Study whether the reasoning process is correct.
Study the relationship between sentences
Which of the following statements is true or false (but not both)?
(a) The only integers divisible by 7 are 1 and 7 itself.
(b) In the universe, the earth is the only living planet.
Please buy two tickets for Friday's concert!
Definition of proposition: A statement is called a proposition if it is true or false (but not both true and false).
P, q proposition
P and q, p∧q, conjunction (and)
P or q, p∞q, disjunction (or)
The truth value of the compound proposition p∧q is represented by the following truth table:
The truth value of the compound proposition p∨q is represented by the following truth table:
The negation of p is expressed as:? P means that the proposition is not p, and the truth table is:
For example, if the math department takes 20 thousand yuan more, it will hire another teacher
If p, then q is called conditional proposition, which is expressed as p → q.
Proposition P is called hypothesis (or antecedent) and proposition Q is called conclusion (or aftereffect).
P: the math department got 20 thousand yuan more.
Q: The Mathematics Department will recruit another teacher.
The truth value of a conditional proposition is defined by the truth table below.
Example: when a, b.
A→B, Sufficient Conditions for A to be B
B→A, a is the necessary condition of B.
A←→B, A is the necessary and sufficient condition of B.
Propositions: p→q, q→p are called inverse propositions. Reading: p contains q,
P If and only if q is called a double conditional proposition, it is expressed as:
Let compound propositions p and q consist of p 1 ... pn. If no matter what the value of p 1 is ... if pn is taken, p and q are always true or false at the same time, it is said that p and q are logically equivalent and the table is p ≡ Q.
De Morgan's law
? (p∨q)≡(？ p)∧(？ Q) and then what? (p∧q)≡(？ p)∨(？ Q)
Commonly used forms:
The inverse proposition (or transposition) of conditional proposition p→q is a proposition (? q)→(？ p .
Conditional proposition p→q (? q)→(？ P) are logically equivalent.
Example: p: n is an odd number, and p is not a proposition. Because the true and false values depend on n.
Let p(x) be a statement containing variable x and let d be a set.
If p(x) is the proposition of every x in d, we call p a propositional function (for d, d is the individual domain (definition domain) of p).
The propositional function itself is neither true nor false, but for each individual ×, p(×) is a proposition.
Statement: all x, p(x) can be written? x，p(x)
Statement: Is there x, p (x) that can be written? x，p(x)
Statement: For each real number x, x? & gt=0, this is a full-name quantifier statement.
A single field is a set of real numbers. This statement is correct.
Statement: For each real number ×, such as × >; 1, then ×+1>; 1。 This statement is correct.
Statement: For some positive integers n, if n is a prime number, then n+ 1, n+2, n+3 and n+4 are not prime numbers. This statement is correct.
Example: Proposition P: For all real numbers x, x? - 1 & gt； 0, then x= 1?
It is proved that p(×) is false for all x. Just find one or several X's that make P dissatisfied.
* ? X P(×) =F ≡ Without X, P(×) ≡? x，P(×)
For example, P is a propositional function, and each pair of propositions in (a) and (b) always has the same true or false value (that is, both are true or false).
(1)? (? x，P(×)； ? x，？ P(×).
(b)？ (? x，？ p(×)； ? x，P(×).
Assuming that a single field is a set of real numbers, the statement
For every x, there is y, and x = 0 is true.
And turn the statement upside down.
There is y, and for each x, x x y=0 is false.
Don't change places at will.
Mathematical system
Axiom-Definition-Undefined Term
The process of proving a theorem to be true is called proof.
The usual form of this theorem is:
For all x 1, X×n, such as
P (x 1, ×, -n), and then q (x 1, × r-n).
We should see three steps: inductive basis, inductive hypothesis and inductive proof.
Suppose that for every positive integer n, we have a statement s(n), and suppose that s( 1) is true;
As for all me
Then s(n) holds for all positive numbers n.
Two forms of mathematical induction:
Strong form: I
-shares: if S(n) is true, then S(n+ 1)
Example: It is proved by induction that 5 n- 1 is divisible by 4.
N= 1 obviously holds. Let 5 n- 1 be divisible by 4.
5n+1-1= 5.5n-1= (5n-1)+4-5n integer divisible by 4 |.
So for n= 1, 2 ..., 5n- 1 can be divisible by 4.
For arbitrary natural numbers x and n, x n-1can be divisible by x- 1
The sum of the first n natural numbers is n(n+ 1)/2. Proof: summarize n.
When n= 1, there is s (1) =1=1× (1+1)/2, so the proposition holds.
Suppose that the sum of the first n natural numbers S(n) is n(n+ 1)/2. Then, the sum of n+ 1 natural numbers is s (n+1) = (n+1) (n+2)/2.
According to the definition of S(n), S(n+ 1)=S(n)+n+ 1,
And by assuming that S(n)= n(n+ 1)/2,
So we can deduce that S (n+1) = n (n+1)/2+n+1= (n+2) (n+1)/2, and the proposition is proved.