Презентация, доклад Methods of proof
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Methods of proof



































Слайды и текст этой презентации
Слайд 2


Описание слайда:
Some terminology
A theorem is a statement that can be shown to be true.
In mathematical writing, the term theorem is usually reserved for a statement that is considered at least somewhat important.
Less important theorems sometimes are called propositions.
A theorem may be the universal quantification of a conditional statement with one or more premises and a conclusion.
Слайд 9


Описание слайда:
Some terminology
A conjecture is a statement that is being proposed to be a true statement, usually on the basis of some partial evidence, a heuristic argument, or the intuition of an expert.
When a proof of a conjecture is found, the conjecture becomes a theorem. Many times conjectures are shown to be false, so they are not theorems.
Слайд 10


Описание слайда:
Methods of proof
In practice, the proofs of theorems designed for human consumption are almost always informal proofs,
where more than one rule of inference may be used in each step, where steps may be skipped,
where the axioms being assumed and the rules of inference used are not explicitly stated.
Слайд 12


Описание слайда:
Methods of proof
The methods of proof discussed here are important not only because they are used to prove mathematical theorems, but also for their many applications to computer science.
These applications include
verifying that computer programs are correct, establishing that operating systems are secure,
making inferences in artificial intelligence,
showing that system specifications are consistent, and so on.
Слайд 19


Описание слайда:
Example 1 Use a direct method of proof to show that if х and у are odd integers, then ху is also odd.
Solution
First, notice that if x is an odd integer then х = 2т + 1, where т is an integer.Similarly, у = 2n + 1 for some integer n.
Then,
ху = (2m + 1)(2n + 1)=
= 4mn + 2m + 2 + 1 =
= 2(2mn + m + n) + 1
Is an odd integer.
Слайд 20


Описание слайда:
Example 2 Let n be a positive integer. Prove, using the contrapositive, that if n2 is odd, then n is odd.
Solution
The negation of n2 is odd is n2 is even, and the negation of n is odd is n is even. Therefore, we proof directly that
if n is even then n2 is even
Since n is even, we can write n = 2m for some integer m. Then, n2 = 4m2 = 2(2m2) is also even.
Слайд 21


Описание слайда:
Example 3 Use a proof by contradiction to show that if x2 = 2 then x is not a fraction.
Solution
By way of contradiction, assume that х is a fraction and write х = m/n where n and m are integers, n is not equal to 0 and n and m have no common factors. Since x2 = 2, we have that (m/n)2 = 2. Therefore, m2 = 2 n2. But this implies that m2 is an even integer. Therefore, т is an even integer. Hence, т = 2р for some other integer р.
Слайд 22


Описание слайда:
Example 3 Use a proof by contradiction to show that if x2 = 2 then x is not a fraction.
Solution
Substituting this information into the equation
m2 = 2 n2 leads to 4 p2 = 2 n2, n2 = 2 p2. But then, n is also an even integer. We have shown that m and n have a common factor (of 2) which contradicts our original assertion that m and n have no common factors.
This contradiction can only be resolved by concluding that if x2 = 2 then x is not a fraction.
Слайд 23


Описание слайда:
Mathematical induction
In computing a program is said to be correct if it behaves in accordance with its specification. Whereas program testing shows that selected input values give acceptable output values, proof of correctness uses formal logic to prove that for any input values, the output values are correct.
Proving the correctness of algorithms containing loops requires a powerful method of proof called mathematical induction.
Слайд 27


Описание слайда:
Mathematical induction
So does the algorithm for all lists of any length n?
Consider an input a1, a2, …, an of length n and let Mk be the value of М after k executions of the loop.
For an input list a1 of length 1, the loop is executed once and M is assigned to be the maximum of 0 and a1,which is just a1. It is the correct input.
If after k executions of the loop, Mk is the maximum element of the list a1, a2, …, ak then after one more loop Mk+1 is assigned the value max(Mk, ak+1 ) which will then be the maximum element of the list a1, a2, …, ak+1.
Слайд 28


Описание слайда:
Mathematical induction
By condition 1) the algorithm works for any list of length 1, and so by condition 2) it works for any list of length 2. By condition 2) again it works for any list of length 3, and so on. Hence, the algorithm works for any list of length n and so the algorithm is correct.
This process can be formalised as follows.
Слайд 31


Описание слайда:
Mathematical induction
Assume now that
1+ 2 + … + k = k(k+1)/2 for some k 1. Then
1+ 2 + … + (k+1) = (1+ 2 + … +k) + (k+1)
= k(k+1)/2 + (k+1)
= (k(k+1) + 2(k+1))/2
= ((k+2)(k+1))/2
= (k+1)(k+2)/2 .
Hence, for all k 1(Р(k) Р(k+1)) is true. Therefore, by induction Р(n) is true for all n 1.
Слайд 32


Описание слайда:
Mathematical induction
Example 2 Prove, by induction, that 7n – 1 is divisible by 6 for all n 1.
Solution
First, note that an integer a is divisible by an integer b if there is some other integer m with
а = mb.
For example, 51 is divisible by 17 since 51 = 3 • 17.
Let Р(n) be the predicate «7n – 1 is divisible by 6».
In the case n = 1,
7n – 1 = 71 – 1 = 7 – 1 = 6,
which is clearly divisible by 6. Therefore, Р(1) is true.
Слайд 33


Описание слайда:
Mathematical induction
Assume now that 7k – 1 is divisible by 6 for some k 1.
Then,
7k+1 – 1 = 7(7k) – 1
= 7(7k – 1) + 7 – 1
= 7(7k – 1) + 6 .
Since 7k – 1 is divisible by 6 it follows that 7(7k – 1) + 6 is also is divisible by 6.
Hence, 7k+1 – 1 is divisible by 6 and so (Р(k) Р(k+1)) for all k 1 is true.
Therefore, by induction Р(n) is true for all n 1.
Слайд 34


Описание слайда:
Mathematical induction
Example 3
A sequence of integers x1, x2, …, xn is defined recursively as follows:
x1 = 1 and xk+1 = xk + 8k for к >= 1.
Prove that
xn = (2n – 1)2 for all n >= 1.
Solution
Let Р(n) be the predicate xn = (2n – 1)2. In the case n = 1, (2n – 1)2 = (2 • 1 – 1)2 = 1. Therefore, Р(1) is true.
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