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But in this more general situation we need a new proof. The following corollaries are word-for-word generalizations of Corollary 2. But we will give a second albeit longer proof that uses prime factorization directly: Then, by Lemma 2. But again we will give a proof that uses prime factorization directly: As the reader will see, our results are much more complete for negative values of D than they are for positive values of D. To keep these straight, in Lemma 2. Note in Lemma 2. We will be using it or its consequences in all of the examples in this section and in the next section.

So once again this provides an example of working hard enough with a single idea and being able to go a long way with it. We use it so often that we give it a name although this name is not standard mathematical language. Now in a UFD, every irreducible is prime Lemma 2. We divide the proof into two cases: But if D 2 , it does not. Hence, by Corollary 2. Let p be the smallest prime dividing D. The Case D 65 by contradiction. We use Proposition 2. Actually, the methods we derive here will also apply to values of D with D i i i i i i i i 68 2.

Once again, we divide the proof into two cases: But it is a result from number theory see Corollary B. We consider case 2b. We begin by noting that, by Lemma 2. The remainder of the proof is identical to the proof of Theorem 2. If b1 and b2 are small, trial and error is as good as any. We will use this in the statement of our results. We wish to apply Theorem 2.

First, we consider condition 1 of that theorem. We consider case b. As in the proof of Lemma 2. Let p2 be an odd prime and let q be any integer relatively prime to p. Note that p1 is odd. Thus, by Corollary 2. Once again we will formulate our results with the help of the Chinese Remainder Theorem. It states that if at least one of p1 and p2 is congruent to 1 mod 4 , then either both of these congruences have a solution or neither does, while if both p1 and p2 are congruent to 3 mod 4 , then exactly one of these congruences has a solution.

Since in Theorem 2. We see how to use this in the next example. In order to satisfy hypothesis 2 of Theorem 6. This is the same as case 1a of Example 2. We proceed by exactly the same logic as before. The techniques in Example 6. Depending on circumstances, either one may be more convenient to use.

We shall show that O D does not have an element of norm p. We prove this by contradiction. For this we need primes congruent to 3 mod 8. Then the hypotheses of Theorem 2. For each value of D we give a pair q1 , q2 that provides the argument. In case q1 is an odd prime, we are using Theorem 2. For many values of D, there is more than one pair q1 , q2 that work. In those cases, we have simply chosen one. In this section we will sum up our work and also report on some interesting results that are beyond our ability to prove here.

Unique Factorization In Theorem 2. See also Example 2. We have the following conjecture, which has been open for years: Complete the proof of Lemma 2. You must prove Lemma 2. Note by Lemma 2. Also by Lemma 2. Exercises 81 Exercise 2. Prove the following properties of a gcd of elements in an integral domain R: Do these exercises by using the results of Sections 2. Prove the analog of Lemma 2.

Factoring large numbers into primes - Famous Math Problems 1 - NJ Wildberger

Suppose that R is a UFD. In the notation of Proposition 2. We make the convention that in this case, we choose the gcd of a set of elements to be positive. Similarly, we choose the lcm of a set of elements to be positive. Exercises 83 Exercise 2. Show that in fact r is an integer. Show that dr is an integer. By this we mean that ai and bi are relatively prime, for each i. Find the lcm of each of the sets of integers in Exercise 2. Exercises 85 Exercise 2. Show that in fact they are not factorizations into irreducibles. Find a factorization of 51 into irreducibles, and show how it yields these three factorizations of Find a factorization of 35 into irreducibles, and show how it yields these three factorizations of Find a factorization of 6 into irreducibles, show how it yields these two factorizations of 6, and use it to construct another factorization of 6.

Verify that Corollary 2. This requires the use of a computer. Do the analogue of Example 2. Here deg denotes the degree of a polynomial. Unique Factorization Exercise 2. Show that R[X] is not a Euclidean domain with the above norm. We make the convention that in this case, we choose the gcd of a set of elements to be monic.


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Similarly, we choose the lcm of a set of elements to be monic. Find the gcd of each of the following sets of polynomials in Q[X], and express the gcd as a linear combination of those polynomials: Consider the following integral domain: Find an explicit element of R that has two distinct factorizations into irreducibles. You may have already done b in doing a. Exercises 89 a Show that I is an ideal of R. We will begin by proving a justly famous theorem of Fermat: Fermat did not write his proof down, but left a hint as to his approach.

The second proof is a twentieth-century proof due to Thue. The third proof uses the fact that the Gaussian integers are a UFD, and, given that fact, is short and easy! The Gaussian Integers 3. One case is easy to rule out. We begin with an observation: That this is true is a simple algebraic fact. It is convenient to introduce the following language.

We will break it up into a number of steps. Then, by Theorem B. But, by Corollary B. The Gaussian Integers Lemma 3. Now p divides N , by assumption, so p divides N b2. Certainly, p divides t2 p. Since p is a prime, it must divide one of the factors. Regard this as a system of two linear equations in the two unknowns s and t and solve. So we may assume x and y are relatively prime as well. Continue in this fashion. We now present our second proof, and begin with a key lemma.

Let p be a prime and let a be any integer relatively prime to p. Then a divides the product bc. Now each of these four factors has norm p, and p is an ordinary prime, so each of these factors is irreducible Lemma 2. In a UFD, primes are irreducible and vice versa Lemma 2. Combining the easy Lemma 3. Let p be a prime. We mention this not because it adds anything to what we have already proved but rather for comparison with the exercises for this chapter. For our purposes, Theorem 3. But not only did Fermat show which primes could be represented as sums of two squares, he also showed which integers could be represented as sums of squares.

There are now two possibilities: This is a direct computation: In this theorem we are making two claims: Both of these claims are consequences of our earlier work. Then p1 must either be 2 or a prime congruent to 1 mod 4 , by Lemma 3. We now show how to factor Gaussian integers into primes. There is a certain amount of trial and error involved, just as for factoring ordinary integers into primes.

We begin with some special cases.

Algebraic_expressions

Let us look at some examples: Note that our factorizations are only unique up to units, so there are many variants. Exercises Remark 3. Looking at Theorem 2. These three kinds of behavior are typical of what happens in general. The Gaussian Integers Let p be a prime. Use this to improve the result of part a to the following: The Gaussian Integers Exercise 3.

Let n be a positive integer. Exercises Exercise 3. A method for solving it, called the cakravala method, was developed by Indian mathematicians in the seventh to twelfth centuries. There is also a twentieth-century approach using Diophantine approximation. Pell, however, had nothing to do with it. Thus, we will present a variant of the cakravala method here, and furthermore we will prove that it always works.

This method is undoubtedly very close to what Fermat had in mind, as you can tell from the similarity between the last chapter and this one. We shall prove all the results in this section by direct computation, but there is a deeper reason why they are true. When a computation produces a particularly nice result, one should always ask why.

This reason can be found in the exercises to this chapter. We shall call a, b the conjugate of a, b. In this case, we say a, b is reduced. Let a, b represent m and c, d represent n. Then e, f represents mn. Representations and Their Composition Definition 4. We now give several properties of composition, which we verify by direct computation. We now combine composition and reduction into a single operation. We now give several properties of reduced composition. Now a and b are relatively prime, as a, b is assumed to be reduced. The same holds for any number of representations: The second claim is that reduced composition is associative, and the third claim is a cancellation result for reduced composition, up to sign.

We call these two representations trivial representations and all other representations of 1 nontrivial representations. So in fact, we are looking for nontrivial representations of 1. Let a, b and c, d both be reduced and suppose that a, b and c, d represent the same integer m. Then e, f represents 1. Then E, F represents m2. Let us begin by seeing that each of E and F are divisible by m. Furthermore, the contrapositive of Lemma 4. Let e, f represent 1 nontrivially i.

We must show that they are all nontrivial and all distinct. We claim that f1 , f2 , f3 ,. We show this claim by direct computation, using induction. Now suppose ek and fk are positive integers. Then we will prove that this method actually works. At the outset, our approach will seem to have nothing to do with the last section, but we will soon see that our method involves a sequence of carefully chosen reduced compositions. However, we are not done yet! Suppose A and B are divisible by e. There are two claims to prove: We prove these in turn.

Thus, we see that the inequality r i i i i i i i i 4. In the situation of Lemma 4. Following the notation of the proof of Lemma 4. Then there is a unique integer is reduced and represents e with e i i i i i i i i 4. That is, we start with a pair a, b , apply Lemma 4. This works for Lemma 4. But since this congruence has a unique solution, this must be the solution. Then, by the contrapositive of Corollary 4. Alternatively, instead of appealing to Lemma 4.

That is, there is a sequence of nonnegative integers s0 i i i i i i i i 4. First, we have a number of tables giving the results of the method for various values of D. The values of D were chosen to provide an illustrative sample of the phenomena encountered. We observe that, in each case, the sequence e0 , e1 , e2 ,.

But we observe that the sequence of triples e0 , a0 , b0 , e1 , a1 , b1 , e2 , a2 , b2 ,. Note in Tables 4. We also give in Table 4. This would not have changed the small period p D , but might have changed the large period. Now we give some results that enable us to speed up our search.

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Small and large periods for values of D between 2 and This is almost identical to the proof of Lemma 4. Now let us look at some examples and we refer the reader to Tables 4. Units in O D Example 4. We cannot apply Lemma 4. In almost all cases this determines c0 uniquely, but if not, i i i i i i i i 4.

But in fact there is such a smallest unit. Note that the statement of the lemma is equivalent to the following statement: We leave the proof of this to the exercises. Hence there is some positive i i i i i i i i 4. If Fermat could do this computation by hand, so can you. The Archimedes cattle problem was a famous problem posed by Archimedes.

Search the Internet for this problem. Show that at step 60 it produces the following solution: Use this to prove Lemma 4. The next few problems give the proof of Lemma 4. In the next two exercises we adopt the notation of Remark 4. All of these possibilities occur, as illustrated in Table 4. But this consideration was just the tip of an iceberg. Our aim in this chapter is to indicate how these results generalize.

As a matter of historical fact, our development here parallels the development of algebraic number theory. Our goal here is to provide a guide to some of the high points of algebraic number theory, especially those most related to the topics we have studied. In order to do so, we will have to introduce some more advanced concepts than we have done so far, so this chapter will require more background of the reader. Also, as this is a guide and not a complete treatment, we will not prove the more advanced results we state. But we hope this chapter will motivate the reader to go on and study algebraic number theory further.

These will provide concrete examples of this general theory, as well as being very interesting in themselves. There is one notational point that we should mention up front. We will be intensively studying ideals in this chapter. Standard mathematical notation, which we shall adopt below, gives parentheses special meaning in the context of ideals.

Parentheses are of course also the standard way to group items for multiplication. In order to avoid confusion, we shall instead, throughout this chapter, except in the last section, use square brackets to group items for multiplication. Towards Algebraic Number Theory 5. Thus, there are rational numbers a0 , a1 ,. Note that this lemma is far from obvious. Referring back to Example 5. It is important to distinguish between the integers in Q and the algebraic integers in K. Thus O K contains a free Z-module of rank n. Hence O K must be a free Z-module of rank n. First we introduce a bit of standard notation.

This notation has the disadvantage that the interesting information is relegated to the subscript. We certainly know how to multiply elements of R and we certainly know what it means for one element of R to divide another. Guided by this knowledge, we shall seek the analogs for ideals. Let us see what that might be. Let I and J be ideals of R.

This is no problem: Let I and J be two ideals of R. The proof of this is similar to the proof of Lemma 2. We leave the proof of the following useful lemma to the reader. Also note that multiplication of ideals is certainly commutative, and that Lemma 5. Now let us think about the analog of divisibility. Ideal Theory Definition 5. But we could have asked for something stronger. This is indeed stronger in general, as the next proposition shows. Let I and J be ideals in an integral domain R.

Let I be a proper ideal in an integral domain R. Being maximal is a stronger condition than being prime, as the next proposition shows. Let I be an ideal in an integral domain R. If I is a maximal ideal, then I is a prime ideal. Assume this is not the case. Here we want the converse to be true as well, so that the prime ideals are exactly the maximal ideals. Every prime ideal of Z is maximal. Dedekind Domains Definition 5. To state the last condition properly, we need to slightly generalize the notion of an algebraic integer. To this end, let Z be any subring of Q.

As we have seen in Example 5. Finally, we can easily generalize the notion of a principal ideal. Let I be a nonzero fractional ideal of K. The following is a key technical lemma, which we shall not prove. But they are equivalent for Dedekind domains. Here is the main general result about Dedekind domains, and the reason why we introduced them. The following are equivalent. Let R be a Dedekind domain. Towards Algebraic Number Theory Proof: In this proof we will be very careful to distinguish between prime elements and prime ideals.

We begin by recalling that an element of a UFD is a prime element if and only if it is an irreducible element. We want to show that every ideal of R is principal. Since, by Theorem 5. Let P be a prime ideal of R. We began by investigating unique factorization of elements. Then R is a Dedekind domain. Then every nonzero ideal I of R can be factored essentially uniquely into a product of prime ideals. This is immediate from Theorem 5. As a matter of notation, for a set S, we let S denote the cardinality of S, i.

But then we can solve for a0: Towards Algebraic Number Theory Definition 5. Proof of Theorem 5. Putting these implications together, we have the following: Let I be a prime ideal of R. Inverses are given by Lemma 5. Clearly, the subset IPrin K is closed under multiplication and taking inverses, so is a subgroup. Algebraic Number Fields and Dedekind Domains Here is one of the fundamental theorems of algebraic number theory.

Thus the ideal class [I] of I is trivial if and only if I is a principal ideal. We conclude this section with a result that we record for future reference. This is a long procedure, and so we proceed in stages. This is because the denominator 2 is relatively prime to p. Let I be any ideal in R. Then S1 is a nonempty set has of positive integers and so has a smallest element n.

Every prime ideal P of R is of the following form: We prove the theorem by ruling out every ideal that is not of one of the above forms. First, suppose g has more than one prime factor. Thus I is not a prime ideal. Suppose that p is in I. In the situation of Proposition 5. Suppose p is odd. Then 2a1 is not divisible by p, and so p is in P P. Then b1 is not divisible by p. If D is not divisible by p then 2b1 D is not divisible by p, and so p is in P P. Thus, if R does not have an element of norm p, P cannot be a principal ideal. Note by Lemma 5.

Towards Algebraic Number Theory Proposition 5. First, suppose that p is an odd prime. Let us assume that 2a1 is in P. Now every element of P has norm divisible by p, so p must divide 2a1 , and hence p divides a1. Conversely, suppose p divides D. Then, by Proposition 5. The following is a complete listing, without duplication, of the prime ideals P in R: There is only one thing left to do. To complete the proof we must show that every ideal of the above form is indeed a prime ideal. To begin with, we note that every element of P has norm divisible by p.

Also, by Lemma 5. Thus at least one of m and n is divisible by p. We shall assume that m is divisible by p. Otherwise interchange I and J. To proceed further, we must break the proof up into several cases. We number the cases as in the statement of the theorem. In the former case, again as in the proof of Proposition 5. Thus, we must have n divisible by p. In the situation of Theorem 5.

From the proof of Lemma 5. Here is a summary of all the possibilities, together with an expression of each of the prime ideals in terms of congruence conditions. Let I be the ideal of Proposition 5. Examples of Ideals in O D Proof: It is immediate from Lemma 2. If D is divisible by p1 p2 p3 , then h D is even. We have proved 1 a in Theorem 2. Thus, by Theorem 5. We use two facts. Here p and q are primes. The argument for this breaks up into several cases, and we do two of them. The others, which are similar, we leave for the reader.

Consequently, h D is divisible by 2k. This is just the easy half of the following much deeper theorem of Gauss: But in these cases C Q D consisted entirely of elements of order 2 plus the identity. Let a be a positive integer with a and q relatively prime. Furthermore, [I] is an element of C Q D of order n. We prove this by induction. Now assume it is true for some value k.

Thus [I] has order j in C Q D for some j dividing n. Then [I] is an element of C Q D of order n. We prove the corollary by contradiction. Since a, b, and q are relatively prime, d cannot be 0, and then c cannot be 0 either. But, as a little algebra shows, this contradicts our hypothesis on q and D.

Towards Algebraic Number Theory Example 5. Here are some instances of this corollary: We saw in Theorem 5. Here we return to our original example of nonunique factorization. Recall the factorizations into irreducible elements: This corresponds to the factorizations into principal ideals: Note I1 and I2 are not principal ideals as R does not have any elements of norm 2 or 3.

We have the following factorizations in R: This corresponds to the factorization of principal ideals: The elements 2, 3, and 6 of R are not irreducible. To be precise, we have the following factorizations into irreducible elements: Then the element 6 has the factorization into irreducible elements: We thus obtain the following factorization of the ideal 6 into a product of prime ideals: We have the following factorizations: Note that I is a nonprincipal ideal as R has no elements of norm 3.

Then, by Theorem 5. Thus, we see that we have the following factorization of the ideal 27 into a product of prime ideals: Then K is an element of C K of order 6. In any event, we have the factorization into prime ideals: It can be shown that R does not have an element of norm 6, from which it follows that K is not a principal ideal, and hence [K] has order 3 in C K. Congruence considerations mod show that R does not have any elements of norm 2, 3, or 6, so I, J, and K are all nonprincipal ideals.

Examples of Ideals in O D Remark 5. The theory of continued fractions, which we do not develop here—it is long, though elementary— enables us to decide this for any n with n relatively small compared to D. We alluded to this in parts 7 , 8 , and 9 of Example 5. The nature of this formula, however, is not such that we can draw general conclusions about the behavior or properties of h D. As you might imagine, the formula is more complicated for positive values of D than for negative values of D.

The class number formula evidently yields an integer for D 0, but it is not even a priori evident that this real number is an integer! Tables of the values of h D for all D, positive and negative, with D i i i i i i i i 5. Towards Algebraic Number Theory late nineteenth century. The explanation is that Gauss investigated binary quadratic forms. In this situation we write IK for RI. Let P be any prime ideal of R. Then P divides p K for exactly one prime number p. By the proof of Lemma 5.

Thus we see that every prime ideal of R lies over some prime number p. Our object in this section is to investigate the prime ideals lying over an arbitrary prime number p. To this end, we consider any prime number p. Here is our basic result. We have numbered the cases as above as they correspond to the cases with the same number in Theorem 5. In case 1 , p is said to ramify in K.

In case 2a , p is said to split in K. In case 2b , p is said to be inert in K. Note that in Theorem 5.

Factorization : Unique and Otherwise

The distinction between cases 2a and 2b is much more subtle, as we see from Remark 5. This is in fact true in general. Towards Algebraic Number Theory Theorem 5. We see that the general theory specializes to the result we had in Theorem 5. We close this section by citing the following theorem: The answer to this question itself involves some interesting mathematics.

Ideal Elements Theorem 5. We adopt a similar notation to that in the previous section. Later on, mathematicians realized that the notion of an ideal was an important one, and have focused on that notion rather than on the notion of an ideal element. Indeed, our proof of Theorem 5. But let us see what the ideal elements of one of the ideals we have considered above are.

We refer to Example 5. Next consider the ideals I2 and I2. It then readily follows that 3 is an ideal generator of both I2 and I2. It is natural to ask whether this procedure must always eventually stop, and the answer is no. It is known that this procedure may go on forever, i. Then every element of K is algebraic, i. Exactly the same as the proof of Lemma 5. Although part of this theorem follows from general considerations, we will give a concrete proof of all of it.

First, let us see that S spans E: Consider an arbitrary element y of E. Next, let us see that S is linearly independent: Thus, k X cannot be a constant polynomial. But k X divides f X , contradicting the hypothesis that f X is irreducible. Let y be an arbitrary nonzero element of E. Let r be the number of real embeddings of K in C and let s be the number of pairs of conjugate complex embeddings of K in C.

Let x in K be as in the proof of Lemma 5. It is a theorem that every irreducible polynomial in Q[X] has distinct roots. The Fundamental Theorem of Algebra tells us that every polynomial of degree n in C[X], and hence every polynomial of degree n in Q[X], has n roots in C. Also, as is well known, the complex i.

Here is the result to which we have been heading. Let UK be the group of units in K. We call a set of generators for FK a fundamental system of units for K. In fact, it is not. Towards Algebraic Number Theory Exercise 5. Let I and J be ideals in a Dedekind domain R. Use the descriptions of the ideals P in Remark 5. Let this polynomial have constant term a.

Note that parts a and b below show that R is not a Dedekind domain. This is true even though every element of R can be written as a product of prime elements. In particular, no power of I is principal. We observed in Example 5. In this appendix we describe it and its variants, and draw some of its particularly useful consequences. Also, suppose that the dominoes are arranged so that if each domino falls, the next one will also fall. What will be the result? Clearly, it is that they will all fall.

To formalize this, let F n be the proposition F n: Stated in this way, there is clearly nothing special about F n , and we may substitute any proposition. This leads us to the principle of mathematical induction. Mathematical Induction Axiom A. Let P n be any proposition about positive integers. Then P n is true for every positive integer n. Occasionally it is the case that verifying 2 is easy but verifying 1 takes work. Rarely it is the case that verifying both 1 and 2 take work.

And it is virtually never the case that verifying both 1 and 2 is easy—that would be getting something for nothing. There is a variant of mathematical induction called complete induction. Again, all the dominoes will fall. There is another possibility. Mathematical Induction and Its Equivalents This second case, translating from dominoes falling to general propositions, gives us complete induction. These two variants of induction are logically equivalent. However, in order to prove this it is convenient to introduce a third variant, well-ordering, which is important in its own right.

First, the domino version: Then among the dominoes that are still standing, there is one with the smallest number. Stated slightly more formally, in this case among the set S consisting of the dominoes that are still standing, there is a domino that has the smallest number. Translated into general terms, this gives the statement of well-ordering. Any nonempty subset S of the set of positive integers has a smallest element. We now show that all three of these variants of induction are logically equivalent. The proof of this is rather subtle and tricky. We need to be clear about what we must prove.


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Mathematical Induction need to show that we can use the method of mathematical induction to obtain the conclusion of complete induction. Let us consider a new proposition Q n. Q n is the proposition Q n: Actually, we will put the precise statement of well-ordering aside for the moment and deal with a related statement instead.

Let S n be the proposition S n: Every subset T of the set of positive integers that contains the integer n has a smallest element. Mathematical Induction and Its Equivalents First we observe that 1 S 1 is true, for if the set T contains the positive integer 1, then 1 is certainly the smallest positive integer in T as 1 is the smallest positive integer, period. There are two possibilities: In the second case, T contains some positive integer k with k i i i i i i i i A. We now apply the well-ordering principle to conclude that S has a least element n0.

There are two cases: But this directly contradicts hypothesis 1 of mathematical induction, which states that P 1 is true. But now we can apply hypothesis 2 of mathematical induction: Thus, our assumption that the conclusion of induction is false leads in any case to a contradiction, so we conclude that the conclusion of induction is true, that P n is true for every positive integer n, and that is what we wanted to prove. We should observe that we have not proved mathematical induction.

Then, by the Well-Ordering Principle, S has a smallest element, which is ak for some k. We claim the sequence stops at ak. This contradicts ak being the smallest element of S. Hence this supposition is false and the sequence stops at ak. The Pigeonhole Principle is an intuitively obvious principle that gets its name from an old but not obsolete technology.

Suppose a postal employee is sorting letters into pigeonholes, and there are more letters than pigeonholes. Then some pigeonhole must contain more than one letter. We have said that the pigeonhole principle is intuitively obvious. But of course that does not mean that it does not require proof. We state it precisely and prove it now, as a consequence of mathematical induction.

We prove this by induction on n. Let P n be the statement of the theorem for a given value of n. We verify the hypotheses of induction: Then, by induction, we conclude that P n is true for every positive integer n. If m objects are sorted into n categories and m i i i i i i i i A. Mathematical Induction 1 P 1 is true: If that category is empty, we are done. Otherwise, the single object is in that category. In that case, every other category is empty.

A decomposition of a positive integer n is a way of writing n as a sum of positive integers in order. Let d n be the number of decompositions of n. Mathematical Induction Exercise A. In algebra, pronumerals are used to stand for numbers. You may or may not know what the value of x is although in this example we do know that x is a whole number. The following table gives us some meanings of some commonly occurring algebraic expressions. In algebra there are conventional ways of writing multiplication, division and indices.

We have been following this convention earlier in this module. It is conventional to write the number first. Instead the alternative fraction notation for division is used. Assigning values to a pronumeral is called substitution. Adding and subtracting like terms. If you have 3 pencil case with the same number x of pencils in each, you have 3 x pencils altogether. This can be done as the number of pencils in each case is the same. The terms 3 x and 2 x are said to be like terms. The terms 2 xy and xy are like terms. Two terms are called like terms if they involve exactly the same pronumerals and each pronumeral has the same index.

The distributive law explains the addition and subtraction of like terms. The terms 2 x and 3y are not like terms because the pronumerals are different. The terms 3 x and 3x 2 are not like terms because the indices are different. There are no like terms for 2 y , so by using the commutative law for addition the sum is. The any-order principle for addition is used for the adding like terms. Because of the commutative law and the associative law for multiplication any-order principle for multiplication the order of the factors in each term does not matter.

Brackets fulfill the same role in algebra as they do in arithmetic. Brackets are used in algebra in the following way. Let x be the number. For a party, the host prepared 6 tins of chocolate balls, each containing n chocolate balls. Each crate of bananas contains n bananas. Three bananas are removed from each crate. Five extra seats are added to each row of seats in a theatre. There were s seats in each row and there are 20 rows of seats. How many seats are there now in total? The following example shows the importance of following the conventions of order of operations when working with powers and brackets.

Multiplying algebraic terms involves the any-order property of multiplication discussed for whole numbers. Simplify each of the following: Arrays of dots have been used to represent products in the module, Multiplication of Whole Numbers. Let n be the number of dots in each row. If an array is m dots by n dots then there are mn dots. The pattern goes on forever. How many dots are there in the n th diagram? The area of a 3 cm by 4 cm rectangle is 12 cm 2. Find the total area of the two rectangles in terms of x and y. The area of the rectangle to the left is xy cm 2 and the area of the rectangle to the right is 2 xy cm 2.

The n th positive even number is 2 n. Write the following using algebra to see what y ou get. Show that the sum of the first n odd numbers is n 2. Quotients of expressions involving pronumerals often occur. We call them algebraic fractions we will meet this again in the modules, Special Expansions and Algebraic Fractions. Dividing by 5 gives. A vat contains n litres of oil. Forty litres of oil are then added to the vat. How much oil is there in each container?

A shed contains n tonnes of coal. An extra tonnes are then added. Expanding brackets and collecting like terms. Numbers obey the distributive laws for multiplication over addition and subtraction. The distributive laws for division over addition and subtraction also hold as shown. As with adding like terms and multiplying terms, the laws that apply to arithmetic can be extended to algebra. This process of rewriting an expression to remove brackets is usually referred to as expanding brackets.

A sound understanding of algebra is essential for virtually all areas of mathematics. It was only in the 17th century that algebraic notation similar to that used today was introduced. For example, the notation used by Descartes La Geometrie , and Wallis was very close to modern notation.