Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities COMPSCI 220 Math Background: Rounding, Logarithms, Proofs, . Lecturer: Miika Hannula 1 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities 1 Rounding: d:::e; b:::c 2 Logarithms 3 Math proofs 4 Inequalities 2 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Rounding Real Numbers: Notation Replacing a real number x with the closest integer: 1 Ceil notation: dxe rounds up to the nearest integer larger than or equal to x: x + 1 > i = dxe ≥ x • Examples: d3:2e = 4; d10:01e = 11; d10:99e = 11 2 Floor notation: bxc rounds down to the nearest integer smaller than or equal to x: x − 1 < i = bxc ≤ x • Examples: b3:2c = 3; b10:01c = 10; b10:99c = 10 3 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Exponential functions: ax Exponential functions ax with the base a and the exponent x: • For an integer i and any real number a, ai is equal to a multiplied i times: ai ≡ a × a × ::: × a. | {z } i factors a • However, i may be and, in fact, often is a real number. • Simple rules for exponential functions: 1 ax a−x = ; ax+y = ax · ay; ax−y = ;(ax)y = ax·y ax ay • Special case: ex where e is Euler's (pronounced \Oil-ah") constant (e ≈ 2:718 :::). • x d x x The function e is its own derivative: dx e = e . • Using the natural logarithm (see Slide 5), ln, all functions aκ are mapped into ex by setting x = κ ln a, because eln a = a. 4 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Logarithms: log y x The inverse of exponential functions: if a = y, then loga y = x. • loga y = x reads \x is the logarithm to base a of y". Logarithm to other base differs by a factor: loga y = loga b · logb y y = aloga y = blogb y = ((aloga b)logb y = aloga b·logb y Thus, the base is often neglected by writing log y instead of loga y. Commonly used logarithms: • \logarithm to base 2" (in computing; lg in this course) • \logarithm to base 10" (in engineering) • \logarithm to base e ≈ 2:718 :::" (the \natural logarithm"): ln x always means loge x. 5 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Logarithms: log y Simple rules for logarithms (from those for exponential functions): loga(x · y) = loga x + loga y ) log2(4 · 32) = log2 4 + log2 32 | {z } | {z } | {z } log 128 = 7 2 5 2 x 128 loga y = loga x − loga y ) log2 = log2 128 − log2 4 4 | {z } | {z } | {z } 7 2 log2 32 = 5 y 10 loga(x ) = y · loga x ) log2(32 ) = 10 log2 32 = 50 log y = 1 ) log 2 = 1 = 0:2 a logy a 32 log2 32 log z log z = y ) log 128 = log2 128 = 7 = 1:4 x logy x 32 log2 32 5 zlogx y = ylogx z ) 128log2 32 = 32log2 128 = 235 6 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Systematic Analysis of Algorithms Some COMPSCI 220 contexts use systematically, step-by-step, the established mathematical tools: definitions, lemmas, and theorems. • A definition is used to make it clear what a certain term means, what we are going to call something, or how we will be using a certain notation. • A theorem is a statement we claim to be true, and it always requires a proof. • A lemma is like a small theorem that we prove to lead us up to the proof of a more extensive theorem. • As a general rule, we do not claim that something is true unless we can also prove it. • To find a statement is true, it should be either deduced (a direct proof), or proved by math induction or by contradiction. 7 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Mathematical Induction Proves that some statement is true for all integers, such that n ≥ n0, where n0 is mostly a non-negative constant. • If the proof should be for all non-negative integers, n0 = 0. • If the proof should be for all positive integers, n0 = 1. Proof by mathematical induction (or simply induction): 1 Basis: Prove that the statement is true for n0. 2 Induction hypothesis: Assume that the statement is true for some n. 3 Inductive step from n to n + 1: If the induction hypothesis holds, prove that the statement is also true for n + 1. The inductive step completes the proof. 8 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Pn n(n+1) Example 1: The Gauss Formula i=1 i = 2 Prove by math induction that for all n ≥ 1, n X n(n + 1) S = 1 + 2 + ::: + n ≡ i = n 2 i=1 1·2 1 Basis: For n0 = 1, the statement is correct: S1 = 1 = 2 . n(n+1) 2 Induction hypothesis: Let Sn = 2 hold for n, i.e., n(n + 1) 1 + ::: + n = 2 3 Inductive step: Then n(n + 1) (n + 1)(n + 2) S = S + (n + 1) = + (n + 1) = n+1 n 2 2 Hence, the same formula holds for n + 1, so that the Gauss formula is valid for all n ≥ 1. 9 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Mathematical Induction: Example 2 2 2 2 n(n+1)(2n+1) Prove that Sn = 1 + 2 + ::: + n = 6 for n ≥ 1. 1·2·3 1 Basis: For n0 = 1, the statement is correct: S1 = 1 = 6 . n(n+1)(2n+1) 2 Induction hypothesis: Let Sn = 6 hold for n, i.e., n(n + 1)(2n + 1) 12 + ::: + n2 = 6 2 3 Inductive step: Then Sn+1 = Sn + (n + 1) = n(n+1)(2n+1) 2 (n+1)[n(2n+1)+6(n+1)] 6 + (n + 1) = 6 = (n+1)[2n2+7n+6] (n+1)(n+2)(2n+3) 6 = 6 i.e., the same formula holds for n + 1. 10 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Mathematical Induction: Example 3 2 n an+1−1 Prove that Sn = 1 + a + a + ::: + a = a−1 for n ≥ 0 and a 6= 1 (partial sums for the geometric series with ratio a). a−1 1 Basis: For n0 = 0, the statement is correct: S1 = 1 = a−1 . an+1 2 Induction hypothesis: Let Sn = a−1 hold for n, i.e., an+1 − 1 1 + ::: + an = a − 1 n+1 an+1−1 n+1 3 Inductive step: Then Sn+1 = Sn + a = a−1 + a an+1 − 1 + an+2 − an+1 an+2 − 1 = = a − 1 a − 1 i.e., the same formula holds for n + 1. 11 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Proving a Statement by Contradiction • A simple, but quite rare way: Find a counterexample. • More general framework: 1 Assume the statement is false. 2 Derive a logical contradiction from this assumption. 3 Conclude the assumption is wrong and the statement is true. Example: Proof of the statement \There is no largest integer". • Assume there exists the largest integer itop. • Consider the number inext = itop + 1. • inext is integer as the sum of two integers. • Obviously, inext > itop, that contradicts the assumption. • Thus, the assumption is wrong, and the statement is true. 12 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Are all cars of the same colour? K + 1 K K \Proof" by a hidden error in using the math induction. • Base case: I have one car (K = 1) and it is green. • Inductive hypothesis. Let any subset of K cars; K ≥ 1, are of the same colour. • Inductive step. Park K cars in line and add my car to the end as the (K + 1)-th car. • Remove the first car: then the remaining intermediate cars and my car (K in total) are of the same colour, i.e., green. • Return the first car, but take away my car: once again these K cars are of the same colour as the intermediate cars. • Return my car to get K + 1 cars of the same colour. What is wrong? (hint: check the inductive step for K = 2) 13 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Inequalities, A Rel B, and their properties Exactly one of three relationships: x < y (less than); x = y (equal to), or x > y (greater than) holds for any two real numbers x and y. Rel Meaning Example x > y \x is greater than y" 5X + 15 > −25 x ≥ y \x is greater than or equal to y" −7X ≥ 28 x < y \x is less than y" 5X − 1 < 0 x ≤ y \x is less than or equal to y" 20X ≤ 1 • Transitivity: If x < y and y < z, then x < z. • Bias: if x < y and c is a real number, then x + c < y + c. • Positive scaling: If x < y and c > 0, then cx < cy. • Negative scaling: If x < y and c < 0, then cx > cy. The same properties hold for x ≤ y, x > y, and x ≥ y. 14 / 16 Outline Rounding: d:::e; b:::c Logarithms Math proofs Inequalities Inequalities, A Rel B, and their properties • Reversal: if x < y, then y > x. • Inverse: When x and y be both positive or both negative, it 1 1 1 1 holds that if x < y then x > y and if x > y then x < y . • Non-negative squares: x2 ≥ 0.