Real Numbers 10th Class

Lesson Page

Math A

Properties of Real Numbers 

Let a, b, and c be real numbers, variables, or algebraic expressions.

  Property Example 1.  Commutative Property of Addition 

a + b = b + a 2 + 3 = 3 + 2

2. Commutative Property of Multiplication a • b = b • a

2 • ( 3 ) = 3 • ( 2 )

3. Associative Property of Addition a + ( b + c ) = ( a + b ) + c 

2 + ( 3 + 4 ) = ( 2 + 3 ) + 4

4.  Associative Property of Multiplication a • ( b • c ) = ( a • b ) • c

2 • ( 3 • 4 ) = ( 2 • 3 ) • 4

5. Distributive Property a • ( b + c ) = a • b + a • c

2 • ( 3 + 4 ) = 2 •  3 + 2 • 4

6. Additive Identity Propertya + 0 = a

 3 + 0 = 3

7.  Multiplicative Identity Property a •  1 = a

3 • 1  = 3

8. Additive Inverse Property a + ( -a ) = 0

3 + (-3) = 0

9.  Multiplicative Inverse Property 

Note: a can not = 0 

 Math A

 Binary Operations

A binary operation is simply a rule for combining two objects of a given type, to obtain another object of that type.  You first learned of binary operations in elementary school.  The

objects you were using were numbers and the binary operations you investigated were addition, subtraction, multiplication and division.  As you will discover in this lesson, binary

operations need not be applied only to numbers.  

 A binary operation on a finite set (a set with a limited number of elements) is often displayed in a table that

demonstrates how the operation is performed.

This table shows the operation * ("star").  The operation is working on the finite set

A = { a, b, c, d }.  The table shows the 16 possible calculations using the elements of set A.

Reading the table:  Read the first value from the left hand column and the second value from the top row.  The answer is in the cell where the row and column intersect.  For example, a * b = b,  b * b = c,                      c * d = b,  d * b = a  and so on.

What is the identity element forthe operation * ?(What single element will always return the original value?)The identity element is a.    a * a = a,   b * a = b,   c * a = c,  d * a = d

Checking for the Identity Element:You will know the identity element when you see it, because all of the values in its

row or column are the same as the row or column headings.

What is the inverse element for b ?(What element, when paired with b, will return the identity element a?)The inverse element of b is d.         b * d = a

Is the operation * commutative?(Does the property x + y = y + x hold for ALL possible arrangements of values?)Start testing values: a * b = b * a  is true since both sides equal b.c * d = d * c  is true since both sides equal b.WOW!  Having to test ALL possible arangements could take forever!  There must be an easier way.........

The operation * is commutative.

Testing for Commutativity Shortcut:It is easy to check whether an operation defined by a table is commutative.  Simply draw a diagonal line from upper left to lower right, and see if the table is symmetric about this line.

Since this table is symmetric about the diagonal (from upper left to

lower right), the operation is commutative.

It would only have taken one instance of lack of commutativity

for this answer to have been "no".  

True or False:(a * b) * c = a * (b * c) ???

This question deals with only one case of the associative property for

this table.  Unfortunately there is NO shortcut for checking

associativity as there is for checking

Perform the operations in the order indicated by the parentheses:(a * b) * c = a * (b * c)       b * c  = a * d            d  =  d   So, yes, this statement is true for this table.

commutativity when working with a table.

 If asked the general question "Does this operation possess the associative property?", you would have to check ALL possible arrangements.  On the other hand, if you find one instance

where associativity fails, you are done and the answer is "NO".

 Math A

Closure Property

The round smiley faces are a closed set.  No matter what happens to the smiley faces, only round smiley faces will be present in their box.

A set is closed (under an operation) if and only if the operation on two elements of the set produces another element of the set.  If an element outside the set is

produced, then the operation is not closed. 

Example:  If you multiply two real numbers, you will get another real number.   Since this process is always true, it is said that the real numbers are "closed under the operation of multiplication".  There is simply no way to escape the set of real numbers when multiplying. 

Closure:  When you combine any two elements of the set, the result is also

included in the set.

Example:  If you add two even numbers (from the set of even numbers),                      is the sum even?

Checking:  10 + 12 = 22 Yes, 22 is even.  6 + 8 = 14 Yes, 14 is even.  2 + 100 = 102 Yes, 102 is even.

Since the sum (the answer) is always even, the set of even numbers is closed under the operation of addition.

Let's check out this question.  If you divide two even numbers (from the set of even numbers), is

the quotient even?

Checking: 12 / 6 = 2 Yes, 2 is even.  24 / 2 = 12 Yes, 12 is even.  100 / 4 = 25 NO, 25 is not even!

When you find even ONE example that does not work, the set is not closed under that operation.  The even numbers are not closed under division.

Example:

The elements in a binary table are displayed horizontally and vertically outside the table (in this table, the elements are 1, 2, 3, and 4).  If the elements inside the table are limited to the elements 1, 2, 3, and 4, the table is closed under the indicated operation.

2 0 0 5

Fundamental Theorem of Arithmetic Before, we can provide this theorem, we will need an initial theorem on

division and one lemma which will we borrow from Euclid.

Theorem 1: Division Algorithm for Integers

This theorem shows that given any two integers, there exist a unique divisor

and a unique remainder.

(1) Let S be a set such that { a - bc : c is an integer, a-bc is greater or

equal to 0}

(2) S is not empty.

Case I: b divides a

In this case, then there exists a c where a - bc = 0. Therefore, 0 is an

element of this set.

Case II: b doesn't divide a

In this case, there exists an c where a - bc has a remainder > 0. This

remainder is then an element of this set.

(3) Then, then there exists d such that d is the smallest element of the set.

[Well Ordering Principle]

(4) We know that d is less than b.

(a) Assume that d is greater than or equal to b.

(b) Then d - b is greater than or equal to 0.

(c) Since

d - b = (a - bc) - b = a - bc - b = a - b(c+1)

(d) We conclude that a - b(c+1) must be an element of S.

(e) But a - b(c+1) is less than d

(f) Which is a contradiction since d is the smallest element. (3)

(5) Finally, for any a,b, the lowest value d and the remainder c are unique.

(a) Assume C,D are integers that can be derived from a,b.

a = bc + d, d is greater than 0 and less than b.

a = BC + D, D is greater than 0 and less than b.

(b) So

bc + d = BC + D --> b(c - C) = d - D

(c) Which means that b divides (d - D).

(d) Since d - D is less than b, we know d - D = 0 and that d = D.

(e) Which gives us b(c - C) = 0 or that c - C = 0.

(f) We can therefore conclude that c = C.

QED

Lemma 2: Euclid's Lemma

This lemma shows that given a prime that divides a product, either it

divides one value or it divides the other.

(1) Let's assume a prime p divides a product ab.

(2) Let's assume that p does not divide a.

(3) Then gcd(p,a)=1. [See blog on Greatest Common Denominators for

details]

(4) There exists s,t such that 1 = as + pt. [See blog on Greatest Common

Denominators]

(5) Multiplying both sides with b gives us:

b(1) = b(as + pt) = b = abs + ptd

(6) Since p divides ab, there exists k such that ab = pk.

(7) Combining with (5), we get

b = pks + ptd = p(ks + td).

QED

We can also tag on a corollary.

Corollary 2.1: Generalized Euclid's Lemma

The idea here is that if a prime divides a product of n elements, then it is

necessarily divides at least one of those elements.

(1) Let's assume that we have a product of n elements.

a = a1 * a2 * ... * an

(2) We can take any one of these elements and get:

a = a1 * (a2 * ... * an)

(3)So, by Euclid's Lemma, the prime either divides a1 or one of the rest of

the elements.

4) So, either it divides a1 or a2 etc.

(5) And we get to the last two elements, we are done by Euclid's Lemma.

QED

Theorem 3: Fundamental Theorem of Arithmetic

This theorem proves that each integer greater than 1 is the product of a

unique set of primes.

(1) Let S be a set S of {primes or products of primes > 1}.

(2) 2 is an element of S since 2 is prime.

(3) From (1), (2) there exists a value n which is 3 or greater where all

values less than n and greater or equal to 2 are in S.

(4) If n is a prime, then it is a member of S.

(5) If n is not a prime, then it is also a member of S.

(a) If n is not a prime, then there exist a,b such that n=ab where a,b are

greater than 1 and less than n. [Definition of a prime]

(b) But a,b must be elements of S since they are less than n. [From (3)]

(c) Then, n must also be a member of S since n is a product of two numbers

which are either primes or products of primes.

(6) Now, all the primes that make up n are necessarily unique.

(a) Let n = p1 * p2 * ... pr = q1 * q2 * ... qs where p,q are all primes.

(b) Let's start with p1

(c) Either p1 divides q1 or q2 or someother value. [Euclid's Generalized

Lemma]

(d) Likewise, each value pr can be matched by some value qs.

(e) In other words, they must necessarily be the same combination.

Chapter: Real Numbers

Reader Friendly Format

Euclid's division Lemma and Algorithm

1. Find the quotient and remainder q and r for the pairs of positive integers a and b given below:(i) 23, 4 (ii) 81, 3 (iii) 12, 5

(i) 23, 4 When 23 is divided by 4 quotient is 5 remainder is 3. Therefore, 23 = 5 4 + 3 q = 5 ; r = 3 and 0  r < 5

(ii) 81, 3 When 81 is divided by 3 quotient is 27 and remainder is 0. Therefore, 81 = 27 3 + 0

So, q = 27 ; r = 0 and 0   r < 27

(iii) 12, 5 On dividing 12 by 5, we have quotient is 2 and remainder 2. Therefore, 12 = 5 2 + 2.

So, q = 2 ; r = 2 and 0  r < 5.

2. State Euclid's Division Lemma.

Euclid 's Division Lemma states that given positive integer a and b, there exist unique integers q and r satisfying a = bq + r ; 0  r < b.

3. Describe  Euclid 's Division algorithm for finding the HCF of two numbers.

Euclid 's division algorithm is used to find the HCF of two numbers by the successive use of Euclid 's division lemma. Let us find the HCF of 60 and 108 using this method.

Step 1: Since 108 > 60 applying Euclid's Lemma to 60 and 108, we have,108 = 60 1 + 48 where 0  48 < 60

Step 2: Since, remainder 48 0

So, again applying the division lemma to 60 and 48, we have,

60 = 48 1 + 12 where 0  12 < 48.

Step 3: Again remainder 12 0 so applying division lemma to

48 and 12, we get, 48 = 12 4 + 0

Here remainder is zero. Therefore, 12 is the required HCF. So,  Euclid 's algorithm can be summarised as shown in the chart.

4. Find the HCF of 1071 and 1029, using  Euclid 's division algorithm.

Since, 1071 > 1029, we apply the division lemma to 1071 and 1029, to get 1071 = 1029 1 + 42

Since, remainder 42   0 so again applying division lemma in 1029 and 42, we get,

1029 = 42 24 + 21 again 21 0

Applying Euclid's Lemma again in 42 and 21, we get, 42 = 21 2 + 0

Since, remainder is zero so HCF is 21.

5. Two bills of Rs 6075 and Rs 8505 respectively are to be paid separately by cheques of same amount. Find the largest possible amount of each cheque.

Largest possible amount of cheque will be the HCF (6075, 8505).

Applying Euclid's division lemma to 8505 and 6075, we have, 8505 = 6075 1 + 2430

Since, remainder 2430  0 again applying division lemma to 6075 and 2430

6075 = 2430 2 + 1215

Again remainder 1215  0

So, again applying the division lemma to2430 and 1215 2430 = 1215 2 + 0

Here the remainder is zero So, HCF = 1215 Therefore, the largest possible amount of each cheque will be 1215.

W E D N E S D A Y , J U L Y 2 7 , 2 0 0 5

Euclid's Method for the Greatest Common Denominator The greatest common denominator (gcd) is the largest common factor

shared by two or more positive integers. In the case of 4,2 the greatest

common denominator is 2. Since all integers are divisible by 1, the greatest

common denominator of any two integers is guaranteed to be at least 1. If

the gcd of two integers is 1, those integers are said to be relatively prime or

coprime. In other words, they do not have any common divisors.

It is perhaps obvious on the face there exists a gcd for any two integers and

there exists a method for finding this value. Here is the proof. The method

for finding the greatest common denominator is found in Euclid's Elements

and is therefore known today as Euclid's algorithm. Here is a link to Euclid's

proof in the Elements.

But what about quadratic integers? What about Gaussian Integers or

Eisenstein Integers? (for those not familiar with quadratic integers, refer to

here).

It turns out that if there is a division algorithm available, then for any two

integers, there is necessarily a greatest common denominator. In other

words, we can prove that Euclid's method for finding the gcd applies. Here

is the proof (note: in the example referenced, it refers to Gaussian Integers,

but the same proof applies to Eisenstein Integers and other types of

quadratic integers that are characterized by a division algorithm).

For this reason, all quadratic integers that are characterized by a division

algorithm are said to be Euclidean.

One result of this is that for any two Euclidean integers (quadratic integers

that have a division algorithm) that are not relatively prime, it is possible to

derive two smaller integers which are relatively prime.

Lemma: gcd(x,y)=d is greater than 1 → there exists X,Y such that x =

Xd, y = Yd and gcd(X,Y)=1.

(1) Assume that gcd(X,Y) = D which is greater than 1.

(2) Then D divides X,Y such that there exists X = DX', Y=DY'

(3) And x = DX'd, y = DY'd

(4) So that Dd divides both x and y.

(5) But Dd > d which is impossible since d is the greatest common divisor.

(6) So we reject (1).

QED

Corollary: This result holds for all quadratic integers that are

Euclidean.

This is true since the result only depends on the gcd which we showed

above holds for all Euclidean Integers.

QED

Lemma: gcd(x,y)=1 → gcd(xn,yn)=1.

(1) Assume that gcd(xn,yn) = d which is greater than 1.

(2) Since d is greater than 1, there must exist a prime p that divides d. [See

here for the proof of Fundamental Theorem of Arithmetic]

(3) if p divides xn, then p divides x. [See here for the proof of Euclid's

Generalized Lemma]

(4) For the same reason, p would also divide y.

(5) But this is a contradiction since gcd(x,y)=1.

(6) So we reject (1).

QED

Corollary: This holds true for all Euclidean Integers

This proof depends on a proof for unique factorization and Euclid's

Generalized Lemma.

(1) Euclid's Generalized Lemma holds true for all Euclidean Integers. [See

here for proof].

(2) Unique Factorization holds true for all Euclidean Integers. [See here for

proof].

Euclid's lemmaFrom Wikipedia, the free encyclopedia

Euclid's lemma (Greek λῆμμα) is a generalization of Proposition 30 of Book VII of Euclid's Elements. The lemma states that

If a positive integer divides the product of two other positive integers, and the

first and second integers are coprime, then the first integer divides the third

integer.

This can be written in notation:

If n|ab and gcd(n,a) = 1 then n|b.

Proposition 30, also known as Euclid's first theorem, states:

If a prime number divides the product of two positive integers, then the prime

number divides at least one of the positive integers.

That can be written as:

If p|ab then p|a or p|b.

Often, proposition 30 is called Euclid's lemma instead of the generalization. A lemma is a "mini" theorem that is proven and used to prove a bigger theorem. Most of the time in mathematics textbooks Euclid's lemma is used to prove the fundamental theorem of arithmetic.

[edit] Proof of Proposition 30

Say p is a prime factor of ab, but also state that it is not a factor of a. Therefore, , where r is the other corresponding factor to produce ab. As p is prime, and also because it is not a factor of a, a and p must be coprime. This means that two integers x and y can be found so that (Bézout's identity). Multiply with b on both sides:

.

We stated previously that , and so:

.

Therefore, p is a factor of b. This means that p must always exactly divide either a or b or both. Q.E.D.

[edit] Example

Euclid's lemma in plain language says: If a number N is a multiple of a prime number p, and N = a · b, then at least one of a and b must be a multiple of p. Say,

,

,

and

.

Then either

or

.

Obviously, in this case, 7 divides 14 (x = 2).

[edit] An example

As an example, consider computing the gcd of 1071 and 1029, which is 21. Recall that “mod” means “the remainder after dividing.”

With the recursive algorithm:

a b Explanations

gcd( 1071, 1029) The initial arguments

= gcd( 1029, 42) The second argument is 1071 mod 1029

= gcd( 42, 21) The second argument is 1029 mod 42

= gcd( 21, 0) The second argument is 42 mod 21

= 21 Since b=0, we return a

With the iterative algorithm:

a b Explanation

1071 1029 Step 1: The initial inputs

1029 42Step 2: The remainder of 1071 divided by 1029 is 42, which is put on the right, and the divisor 1029 is put on the left.

42 21 Step 3: We repeat the loop, dividing 1029 by 42, and get 21 as remainder.

21 0Step 4: Repeat the loop again, since 42 is divisible by 21, we get 0 as remainder, and the algorithm terminates. The number on the left, that is 21, is the gcd as required.

Observe that a ≥ b in each call. If initially, b > a, there is no problem; the first iteration effectively swaps the two values.

[edit] Proof

Suppose a and b are the natural numbers whose gcd has to be determined. Now, suppose b > 0, and the remainder of the division of a by b is r. Therefore a = qb + r where q is the quotient of the division.

Any common divisor of a and b is also a divisor of r. To see why this is true, consider that r can be written as r = a − qb. Now, if there is a common divisor d of a and b such that a = sd and b = td, then r = (s−qt)d. Since all these numbers, including s−qt, are whole numbers, it can be seen that r is divisible by d.

The above analysis is true for any divisor d; thus, the greatest common divisor of a and b is also the greatest common divisor of b and r. Therefore it is enough if we continue searching for the greatest common divisor with the numbers b and r. Since r is smaller in absolute value than b, we will reach r = 0 after finitely many steps.

[edit] Relation with continued fractions

The quotients that appear when the Euclidean algorithm is applied to the inputs a and b are precisely the numbers occurring in the continued fraction representation of a/b. Take for instance the example of a = 1071 and b = 1029 used above. Here is the calculation with highlighted quotients:

1071 = 1029 × 1 + 42

1029 = 42 × 24 + 21

42 = 21 × 2 + 0

Consequently,

.

This method applies to arbitrary real inputs a and nonzero b; if a/b is irrational, then the Euclidean algorithm does not terminate, but the computed sequence of quotients still represents the (now infinite) continued fraction representation of a/b.

The quotients 1,24,2 count certain squares nested within a rectangle R having length 1071 and width 1029, in the following manner:

(1) there is 1 1029×1029 square in R whose removal leaves a 42×1029 rectangle, R1;

(2) there are 24 42×42 squares in R1 whose removal leaves a 21×42 rectangle, R2;

(3) there are 2 21×21 squares in R2 whose removal leaves nothing.

The "visual Euclidean algorithm" of nested squares applies to an arbitrary rectangle R. If the (length)/(width) of R is an irrational number, then the visual Euclidean algorithm extends to a visual continued fraction.

[edit] Generalization to Euclidean domains

The Euclidean algorithm can be applied to some rings, not just the integers. The most general context in which the algorithm terminates with the greatest common divisor is in a Euclidean domain. For instance, the Gaussian integers and polynomial rings over a field are both Euclidean domains.

As an example, consider the ring of polynomials with rational coefficients. In this ring, division with remainder is carried out using long division, also known as synthetic division. The resulting polynomials are then made monic by factoring out the leading coefficient.

We calculate the greatest common divisor of

x4 − 4x3 + 4x2 − 3x + 14 = (x2 − 5x + 7)(x2 + x + 2)

and

x4 + 8x3 + 12x2 + 17x + 6 = (x2 + 7x + 3)(x2 + x + 2).

Following the algorithm gives these values:

a b

x4 + 8x3 + 12x2 + 17x + 6 x4 − 4x3 + 4x2 − 3x + 14

x4 − 4x3 + 4x2 − 3x + 14

x2 + x + 2

x2 + x + 2 0

This agrees with the explicit factorization. For general Euclidean domains, the proof of correctness is by induction on some size function. For the integers, this size function is just the identity. For rings of polynomials over a field, it is the degree of the polynomial (note that each step in the above table reduces the degree by one).

Least common multipleFrom Wikipedia, the free encyclopedia

In arithmetic and number theory, the least common multiple or lowest common multiple (lcm) or smallest common multiple of two integers a and b is the

smallest positive integer that is a multiple of both a and b. Since it is a multiple, it can be divided by a and b without a remainder. If there is no such positive integer, e.g., if a = 0 or b = 0, then lcm(a, b) is defined to be zero.

For example, the least common multiple of the numbers 4 and 6 is 12.

When adding or subtracting vulgar fractions, it is useful to find the least common multiple of the denominators, often called the lowest common denominator. For instance,

where the denominator 42 was used because lcm(21, 6) = 42.

Calculating the least common multiple

If a and b are not both zero, the least common multiple can be computed by using the greatest common divisor (gcd) of a and b:

Thus, the Euclidean algorithm for the gcd also gives us a fast algorithm for the lcm. To return to the example above,

Because (ab)/c = a(b/c) = (a/c)b, one can calculate the lcm using the above formula more efficiently, by first exploiting the fact that b/c or a/c will be easier to calculate than the quotient of the product ab and c, because the fact that c is a factor of both a and b entails that in either fraction, a/c or b/c, one can completely cancel the c. This can be true whether the calculations are performed by a human, or a computer, which may have storage requirements on the variables a, b, c, where the limits may be 4-byte storage - calculating ab may cause an overflow, if storage space is not allocated properly.

Using this, we can then calculate the lcm by either using:

or

Done this way, the previous example becomes:

Even if the numbers are large and not quickly factorable, the gcd can be calculated quickly with Euclid's algorithm.

[edit] Alternative method

The unique factorization theorem says that every positive integer number greater than 1 can be written in only one way as a product of prime numbers. The prime numbers can be considered as the atomic elements which, when combined together, make up a composite number.

For example:

Here we have the composite number 90 made up of one atom of the prime number 2, two atoms of the prime number 3 and one atom of the prime number 5.

This knowledge can be used to find the lcm of a set of numbers.

Example: Find the value of lcm(8,9,21).

First, factor out each number and express it as a product of prime number powers.

The lcm will be the product of multiplying the highest power in each prime factor category together. Out of the 4 prime factor categories 2, 3, 5, and 7, the highest powers from each are 23, 32, 50, and 71. Thus,

[edit] Viewing this method via Venn diagrams

One can find the least common multiple of two numbers by using a Venn diagram as follows. Find the prime factorization of each of the two numbers. Put the prime factors into a Venn diagram with one circle for each of the two numbers, and all factors they share in common in the intersection. To find the LCM, just multiply all of the prime numbers in the diagram.

Here is an example:

48 = 2 × 2 × 2 × 2 × 3,

180 = 2 × 2 × 3 × 3 × 5,

and what they share in common is two "2"s and a "3":

This also works for the greatest common divisor (GCD), except that instead of multiplying all of the numbers in the Venn diagram, one multiplies only the prime factors that are in the intersection. Thus the GCD of 48 and 180 is 2 × 2 × 3 = 12.

[edit] The lcm in commutative rings

The least common multiple can be defined generally over commutative rings as follows: Let a and b be elements of a commutative ring R. A common multiple of a and b is an element m of R such that both a and b divide m (i.e. there exist elements x and y of R such that ax = m and by = m). A least common multiple of a and b is a common multiple that is minimal in the sense that for any other common multiple n of a and b, m divides n.

In general, two elements in a commutative ring can have no least common multiple or more than one. However, any two least common multiples of the same pair of elements are associates. In a unique factorization domain, any two elements have a least common multiple. In a principal ideal domain, the least common multiple of a and b can be characterised as a generator of the intersection of the ideals generated by a and b (the intersection of a collection of ideals is always an ideal). In principal ideal domains, one can even talk about the least common multiple of arbitrary collections of elements: it is a generator of the intersection of the ideals generated by the elements of the collection.