side of the triangle. Thus, along the way, we are going to
prove that
1 + 2 + ... + n = n(n+1)/2.
Another diagram will make this statement obvious
Indeed, two triangles, each with n crosses on the side, together form an n  n  1
rectangle. You may question whether pointing to the diagram proves anything. No, of
course by itself the diagram does not prove anything. But it does help to visualize and
perhaps discern the proof. Indeed, we seek a formula for the sum of integers from 1
through n. This sum represents the number of crosses in a triangle. Two triangles of which one is
rotated 180o form a rectangle, i.e. a shape in which all rows have the same number of crosses. Rows
in the first triangle count crosses 1,2,3... in the increasing order whereas in the second we have rows
with n, n-1, ... crosses in the decreasing order. Let us mimic this procedure in an algebraic manner.
First take two sums (one in increasing, another in decreasing order) (1 + ... + n) + (n + ... + 1). Now,
regroup the terms by combining first terms in both sums, then second terms and so on:
[1+n] + [2+(n-1)] + [3+(n-2)] + ... + [(n-1)+2] + [n+1] = n(n+1)
Quite rigorously. Q.E.D.
It is obvious that all Euclid's perfect numbers are triangular.
24
Multiplication of Negative Numbers: (-1)(-1) = 1
We know the associative property of addition and the distributive property of multiplication over
addition. We also know the commutative properties of addition and multiplication of real numbers.
We know the identity properties of addition and multiplication: a  0  0  a  a and a 1  1 a  a
for all real numbers. Finally, we know the inverse property of addition: For each real number a
there is a real number (a) such that a  (a)  (a)  a  0 . Using these basic arithmetic
properties we are able to prove that (1)(1)  1 .
We have already established that a 0  0 for all real numbers a; thus ( 1)  0  0 . Also
(1)  1  1  (1)  0. So applying the multiplicative identity property and the distributive property
we obtain the equation
(1)  (1)(1)  (1) 1  (1)(1)  (1) 1  (1)   (1)  0  0 .
So we have shown that
(1)  (1)(1)  0 .
Now add 1 to both sides of the above equation to obtain
1  [(1)  (1)(1)]  1  0  1
To this latter equation apply the associative property of addition to obtain
(1)(1)  0  (1)(1)  [1  (1)]  ( 1)( 1)  1  [( 1)  ( 1)( 1)]  1 .
25
Zero Property of Multiplication: a∙0 = 0∙a = 0 for all real numbers a
We know the associative property of addition and the distributive property of multiplication over
addition. We also know the commutative properties of addition and multiplication of real numbers.
We know the identity properties of addition and multiplication: a  0  0  a  a and a 1  1 a  a
for all real numbers. Finally, we know the inverse property of addition: For each real number a
there is a real number (a) such that a  (a)  (a)  a  0 . Using these basic arithmetic
properties we are able to prove that a  0  0  a  0 for all real numbers a.
Let a be an arbitrary real number. Then using the identity and distributive properties we obtain the
equation
(1) a  0  a(0  0)  a  0  a  0
By the inverse property for addition, there is a real number  (a  0)  such that
a  0   (a  0)    (a  0)   a  0  0 .
So from equation (1) we obtain
a  0   (a  0)   [a  0  a  0]   (a  0)  .
Applying the associative property of addition, we obtain the equation
a  0   (a  0)   a  0  [a  0   (a  0) ]
Applying the inverse property of addition, we obtain the equation
0  a0 0
We conclude that a 0  0 . Since multiplication is commutative we also have that 0  a  0 .
26
To show that  a  b   a 2  2ba  b 2 (an alternate approach – see page 7)
2
a-b
No harm is done if we assume that a  b .
The figure on the right is a square made up
of four Regions: I, II, III & IV.
b
I
II
III
IV
a-b
Each side of the large square has length
a units.
b
Area of the large square, composed of the four regions is
a 2 square units.
Area Region I:
A   a  b  a  b   (a  b)2 square units
Area Region II:
A   b  a  b   ba  b2 square units
Area Region III:
A   b  a  b   ba  b2 square units
Area Region IV:
A  b b   b2
square units
Area of the Square = Area Region I + Area Region II + Area Region III + Area Region IV
Therefore we obtain the equation a 2  (a  b)2  (ba  b2 )  (ba  b2 )  b2 . Now simplify this
equation using associative and commutative properties of addition to obtain the equation
a 2  (a  b)2  2ba  b2 . We can now solve for (a  b)2 to obtain
(a  b)2  a 2  2ba  b2 .
27
Use the commutative, associative and distributive properties and the four operations to justify
each step of the following calculation:
[3 × (4 – 7) + (12 ÷ 3 × (4 – 2) + 8)] – [(2 × ((3 + 8) + 4) ÷ (7 – 3)]
= [3 × (4 – 7) + (12 ÷ 3 × (4 – 2) + 8)] – [(2 × ((8 + 3) + 4) ÷ (7 – 3)]
Commutative Property
= [3 × (4 – 7) + (12 ÷ 3 × (4 – 2) + 8)] – [(2 × (8 + (3 + 4)) ÷ (7 – 3)]
Associative Property
= [3 × (4 – 7) + (12 ÷ 3 × (4 – 2) + 8)] – [(2 × (8 + 7) ÷ (7 – 3)]
Addition
= [[(3 × 4) – (3 × 7)] + (12 ÷ 3 × (4 – 2) + 8)] – [(2 × (8 + 7) ÷ (7 – 3)]
Distributive Property
= [([(3 × 4) – (3 × 7)] + 12 ÷ 3 × (4 – 2)) + 8] – [(2 × (8 + 7) ÷ (7 – 3)]
Associative Property
= [([(3 × 4) – (3 × 7)] + 12 ÷ 3 × (4 – 2)) + 8] – [((2 × 8) + (2 × 7)) ÷ (7 – 3)] Distributive Property
= [([12 – 21] + 12 ÷ 3 × (4 – 2)) + 8] – [(16 + 14) ÷ (7 – 3)]
Multiplication
= [( – 9 + 12 ÷ 3 × (4 – 2)) + 8] – [(16 + 14) ÷ 4]
Subtraction
= [( – 9 + 4 × (4 – 2)) + 8] – [(16 + 14) ÷ 4]
Division
= [( – 9 + [(4 × 4) – (4 × 2)]) + 8] – [(16 + 14) ÷ 4]
Distributive Property
= [([(4 × 4) – (4 × 2)] + (-9)) + 8] – [(16 + 14) ÷ 4]
Commutative Property
= [[(4 × 4) – (4 × 2)] + ((-9) + 8)] – [(16 + 14) ÷ 4]
Associative Property
= [[16 – 8] + ((-9) + 8)] – [(16 + 14) ÷ 4]
Multiplication
= [[16 – 8] + ((-9) + 8)] – [(14 + 16) ÷ 4]
Commutative Property
= [8 + ((-9) + 8)] – [(14 + 16) ÷ 4]
Subtraction
= [8 + ((-9) + 8)] – [30 ÷ 4]
Addition
= [8 + (-1)] – [30 ÷ 4]
Addition
= [8 + (-1)] – [15/2]
Division
= 7 – [15/2]
Addition
= –½
Subtraction
28
Prove that multiplication is commutative by showing as a typical example that
29 × 58 = 58 × 29. Use only the multiplication algorithm, expanded form, factoring, and the
distributive property.
Proof.
29 × 58 = [20 + 9] × 58
Expanded form
= [20 × 58] + [9 × 58]
= [20 × 58] + [9 × (50 + 8)]
= [10 × 2 × 58] + [9 × 50] + [9 × 8]
= [10 × 2 × (50 + 8)] + [9 × 50] + [9 × 8]
= [10 × 2 × 50] + [10 × 2 × 8] + [9 × 5 × 10] + 72
Distributive property
Expanded form
Distributive property and factoring
Expanded form
Distributive property,
factoring, and multiplication
= [10 × 2 × 5 × 10] + [10 × 16] + [45 × 10] + 72
= [10 × 10 × 10] + 160 + 450 + 72
= 1000 + 160 + 450 + 72
= 1682
Multiplication and factoring
Multiplication
Multiplication
Addition
58 × 29 = [50 + 8] × 29
Expanded form
= [50 × 29] + [8 × 29]
= [50 × 29] + [8 × (20 + 9)]
= [10 × 5 × 29] + [8 × 20] + [8 × 9]
= [10 × 5 × (20 + 9)] + [8 × 20] + [8 × 9]
= [10 × 5 × 20] + [10 × 5 × 9] + [8 × 2 × 10] + 72
Distributive property
Expanded form
Distributive Property and factoring
Expanded form
Distributive Property,
factoring, and multiplication
= [10 × 5 × 2 × 10] + [10 × 45] + [16 × 10] + 72
= [10 × 10 × 10] + 450 + 160 + 72
= 1000 + 450 + 160 + 72
= 1682
Multiplication and factoring
Multiplication
Multiplication
Addition
Since both products are equal to 1682, they must be equal to each other.
Thus multiplication is commutative in this example. But since there was nothing special about this
example, we can conclude that multiplication is always commutative.
29
Prove that a decimal is a terminating decimal if and only if it is equivalent to a fraction whose
denominator contains no prime factors other than 2 and 5.
Proof. Suppose the denominator of a fraction contains only the factors 2, 5 or both of them. Then
we can multiply that denominator by additional 2s and 5s so that there is an equal number of 2s and
5s. We also multiply the numerator of the fraction by the same number of 2s and 5s. This produces
a fraction equivalent to the original one whose denominator consists of an equal number of 2s and
5s. But this means that the denominator is a power of 10. A fraction whose denominator is a power
of 10 corresponds to a terminating decimal. Thus, if the fraction with a denominator that is a power
of 10 is the fraction a1a2a3….an /10n, then the decimal that is equivalent to this fraction is the
terminating decimal 0. a1a2a3….an .
Now suppose that we begin with a terminating decimal, say 0.a1a2a3….an . Then this decimal is
equivalent to the fraction a1a2a3….an /10n. The denominator of this fraction contains only the
prime factors 2 and 5.
30