Determine the location (x, y) of the centroid of the wire.

Midterm Exam Study Guide by Ariel Harris Counting numbers, Natural numbers Represented by a hollow n = ℕ The natural numbers start with one and continue on for an infinite amount {1, 2, 3, …} ℕ = {1, 2, 3, …} aka is 357 ∈ ℕ ∈ means "an element of" Whole numbers No character representation, the whole numbers The whole numbers start with zero and continue on for an infinite amount {0, 1, 2, 3, …} The whole numbers can be presented like this: ℕ U{0} Where U means union if a ∈ ℕ U {0} then either a ∈ ℕ -OR- a ∈ {0} ∞: Infinity is not a number, but a concept, infinity does not follow the common sense rules that our numbers do Question: Are there more whole numbers than natural numbers ℕ = {1, 2, 3, 4, …} ℕ U {0} {0, 1 , 2, 3, …} Answer: no, the exact same amount, because there is a one-to-one (1-1) correspondence Why because we have established a 1-1 correspondence between the elements of each set One-to-One correspondence- every item in each set has one and only one buddy (class application when we all lined up across from each other) Finite- with ending If n ∈ ℕ then n's buddy in the set of whole numbers is n-1. Definition: A set is called countably infinite if its elements can be put into a 1- 1 correspondence Even and Odd numbers: these occur to us naturally. How do we visualize this . . . . . . . . (Everyone has a buddy) A number that is even equals 2 times a natural number 12 is a multiple of 2 . . . . . . . . . . . . Justify with math that 8 is even Now 12 Now 64 8 is even because 12 is even because 64 is even bc 8=2*4 12= 2*6 64=2*32 The definition of an even natural number, n, is even iff (if and only if) Definition of an even number: n = k ∈ ℕ (k is some natural number) Definition of an odd number: n= 2*k +1 where k ∈ ℕ ^--something even, plus the "odd guy out" How to visualize that 9 is odd: . . . . . . . . . this is the odd man out, he doesn't have a "buddy" Is 1 an even number NO gut instinct: 1 is odd, but we can't represent 1 as odd with our definition because there is NO k ∈ ℕ (1 ≠ 2K+1) we would need k=0, but 0 ≠ ℕ We can fix this problem by changing our definition slightly - we need 0 to play k ∈ ℕ U {0} With this change we can mathematically justify that 1 is odd because 1=2*0+1 What about 0 Gut instinct: 0 is neither, but with a changed definition, 0 is even. - even : n=2k for some k ∈ ℕ U {0} – with this definition zero is now even zero is even because 0=2(0) and 0 ∈ ℕ U {0} Is it possible for 0 to be odd Is there a whole number (k) such that (s.t.), 0 =2k+1 (k would need to be -.5) no, 0 is not odd! NEGATIVE NATURAL NUMBERS: We quickly realize we need to enlarge our number system to talk about items owed, something being below zero, a negative: {… -3, -2, -1, 0, 1, 2, 3, …} The set of integers, there are infinitely many The integers: notation: ℤ The countability of this set is 1-1, this new number system allows us to further understand the concept of even and odd | n is even iff n=2k for some k∈ ℤ | |n is odd iff n=2k+1 for some k ∈ ℤ | -4 is even because -4 = 2*(-2) where -2 ∈ ℤ -9 is odd because -9= 2*(-5) +1 where -5 ∈ ℤ In class way to create a 1-1 correspondence between the integers and the natural numbers union 0 (ℤ and ℕ U {0}) ℤ: {…-3, -2 ,-1 , 0 , 1 , 2 , 3…} ℕU{0}: {0 , 1 , 2 , 3 , 4 , 5 , 6…} ℕ U {0}: = {1,3,5,7, …} U {2,4,6, …} Prove that odd + odd=even PROVE: The sum of ANY two odd integers is even -OR- if two integers are each odd then their sum is even. If I know this: If two integers are each odd, Then I can prove this: Then their sum is even. Prove: the sum of any two odd integers is even. Proof: Let m and n be odd integers, so m and n must have the form: m=2k+1 and n=2j+1 where k and j ∈ ℤ adding m and n, we have m+n= (2k+1) + (2j+1) = 2k+1+2j+1 = 2k + 2j + 2 we can now factor out a 2 of each =2(k + j +1) = 2b where b = k + j + 1 Note that b ∈ ℤ by closure of the integers under addition. What is closure -The idea that you don't leave the set, the integers are closed under addition Integers represent a set. Addition represents and operation (+) This means that if we add integers the result will still be an integer. CLOSURE: involves a set and an operation I.E: PROVE: the sum of two odd integers is even. PROOF: let m and n be odd integers therefore (∴) m=2k+1 and n=2j+1 where m, n ∈ ℤ adding n + m: (2k+1)+ (2j+1) = 2k+1+2j+1= 2k+2j+2= (we can now pull a factor of two from this equation) 2(k+j+1) = 2(b) = where b= k+j+1, note that b ∈ ℤ ∴ by definition of even m+n is even∎ (∎ Means, "end of proof") PROVE: the products (indicating multiplication) of two odd integers are odd. I.E: is also an odd integer must have the form 2*(integer) +1 PROOF: let m and n be odd integers. ∴ m=2k+1 and n=2j+1 where k, j ∈ ℤ. We are multiplying in this proof* m*n= (2k+1)*(2j+1) (we now need to FOIL (First Outer Inner Last) this equation) F O I L F I OL (2k + 1) * (2j + 1) F= (2k*2j) O= (2k)*(1) I= (1)(2j) L=(1*1) We then add all of these equations together because the operation within The final FOILed equation looks like this: (2k*2j) + (2k(1)+(1)(2j)+(1)(1) This can then be simplified after combining like terms to look like this: 4kj + 2k + 2j +1 What is exciting about this simplified equation is the plus one at the end, because the end result is hopefully going to look like the odd equation: 2k + 1 we can then factor out a 2 from the terms that contain a factor of 2: 2(2kj + k+ j) +1 = 2b+1 Note that b is an integer under addition and multiplication ∴ by definition of odd, m*n is odd. Rational numbers: ℚ 2 Ration examples: 2 to 1. 2:1, 2 ℤ 1 3 IE: 3 = 1 By this example, every number is a rational number. ℕ= {1, 2, 3, …} Whole= {0,1,2,3,…} ℤ={…-3, -2, -1, 0, 1, 2, 3, …} Rational = {} Are there rational numbers that aren't integers YES! 1 2 −3 3 −3 1 −1 1 2 IE: 2 , 3 , 6 , −6 = 6 = - 2 = 2 = −2 = −4 = … p Rational numbers can by represented by this equation:q{ | (this long line symbolizes: "such that, where it is true that) p , q, ∈ ℤ the set of all numbers p that have the form q where q ≠ 0} p The actual equation looks like:q{ | p , q ∈ ℤ , the set of all numbers that have p the form q where q ≠ 0} (This set is also countably infinite and has a 1-1 correspondence) The rational numbers are not closed under division. n REASON: Closed under division: given any two n, m, ∈ ℚ m must also ∈ ℚ Answer: No, since 0 ∈ ℚ, we cannot divide by 0 and still get a rational number. Counter Example: 1 0 Let n = 2 so n ∈ ℚ Since 1, 2 ∈ ℤ and 2 ≠ 0, let m = 1 so m ∈ ℚ since 0,1 ∈ ℤ and 1 ≠ 0 n But, m = 0 so dividing ∉ ℚ since m=0 m If we make ℚ - {0} Then this represents the set of all rational numbers excluding 0. THEN the rational numbers are closed under division. a b } a, b, c, d ∈ ℤ b, d ≠ 0 and c ≠0 c d 0.25 4 1.000 0.8 1 0.20 4 0.20 0.00 = 0.25 Why Long division! The decimal expansion terminates (the long division finally reaches zero.) 1 1 1 Do you agree that 3 3 3 =1 1+1 2 1 3 3 = 3 =3=13 1 ´ But we also just showed that 3 is just the same as 0.3 . Just to make it 1 more explicit 3 =0.3333333333333333… What if we do: 0.3 3 3 3 3 3 3 3 3 3 0.3 3 3 3 3 3 3 3 3 3 + 0.3 3 3 3 3 3 3 3 3 3 0.9 9 9 9 9 9 9 9 9 9 ´ So now we've learned that 1=0. 9 A quarter of a quarter (of a whole) "of a whole" is implied, but not written. We first need to look at the quarter of the whole. The best way to show a whole is to use a circle because it is easier to cut into multiple pieced evenly. This circle indicates a whole this indicates the whole in quarters. The sentence is asking for a quarter of a quarter. That indicates one fourth of the whole. 1/4 of 1/4 of 1 Now we need to cut the fourth into fourths. Therefore, the whole piece would look like this (except a lot better Because I am really bad at shapes in word) 1/4 of 1/4 = 1/16 aka: 1/4 * 1/4= 1/16 Recall: fraction multiplication a p b * q = Where a,b,p,q are ∈ℤ and b,q are ≠0 a p a∗p * = b q b∗q Are rational numbers closed under multiplication a∗p b ap } These integers are closed under multiplication q = bq Yes, the rational numbers are closed under multiplication, addition, and subtraction. 2− 1 Example: 3 4 Find a common denominator. The common denominator is known as re-writing each number in a more convenient form from its infinitely many ways to be written so that we're putting each into the same number of pieces. - = 2 ∗4 2 =3 = 8 3 4 12 2∗4 1∗3 2 1 3 4 3 −4= 4 − 3 8 3 5 ¿12 −12= 12 The general abstract version looks like this for addition: a,b,p,q,∈ ℤ, b, q ≠0 a+p b q a p ∗q ∗b ¿ b + q q b aq pb ap+pb ¿ bq qb= qb Note that bq = qb by commutative law of multiplication for integers Rational numbers have been around for thousands and thousands of years. We also become aware of decimal representations for rational numbers. 1=¿ Example: 4 0.25 (terminating decimal) 1 Example: = 3 0.3333333333333333333333333333333333333333333 1 ´ 3 = 0. 3 The over bar tells us to repeat this forever. ℕ = natural numbers/ counting numbers (1,2,3,4,…) Whole numbers= (0,1,2,3,…) ~ℕ U {0} ℤ = integers, everything with the whole numbers, extended mathematical concepts ℕ Whole= ℕ U {0} ℤ = Whole and negative numbers ℚ = Rational numbers ℚ = Rational numbers, these numbers encapsulate every set before it, but a 1-1 correspondence between each set can still be found between each set and the rational numbers. It was thought for a long time that this was all there was. The first crisis in mathematics: The crisis of the Irrational Irrational- not rational, numbers that are not rational IE: a real number that cannot be represented as a ratio of integers Recall, the definition of a rational number is a number, say, p n = q where q ≠ 0 and p, q ∈ ℤ p so if n is irrational, all we can say is that n ≠p, q, ∈ ℤ and q ≠ 0 q i ℕ Whol e ℤ ℚ Irrational numbers all you can say about it is what it isn't i: set of irrational numbers (they don't encompass anything) -a totally separate and distinct thing from our previous number systems This leads to the new mathematical number system world of real numbers so the definition of the Real Numbers is that the set of real n,ℝb,is ℝ = ℚ U i, or in words: the real numbers are the union of the rational numbers with the irrational numbers. If x ∈ ℝ then either x ∈ ℚ, or i(but clearly not both) ℝ ℚ ℤ Whole numbers ℕ i THE PYTHAGOREAN THEOREM c c b b c a b a a The Pythagorean Theorem is more than arithmetic, it is geometric. It's about the squares that can be made on all sides. A unit Squa1e The area of this square is 1*1=1 sq unit 1 1 1 Suppose I want to have a square where the area is twice the area of the unit square S S= side length S Simple question: What is the length of the side of such a square p If we only had rational numbers, we're asking whatq where p, q ∈ ℤ, q ≠ 0 is this length p p ∗p So for what does q = 2 q q s*s=2 Almost= key word (approximation) S Side of the second 7 1 sSide of the first square ≈ 5 7 p S ≈ 5 } rational number because q are integers A square is generated from a root side Square root of 2 is the shortcut for "the root of the square who's area is 2" Modern Notation: In our modern notation this makes more shortcuts 49 7 7 ∗7 = 5 (7∗7) 49 √ 25 5 5 = (5∗5)= 25 2 2 2 Going back to the simple question:a +b =c unit square + unit square The area must be exactly 2 by Pythagorean Theorem 12 12 c2 area + area = 2 1*1 + 1*1 = c c2 1 + 1 = 2 2 = c SPEND LESS TIME ON THE PROOF BY CONTRADICTION, IT WONT BE ON THE MIDTERM SO: our goal becomes PROVE: that the √2 is not a rational number. That is, prove 2 is irrational PROVE: that the length of the side of a square with the area 2 is not a rational number Our big question is: HOW are we going to prove this So far, all we have as a method is a direct proof. 2 So suppose we wanted to use a direct proof to prove that √ 2 PROOF: let s be a number s.t. s = 2 so s*s = 2 This led to a new proof argument "reduction to absurdity" AKA: proof by contradictions In this type of proof you start by assuming the opposite (negate) of the thing you want to prove is true. Then you do logical steps, each following from the other with solid reasoning until you reach a contradiction to some given true/known fact! THEN, you can say with certainty that the only place the argument could have gone wrong is in your original assumption, so that assumption must be false. So to: PROVE √2 is irrational Proof by Contradiction 1) Assume (to the contrary) that the2 is not rational. √2 2) Then must be a rational number 2 2 3) So by definition of ℚ, it must be that we can writin the form √ = p q where p, q, ∈ ℤ, q ≠ 0. p 4) we also require that the fractqon be reduced to its lowest terms so that p 28 4∗7 7 and q have no common factor other than 1 = = Lowest 36 4∗9 9 terms