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Direct proof technique

- The goal is to explain why some assertion is true.
- Formal proof:
- Sequence of statements
- Starts with some axiom (universally agreed-upon premise).
- Remaining statements are further axioms, or derived statements based on earlier statements in the proof according to rules of valid argument/inference. (You may also use results of theorems proved elsewhere.)
- Last statement is the theorem to be proved.
- Our proofs will not be quite this formal.
- Nevertheless: we need to begin by looking at relevant definitions, axioms and given information.

Technique

- Follow advice in book
- Steps:
- Begin with given information
- Apply definitions
- Compare what you have with what you want. Experiment by manipulating this information, use related facts, with the goal in mind.
- End with satisfying the desired property.
- Let’s first look at some simple definitions
- Odd and even numbers
- Hint: if something looks hard to prove, maybe it’s false!

Example direct proof

- “If the sum of 2 integers is even, then their difference is even.”
- Incidentally, in symbols, the statement looks like this:

x y (even(x + y) even(x – y))

- Here we go!

Let x and y be integers such that x + y is even. By the definition of even, k such that x + y = 2k.Then, x – y = x + y – 2y

= 2k – 2y = 2(k – y)

and observe that k – y is an integer too. So, by definition, x – y is even because it’s 2 * (integer).

Divisibility

- The symbol “|” means “divides”. We say that a | b if there exists some integer k such that ak = b.
- Or we can say: b is a multiple of a.
- Let’s use the direct proof technique for divisibility problems. It will sound like how we worked with even numbers, but more general.
- Prove or disprove:
- 3 divides the sum of any 3 consecutive integers
- If a | b and b | c, then a | c.
- If a | b and a | c, then a | (b + c).
- If a | bc, then a | b or a | c.
- If x and y are odd, then 6 | (3x + 3y).

Rational

- A number is rational if it can be written of the form a / b, where
- a and b are integers
- a and b have no common factors
- b is not 0
- Example: 7.5 is rational because we can let a = 15 and b = 2.

Rational opposite

- Let’s show that: if x is rational, then –x is rational.
- Since x is rational, x = a/b where a and b are integers with no common factors, and b is not 0.
- By arithmetic, –x = –a/b.
- We have to explain why –x satisfies the definition of rational.
- Notice that both –a and b are integers.
- If a and b have no common factors, then –a and b couldn’t have any common factors either.
- b is still not zero.
- We conclude that –x is rational.

Rational addition

- If x and y are rational, then x+y is rational
- According to the definition, let x = a/b and y = c/d.
- x+y = (ad + bc) / bd
- Since a, b, c and d are all integers, we can add and multiply integers to obtain more integers. Thus, (ad +bc) and bd are integers.
- There is no common factor between (ad+bc) and bd.
- Neither b nor d is zero. Thus, bd is not 0.
- Conclusion: x+y satisfies the definition of rational number.

Proof practice

- Floor and ceiling functions
- Proof by contradiction
- Iterative algorithms
- What is the termination condition?
- Application: Sequences and series

Floor and ceiling

- Used to calculate phone bill, postage, and loop iterations
- floor(x) = round down to the next lower integer
- ceil(x) = round up to the next higher integer
- Examples
- floor() = 3, floor(–) = –4, floor(17) = 17
- ceil() = 4, ceil(–) = –3, ceil(17) = 17
- For which real numbers x is floor(x) equal to 2?
- For which real numbers x is ceil(x) equal to 6?

Fractions

- Often we need to take floor/ceil of a fraction. Can we derive formulas for floor(a/b) and ceil(a/b) ?

Floor(a/b) = (a – a % b) / b

Ceil(a/b) = (a + (b – a) % b) / b

Loop iteration formula

- A for-loop (in C, C++ or Java) often has this general format:

for (i = a; i <= b; i += s)

- The number of loop iterations is

floor((b – a)/s) + 1

- Example:

for (i = 3; i <= 18; i += 5)

// We execute when i = 3, 8, 13, 18.

Indirect proof

- Common proof technique
- Essentially, we show that the negation of the given statement is false.
- Begin the proof by saying, “Suppose not.” The negation of the given statement gives us 1 extra piece of information.
- During the proof, we arrive at a contradiction. Thus, we conclude that ~P is false; thus P must be true.
- Example:
- If I give 100 marbles to 23 people, then somebody must get at least 5.

Another example

- “For all integers n, if n2 is even, then n is even.”
- Proof by contradiction.
- Assume the statement is false. Then there exists an integer n such that n2 is even and n is odd. Since n is odd, it may be written as n = (2k+1). Then n2 = (2k+1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Since 2k2 + 2k is an integer, then n2 satisfies the definition of an odd number. We have reached a contradiction, since earlier we said n2 is even. Thus the original statement must be true.
- Alternatively, can prove by contrapositive. The equivalent statement is: “… if n is odd, then n2 is odd.”

Irrational addition

- If x is rational and y is irrational, then x+y is irrational.
- Suppose not!
- Then, we have that: x is rational, y is irrational, AND x+y is rational.
- We already know that the sum of two rationals is rational, and that the opposite of a rational is rational.
- Subtraction means adding the opposite. Thus, the difference of two rationals must be rational.
- Notice that y = (x+y) – x. The two numbers (x+y) and x are rational. Therefore the difference, y, is rational. But this contradicts our earlier assumption that y is irrational!
- Conclusion: The original statement must be correct.

Algorithm

- Heart of any computer program
- 2 kinds
- Explicit formula: area of triangle, compute loan payment
- Iterative computation: many examples!

sorting an array

listing prime numbers

factorial

GCF

Square root

Fibonacci

Closer look

- Euclidean algorithm to compute GCF
- Start with 2 numbers
- Loop:

Divide the smaller into the larger

if remainder = 0, answer is the smaller

else, continue with the smaller number and remainder

- Square root of “a”
- Start with an initial guess x0
- xn+1 = average of xn and a/xn
- Stop when difference between 2 consecutive xi are sufficiently small.

Sequence and series

- A sequence is essentially a list or array of values
- A series is the sum of these values
- Terms separated by “+” instead of “,”
- Application: an operating system often needs to know the execution time of a computer program in order to best schedule that task.
- A program’s execution time is largely dependent on loops! Especially nested loops.
- We’ll use series formulas to calculate the number of iterations.

Notation

- Sequence notation (like an array)
- Can be explicit formula, as in: an = 4 + 3n
- Or can be recursively defined, as in: a1 = 5

an+1 = 2an, n 1

- Series notation
- a1 + a2 + a3 + a4 Too inefficient
- Better to use Sigma notation

Applications

- Sequence vs. series
- Meaning of sigma notation
- Common series formulas
- Mathematical induction

Bernoulli formulas

- Help us evaluate series, e.g. to count loop iterations
- Nested loop can give rise to nested sigma expression

Example

- Use Bernoulli formulas to determine the first hundred terms of

(3)(7) + (5)(12) + (7)(17) + (9)(22) + …

- First, write each term in terms of the term number, i

Term formula = (2i + 1)(5i + 2) = 10i2 + 7i + 2

- Apply Bernoulli formulas.

How to check

- If we say: sum(f(i)) = S(n), then we can check our sum answer this way: S(n) – S(n – 1) = f(n)
- For example, consider f(i) = 6i.
- The Bernoulli formula tells us that sum(f(i)) = 3n(n+1).
- So, S(n) = 3n(n+1)
- Work out S(n) – S(n – 1) = 3n(n+1) – 3(n – 1)n = 3n(n+1-(n – 1)) = 6n
- The answer checks because f(n) = 6n equivalently means f(i) = 6i.
- What would it look like if our summation formula is wrong?
- The f(n) would not match the original term formula.
- For example, we might have f(n) = 15n, but we originally wanted the summation of 6i, not 15i.

Inverse Bernoulli

- Review: Bernoulli formulas are used to help us convert from a term formula to a summation formula for a series.
- The way of checking summation formulas gives us an elegant way to calculate a term formula if we are given the summation formula.
- Ex. Finding the sum of i32i is not easy.
- Ex. But it is easy to find the term formula if we know if a series sum formula is n32n. Just plug in n and n – 1 and subtract!
- In fact, doing S(n) – S(n – 1) can be used to guess an unknown summation formula.

Nested loop

- Let’s count the operations in this nested loop.

for (i = 1; i <= n; ++i)

for (j = 1; j <= n; ++j)

// assume 3 stmts in body

- Outer loop:
- We do the i=1 once
- We do the i<=n (n+1) times
- We do the ++i n times.
- Inner loop:
- We enter the inner loop n times. Each time we have:

(1 + (n+1) + n + 3n).

- Total = 5n2 + 4n + 2 operations.

Nested loop #2

- Let’s change the inner loop bound to i

for (i = 1; i <= n; ++i)

for (j = 1; j <= i; ++j)

// assume 3 stmts in body

- Outer loop is still: 1 + (n+1) + n operations
- Inner loop:
- We enter the inner loop i times, where i could be 1 to n. For each i, we have: (1 + (i+1) + i + 3i).
- So, we have to sum: 5i + 2.
- Total = (5/2)n2 + (13/2)n + 2 operations.

Loop correctness

- Eventually, we want to be able to show that our loops are correct
- Powerful technique: Principle of Mathematical Induction
- Useful for proving many assertions
- Sum formulas
- Linear-combination formulas
- Divisibility questions
- Inequalities
- Correctness of loops

Induction

- P(n) is a statement that should be true for all positive integers n, or more generally for integer values (some integer).
- How to prove:
- Show that P(1) is true. This is the “base case.”
- Show that if P(k) is true for some k 1, then P(k+1) must also be true. This is the “inductive step.”
- If you can do these two steps, then you can say:

“Since P(1) is true and P(k) P(k+1) for an arbitrary k 1, then P(n) is true for all n 1.”

- “domino effect”

Type I: Summations

- Let’s first apply induction to verifying summation formulas. The key is to add the next term to both sides of the equation.
- We’ll work out these examples:
- n 1, 1 + 3 + 5 + 7 + … + (2n – 1) = n2
- n 1, 12 + 22 + 32 + … + n2 = n(n+1)(2n+1)/6
- n 1, 1(1!) + 2(2!) + 3(3!) + … n(n!) = (n+1)! – 1

Induction

- Types of statements you’ll often see
- Summations
- Linear combination
- Divisibility
- Inequalities
- Loop correctness

Type II: Linear formulas

- n 24, x, y 0 such that n = 5x + 7y.
- “Any integer n 24 can be expressed …”
- Base case: if n = 24, then choose x = 2 and y = 2.
- Next, assume P(k) is true. That is, k = 5x + 7y.
- Now we need a similar formula for k+1. A table of values may help.
- We see that if y 2, then k + 1 = 5(x + 3) + 7(y – 2)

else k + 1 = 5(x – 4) + 7(y + 3)

Two ways to add a penny:

Trade away two 7’s for three 5’s, or

Trade away four 5’s for three 7’s.

Try another

- n 14, x, y 0 such that n = 3x + 8y.
- “Any amount of postage 14 cents or higher can be achieved by using some combination of 3 and 8 cent stamps.”
- The goal of the inductive step is to write alternate formulas for k+1:

k + 1 = a(x + ___) + b(y – ___)

k + 1 = a(x – ___) + b(y + ___)

We have to give two formulas. Choose the appropriate formula to avoid a negative.

Type III: Divisibility

- We want to show that: a | f(n)
- Recall what our goal is: to go from P(k) to P(k+1)
- One way to bridge the gap is to “subtract” P(k) from P(k+1) and verify that what results is true. Then add this residual to P(k) to conclude P(k+1).
- For example, we may need to show that 5 | 6n – 1. The crux of the proof centers on f(n) = 6n – 1. Then:

f(k) = 6k – 1 and f(k+1) = 6k+1 – 1, so that:

f(k+1) – f(k) = (6k+1 – 1) – (6k – 1) = 6k+1 – 6k = 6k (6 – 1)

= 5 * 6k. Note that this is divisible by 5.

*** see handout

Alt approach

- Alternative approach: write P(k+1) of the form: P(k+1) = x P(k) + y, where x and y are clearly multiples of a.
- Often handy for multiple exponential terms.
- Ex. For all positive integers, 4 | 3n + 7n – 2.

In inductive step, we can write

3k+1 +7k+1 – 2 = 3*3k + (3+4)7k – 2

= 3*3k + 3*7k + 4*7k – 6 + 4

= 3(3k + 7k – 2) + 4(7k + 1)

Examples

- For all positive integers n, 3n is odd.

Sometimes we have a formula, and we want to be sure that it always evaluates to an integer. This turns out to be a divisibility problem in disguise.

- For all positive integers n, (2/5)4n +(3/5)(–1)n is an integer.

Type IV: Inequalities

- Purpose: Sometimes we want to know if one algorithm is more efficient than another.
- The statement P(n) is of the form f(n) < g(n).
- The “<“ could be some other relational operator, e.g. “”.
- During the inductive step, to leap from P(k) to P(k+1), we need to verify either:
- f(k+1) – f(k) < g(k+1) – g(k) or
- f(k+1) / f(k) < g(k+1) / g(k) whichever is easier
- Once you have established that inequality, then add or multiply to f(k) < g(k), as appropriate to conclude with f(k+1) < g(k+1).
- Be careful to structure the proof so that you are not assuming what you are trying to prove.

Type V: loop correctness

- Given a loop, and certain information about the loop:
- Precondition
- Invariant
- Postcondition
- For a loop to be correct, verify the following:
- The precondition is true when the loop starts.
- The invariant is true as we go from iter k to iter k+1.
- The loop terminates.
- The postcondition is true when the loop is done.
- ***e.g. See handout (adding x 10 times yields 10x)
- e.g. a loop that computes a summation of i2

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