CHAPTER 1. REVIEW: NUMBERS

Transcription

1 CHAPTER. REVIEW: NUMBERS Yes, mathematics deals with numbers. But doing math is not number crunching! Rather, it is a very complicated psychological process of learning and inventing. Just like listing to music, to appreciate a good piece mathematics, you need to understand it. Very often, listening to a good piece of music for the first time, you are unable to enjoy because you still do not understand. But after a few more times, you start to like it because you begin to understand. When you thoroughly understand this piece of music, you are totally absorbed into it and you can hum every note of it (there is no need to make an extra effort of memorizing). These notes are not unrelated items stored into your memory; they are concrete representation of melodic ideas and overall musical structure in your mind. You can almost say the same thing for learning mathematics! Number crunchers are calculators and computers, not us! We are superior beings who can understand and enjoy math. Anyway. let s start with numbers. The natural numbers are, 2, 3, 4,... They are also called positive integers. The negative integers are, 2, 3, 4,... An integer is either positive, negative, or zero. Thus the set of all integers consists of..., 3, 2,, 0,, 2, 3,... Sometimes, to emphasize a number being positive, we put the plus sign + in front by writing, for example, +2 instead of 2. + and in front of a number is called the sign of that number. Thus a number bearing the negative sign is a negative number. If we drop the sign of a number, say a, what we get is called the absolute value of a, which is denoted by a. For example, 2 = 2 and 2 (= +2 ) = 2. By convention, the absolute value of zero is zero: 0 = 0. We assume that the reader is familiar with the usual arithmetics of integers and basic facts about them, such as the distributive law a(b+c) = ab+ac.

2 Notice that if a nonzero number is multiplied by, its sign is changed while its absolute value remains the same. Review Exercise. Simplify each of the following expressions: 2 3, ( 2) + ( 3), ( 2) ( 3), 2 ( 3), 0 3, 2 ( 3), ( 2) 3, ( 2) ( 3), 0 ( 3). Often we would like to talk about any numbers. We need to introduce letters to designate these arbitrary numbers. For example, for arbitrary numbers a, b, c, the associative law says a(b+c) = ab+ac. Now, try to use words instead of letters and symbols to state this law. This is very hard even you can find a way to express it in words, your friend may not be able to understand you. Slightly more complicated laws such as a 2 b 2 = (a b)(a+b), (a+b) 2 = a 2 +2ab+b 2 are almost impossible to be expressed in words. Therefore, using letters and symbols in mathematics is necessary in many circumstances. It is important to understand symbols properly and to learn how to use them efficiently, although many important ideas often should be described in words in order to feel and to see it. Review Exercise 2. We write a > 0 to stand for a is positive, a < 0 for a is negative, and a 0 for a is positive or zero. Match each part in List A to a part in List B: List A: (a) a is positive (b) a is nonnegative (c) a is nonzero (d) a is less than or equal to zero (e) a is negative. List B: () a 0 (2) a < 0 (3) a 0 (4) a > 0 (5) a 0. Review Exercise 3. We write a < b (read as a is less than b ) or b > a (read as b is greater than a ) if b a > 0. Also we write a b or b a if b a 0 (that is, b a is nonnegative). (a) Verify:. If a < b, then b < a, and 2. If a < b, then a+c < b+c. (b) True or false: if a < b, then a 2 < b 2. Next we review some familiar terminology and facts about integers. The numbers 0, 2, 4, 6, 8,... are called even numbers: they are multiples of 2. (Notice that 0 is also regarded as an even number.) Here is the formal definition of 2

3 even numbers: by an even number we mean a number which can be written in the form 2k, where k is a nonnegative integer. The numbers, 3, 5, 7, 9,,... are called odd numbers. Notice that each of them can be obtained by adding to each of the above list of even numbers. Hence we can give the formal definition of odd numbers as follows: by an odd number we mean a number which can be written in the form 2k + for some nonnegative integers. As we know, the sum of an even number and an odd number is an odd number. Why? Here is the proof. Let a be an even number and b be an odd number. Then we can write a = 2m and b = 2n+ for some nonnegative integers m and n. Hence a + b = 2m + 2n + = 2(m + n) +. Thus m + n = 2k +, where k = m + n is a nonnegative integer, because it is the sum of two nonnegative integers. This shows that a + b is an odd number. (This proof is easy. The important thing to learn here is how to set up a proof and proceed properly.) Review Exercise 4. Being even or being odd is called the parity. We say that two (nonnegative) numbers have the same parity if either both of them are even or both of them are odd. What we have proved above is that the sum of two numbers with different parity is odd. Nowyouare askedto prove thefollowing statement: the sum of two numbers of the same parity is even. Review Exercise 5. Prove that the product of two odd numbers is odd. The natural numbers 2, 3, 5, 7,, 3, 7, 9, 23,... are called prime numbers, because each of them cannot be written as the product of smaller natural numbers. On the other hand, numbers such as 4, 6, 8, 9, 0, 2 etc., which are not prime, are called composites. Indeed, we have 4 = 2 2, 6 = 2 3, 8 = 2 4, 9 = 3 3, 0 = 2 5, 2 = 3 4, etc. One basic property of prime numbers is the following: if p is a prime number and if p divides the product ab of two integers a and b, then p divides a or divides b. (We will take this property for granted.) Review Exercise 6. Give a simple example to show that the primality assumption on p in the last assertion is essential: if we drop the assumption that p is prime, then this assertion fails. Review Exercise 7. Let a, b be natural numbers. Write down the formal definition for a is a factor of b (or a divides b ). 3

4 By a rational number we mean a number which can be expressed as a fraction of integers. In other words, a rational number can be written as m/n, where m and n are integers, with n 0. An integer n is always a rational number, because it can be written as n/, (although n/ looks silly). Review Exercise 8. Simplify each of the following expressions: (a) (b) (c) (d) 2 3. Next we consider the expression b n (read as the nth power of b), where n is an integer. When n =, b n is simply b. For n = 2,3,4 etc., b n becomes: b 2 = b b, b 3 = b b b, b 4 = b b b b b etc. Next, We consider the case that n is zero or a negative integer. In this case we have to assume b 0. For n = 0, 2,, 3 etc., b n becomes: b 0 =, b = b, b 2 = b 2 b b, b 3 = b3, etc. Remember that the expressions 0 0, 0 etc. do not make sense and hence the assumption b 0 is necessary in this case. When m, n are integers and when b 0, we have b m b n = b m+ n (b m ) n = b mn Review Exercise 9. Verify that ( ) n = + if n is an even number and ( ) n = if n is an odd number. Hint: if n is an even number, we can write n = 2k for some integer k. and if n is an odd number, we can write n = 2k + for some integer k. There are real numbers which are not rational and naturally we call them irrational numbers. If a positive integer n is not a perfect square (e.g. n = 2,3,4,6,7), n is known to be irrational. The number π, which is the ratio of the length of a circle and its diameter, is also irrational. Challenging Exercise 0. Prove that 2 is an irrational number. All real numbers can be approximately represented by decimal numbers. 4

5 For very small or very large numbers, it is convenient to use powers of 0 to indicate its order of magnitude. For example, one hundred = 00 = 0 2, one thousand = 000 = 0 3, one million =,000,000 = 0 6, one billion = 000,000,000 = 0 9, one trillion = 0 2. Thus, 3.4 billion is On the other hand, km (kilometer) is 0 3 meters, cm (centimeter) is 0 2 m, mm(millimeter) is 0 3 meter, µm(micrometer) is0 6 meter, nm (nanometer) is 0 9 meter. Also, nanosecond is 0 9 second. The size of a bacteria is often measured in µm s, and nm is for atomic size. A term in the news about the present state of art in high tech in recent years is called nano technology. Now you know how small nano stands for! Review Exercise. The mass of the Earth is approximately kg (kilogram) and its volume is approximately m 3. Find its average density in kg/m 3. Using powers of 0, we can give a correct interpretation of decimal numbers. For example, the number is Indeed, = = = Review Exercise 2. Use powers of 0 to express and 3.459, as indicated in the above example. Real numbers can be geometrically represented as points on a line called the real axis. Usually we draw this line horizontally. The point on it denoted by O represents zero. The point representing is on the right side of it. Review Exercise 3. Plot the locations of the points on the real axis approximately representing the following numbers: 3, 3/2, 5/2, 0/3, 6/3, 2π. Given real numbers a and b, the absolute value of their difference, namely a b, or b a equivalently, is equal to the distance between the points on the real axis representing a and b. Please draw the points representing 3 and 5 on the real axis and convince yourself that 3 5 is indeed the distance between these points. We may use percentage to represent a number between 0 and. For example, 0.5 is 50% and is 54.6%. 5

6 Review Exercise 4. If 0.02% is.2 0 n, what is n? As you may know, if you divide 0 by 3 (by hand or by a calculator), you will get (which is often written as 3. 3). In this decimal expansion, 3 is repeated indefinitely. Conversely, = = ( ). ( ) The right hand side of ( ) is 3(+a+a 2 +a 3 +a 4 + ) for a = /0. This leads us to the general question: what is the meaning of the infinite series +a+a 2 +a 3 + (called a geometric series) and what is its value? We give the answer in the following theorem. Then we give some examples, leaving the explanation of this answer to the last. Theorem.. W hen a <, we have +a+a 2 +a 3 + = lim n (+a+a 2 + +a n ) = a. Applying the above theorem to a = /0, the right hand side of (*) becomes 3 ( ) = 3 Certainly, this is what we have started with. 0 = = 0 3. Example. We are asked to find the fraction whose decimal expansion is given by We proceed as follows: = = 0.26( ( = ) = = = Here you have to be a bit careful about the spacing of digits. Exercise 5. Find the value of each of the following geometric series: (a)

7 (b) (c) Hint: Rewrite it as 2 ( + ( 2 ) + ( 2 ) 2 ( ) ). Exercise 6. Express each of the following decimal numbers as usual fractions: (a) 0. 5 (b) 0. 6 (c) Now we get back to the verification and basic understanding of the theorem. We have to investigate the infinite series +a+a 2 +a 3 +. The starting term of this series is. Adding the next a to it, we get +a. If we continue to add more terms, we will get +a+a 2, +a+a 2 +a 3, +a+a 2 +a 3 +a 4, etc. In general, in the nth step, we will get +a+a 2 + +a n. It is desirable to get a closed form for the last expression. This is hard, because at the present stage we have no clue how to start. But at least two things we can do. First, observe the pattern of this expression: it is a sum of increasing powers of a. Second, we give a name to it: call it S n. So S n = +a+a 2 +a 3 + +a n. (You know the importance of knowing the right names of people in human relationship. If you don t know a person s name and you want to talk about him with your friends, you ll probably make up a nickname. In mathematics too. It is important to know the right names in standard definitions or assigning an appropriate name for your investigation or discussion.) Now, as the name S n is given and the pattern of S n is recognized, we try to do something about it, say, multiply S n by a to see what will happen: as n = a(+a+a 2 +a 3 + +a n ) = a+a 2 +a 3 +a 4 + +a n+. Comparing with the original S n = +a+a 2 + a 3 + +a n, we find that the starting term is missing but an extra term a n+ is gained. Thus +as n = +(a+a 2 +a 3 +a 4 + +a n+ ) = (+a+a 2 +a 3 +a 4 + +a n )+a n+ = S n +a n+. 7

8 By rearranging +as n = S n +a n+, we have S n as n = a n+, or ( a)s n = a n+. If a, we are allowed to divide both sides of the last identity by a to obtain S n = ( a n+ )/( a). We conclude: Theorem.2. W hen a 0, we have +a+a 2 + +a n = an+ a. Now assume that a <, that is, the size is a is strictly less than. What happens to S n +a+a 2 + +a n = an+ a when n tends to infinity? In that case, the size of a n+, namely a n+ a n+, shrinks to zero and hence the right hand side ( a n+ )/( = a) tends to /( a). Thus the left hand side converges to /( a) as n approaches to infinity, which is naturally interpreted as the value of the infinite series +a+a 2 +a 3 +. This of course tells us why Theorem. is true and what is its exact meaning. Note: The material in the rest of this chapter is optional. As you may know, the integer 2345 is divisible by 3 because the sum of all digits = 5 is divisible by 3, and is divisible by 9 because = 36 is divisible by 9. A natural question is, how do we come up with all sorts of criteria about divisibility of digital integers? The math behind all these may be called reduction by modulo. For two integers m and n, we write m = n(mod 3) (read as m is equal to n, modulo 3 ) if their difference m n is divisible by 3. For example, we have 5 = 7(mod 3) because 5 7 = 2 is divisible by 3: 2 = 3 ( 4), where 4 is also an integer. The relation (mod 3) has the following elementary properties: (a) If m = n(mod 3), then km = kn(mod 3). (b) If m = n (mod 3) and m 2 = n 2 (mod 3), then m +m 2 = n +n 2 (mod 3). (c) If m = n (mod 3) and m 2 = n 2 (mod 3), then m m 2 = n n 2 (mod 3). (d) m = 0(mod 3) if and only if m is divisible by 3. 8

9 (e) If m is divided by 3 with r as the remainder, the m = r(mod 3). (Here, m, k, m, m 2, n, n 2, r are some integers.) Let us quickly see why. (a): By assumption, m n is divisible by 3 and hence so is k(m n). Now k(m n) = km kn, showing that km kn is divisible by 3. So km = kn(mod 3). (b): By assumption, we know that m n and m 2 n 2 are divisible by 3. So their sum (m n )+(m 2 n 2 ) is also divisible by 3. From (m n )+(m 2 n 2 ) = m n +m 2 n 2 = (m +n 2 ) (n +n 2 ) weseethatboth (m +n 2 ) (n +n 2 ) isdivisibleby 3, thatis, m +m 2 = n +n 2 (mod3). (c): By part (a), we know that m m 2 = n m 2 (mod 3) as well as n m 2 = n n 2 (mod 3). Hence m m 2 = n n 2 (mod 3). (Note: One can easily see that (mod 3) is a transitive relation, that is, a = b(mod 3) and b = c(mod 3) imply a = c(mod 3).) (d): That m = 0(mod 3) means that m 0 is divisible by 3, that is, m is divisible by 3. (e): Let q be the quotient of the division. Then m = 3q +r. So m r = 3q, which is divisible by 3. So m = r(mod 3). Now we go back to the discussion about the well known criterion for divisibility by 3. Take a positive integer a with digits a 0, a, a 2, a 3,..., a n : a = a n a 3 a 2 a a 0 = a 0 +a 0+a a a n 0 n. Since 0 = 9 is divisible by 3, we have 0 = (mod 3). Multiplying both sides by 0, we get 0 0 = 0(mod 3); (this is possible in view of property (b) mentioned above). So, modulo 3, we have 0 2 = 00 = 0 0 = 0 = (mod 3). We can continue in this manner: 0 3 = = 0 = (mod 3). 0 4 = = 0 = (mod 3), etc. In general, any power of 0 is equal to modulo 3. Therefore a n a 3 a 2 a a 0 = a 0 +a 0+a a a n 0 n = a 0 +a +a 2 +a 3 + +a n (mod 3). Thus a n a 3 a 2 a a 0 = 0(mod 3) if and only if a 0 +a +a 2 +a 3 + +a n = 0(mod 3). We have proved that an integer in decimal expression is divisible by 3 if and only if the sum of its digits is divisible by 3. We can replace 3 by 9 throughout the above discussion, in view of the fact that 0 = (mod 9) also holds. This leads us to the conclusion that an integer in decimal expression is divisible by 9 if and only if the sum of its digits is divisible by 9. 9

10 Exercise 7. Verify that a n a 3 a 2 a a 0 a 0 +0a +0 2 a a n a n is divisible by if and only if the alternating sum of digits a 0 a +a 2 + +( ) n a n is divisible by. For example, 9839 is divisible by because = 22 is divisible by. Exercise 8. (a) Verify that a n a 3 a 2 a a 0 is divisible by 25 if and only if a a 0 a 0 +0a is divisible by 25. (b) Verify that a n a 3 a 2 a a 0 is divisible by 6 if and only if a 4 a 3 a 2 a a 0 a 0 +0a +0 2 a a a 4 is divisible by 6. The computer age brings us the practice of frequent use of binary expressions. A binary number uses only two digits, 0 and, instead of ten digits for decimal numbers. The scale for decimal numbers, in the increasing direction, is 0 0 =, 0 = 0, 0 2 = 00, 0 3 = 000 etc., and in the decreasing direction is 0 0 =, 0 = 0., 0 2 = 0.0, 0 3 = 0.00, 0 4 = 0.000, etc. Similarly, the scale for binary numbers, in the increasing direction, is 2 0 =, 2 = 0, 2 2 = 00, 2 3 = 000, etc. (in binary) and in the decreasing direction is 2 0 =, 2 = 0., 2 2 = 0.0, 2 3 = 0.00, 2 4 = 0.000, etc (in binary) Thus, in the binary system, a number in n digits is of the form a n a 2 a a 0 = 2 0 a 0 +2 a +2 2 a a 3 +2 n a n. For example, 0 (in binary) is equal to = 3 (in decimal). To convert 3 back a binary expression, we have to rewrite it as a sum of powers of 2 as follows: 3 = 2+ = 3 4+ = = (2+) = The powers appearing are 0, 2, 3, but not. Hence the binary expression for 3 is 0. For non integers, we can also use the binary system: 0.b b 2 b 3 b 4 (binary) = b 2 + b b b For example, = = 2 2 ( ) = 4 2 = 2. 0

11 Again, as you see the theorem about geometric series is very useful for binary numbers. Exercise 9. Convert the following binary numbers to the usual decimal numbers: (a), (b) 0, (c) 000. Exercise 20. Convert the following decimal numbers into the binary numbers: (a) 32, (b) 62, (c) 32. Exercise 2. Convert the following binary numbers to the usual fractions: (a) 0. 0, (b) Exercise 22. Prove that a binary integer a n a a 0 is divisible by 3 if and only if a 0 a +a 2 a 3 + +( ) n a n is divisible by 3.

12 Answers to Review Exercises. 2 3 =, ( 2)+( 3) = 5, ( 2) ( 3) = 2+3 =, 2 ( 3) = 2+3 = 5, 0 3 = 3, 2 ( 3) = 6, ( 2) 3 = 6, ( 2) ( 3) = 6, 0 ( 3) = a is positive (4) a > 0; a is nonnegative (5) a 0; a is nonzero () a 0; a is less than or equal to zero (3) a 0; a is negative (2) a < (a). That a < b means b a > 0, which can be rewritten as ( a) ( b) > 0 and hence b < a. 2. That a < b means b a > 0, which gives (b+c) (a+c) > 0 and hence a+c < b+c. (a) False. Indeed, take a = 2 and b = 0. Then a < b, but a 2 = 4 and b 2 = 0, indicating the failure of the inequality a 2 < b Suppose that m and n have the same parity. We are going to show that their sum is even. We deal with the following two caes: both of them are even, and both of them are odd. When both of them are even, we can write m = 2k and n = 2l for some integers k and l. In this case we have m+n = 2k +2l = 2(k +l) which is even. When both of them are odd, we can write m = 2k + and n = 2l+ for some integers k and l. In this case we have m+n = 2k++2l+ = 2(k+l+) which is also even. 5. Let m, n be odd numbers. Then we can write m = 2k+ and n = 2l+, where k and l are some integers. Thus mn = (2k +)(2l+) = 4kl+2k +2l+ = 2(2kl+k +l)+ = 2j + where j = 2kl+k +l is an integer. Therefore mn is an odd number. 6. For example, 6 divides 2 = 3 4, but 6 divides neither 3 nor Formal definition of a divides b : Given positive integers a and b, we say that a divides b if there exists a positive integer c such that b = ac. (Note: words used in a defintion are as important as mathematical expressions such as b = ac here.) 8. (a) (c) = = = = 0 (d) 2 3 = 2 6 = 3. (b) = 0. = /3 = 3. 2

13 9. When n is even, we can write n = 2k, where k is another integer, and hence ( ) n = ( ) 2k = (( ) 2 ) k = k = using the fact that ( ) 2 = ( )( ) =. When n is odd, we can write n = 2k +, where k is another integer, and hence ( ) n = ( ) 2k+ = ( ) 2k ( ) = k ( ) =. We conclude: ( ) n = { if n is even if n is odd. (This conclusion is considered to be elementary in mathematics and hence is often used freely without explanation.) 0. Here we prove the assertion by contradiction. Suppose the contrary that 2 is a rational number. This means that we can write 2 = m/n for some integers m and n. Here we may (and we do) assume that m and n are coprime positive integers. Squaring both sides of 2 = m/n, we obtain 2 = m 2 /n 2, or m 2 = 2n 2. Since 2 occurs as a factor on the right hand side, we know that 2 divides m 2. Hence 2 divides m as well; (recall that if a prime number divides the product ab of two integers a and b, then p divides a or b). So we may write m = 2m for some integer m. Letting m = 2m in m 2 = 2n 2, we obtain 2m = n 2. Now we can repeat the argument to prove that 2 divides n. Thus 2 divides both m and n, contradicting the earlier assumption that a and b are coprime to each other.. The average density of the earth is approximately = = kg/m Skipped = ; = % = 0.02/00 = = = So n = 4. 3

14 5. (a) The sum is 2 = 2; (b) The sum is 3 3 (+ + + ) = = 2 ; (c) The sum is ( 2 ) = (a) 5/9, (b) 2/3, (c) 5/66. 4