High School Mathematics Full Quantifiers, High School Mathematics Full Quantifiers and the Negation

A full quantifier is a word that contains the phrases "full", "each", "any", "everything", etc., which are all within the specified range, indicating the meaning of all objects within the specified range or the whole of the specified range. A proposition that contains a full quantifier is called a full proposition. The negation of the full quantifier is the presence of the quantifier.

In some universal propositions, sometimes the universal quantifier can be omitted. For example, a prism is a polyhedron, which means that "any prism is a polyhedron".

1. Words such as "to the full" and "to the arbitrary" are called full quantifiers in logic, and are recorded as " "A proposition containing a full quantifier is called a full proposition.

For any x in m, p(x) is true, denoted as x m, <>

Reads as: For any x belonging to m, there is p(x) that makes p(x) true.

2. Words such as "there is one" and "at least one" are called existential quantifiers in logic, and are recorded as " "propositions containing existential quantifiers are called special propositions.

The presence of at least one x in m makes p(x) true and denoted as x m, <>

Reads as: Reads as: The existence of an x belongs to m, making p(x) true.

Negation: 1. For the full proposition p: x m, the negation of p(x) for the full proposition p: x m, <>

2. For a special proposition p: x m, the negation p of p(x) is: x m, <>

In this range, if there is a value equal to 0, it means that there is a value in the range that intersects the x-axis, one above and one below, so the multiplication is less than 0

1. For the full proposition p:, which contains a quantifier"Arbitrary"The negation of x m,p(x) p is:"exists"x∈m，┐p(x)。

2. For a special proposition p:, which contains a quantifier"There is one"The negation of x m,p(x) p is:"All of them"x∈m，┐p(x)。

Full name proposition Special name proposition.

1.For all x a, p(x) holds 1The presence of x a makes p(x) hold.

2.For everything x a, p(x) holds 2There is at least one x a for p(x) to hold.

3.For each x a, p(x) holds 3For some x a, make p(x) hold.

4.Choose any x a, p(x) to form 4For a certain x a, make p(x) true.

5.Where x a, p(x) is established 5There is an x a, which makes p(x) hold.

Also: The negation of a proposition is a total negation, not a partial negation.

When negating the universal proposition, special attention should be paid to some propositions that omit the universal quantifier, such as the absolute value of the real number is positive. It would be wrong to write "the absolute value of a real number is not positive", and the correct denial would be: "The absolute value of a real number is not a positive number." ”

Commonly used "all" to indicate the full name of the affirmation, its existence negation is "not all", the two are mutual negation, with "neither" to indicate the full name of the negation, its existence affirmation can be expressed by "at least one is"...

In short, it is to remember that the negation of a proposition is a complete negation, not a partial negation. If you grasp this, you will basically not be wrong.

Let the original proposition liquid rise to be auspicious: if p then q

The negative proposition is: if it is not p, it is not q (double negation).

Negation of propositions: If p is not q (only the conclusion is negated).

Negative proposition: There is an x, such that x<5;

The negation of the proposition:

For any old x, there are x<5;

No proposition: x exists, and x is less than 5

Negative: Arbitrary x, x less than 5

The idea is correct, but the idea is off-track, and the problem is to find the range of m values if the conditions are met. The focus is on the understanding of the two conditions and turning them into mathematical formulas. Therefore, it is not a judgment of the truth or falsity of a proposition, and the relationship between sum or in this case a condition.

What we need to know is the range of values of x when determining true and false, so as to know the range of m values that make condition 1 true.

1) By condition 1, when x is less than 1, the g function is never less than 1, and condition 1 is true.

When x is greater than or equal to 1, the g function is greater than or equal to 0, and for condition 1 to be true, the f function must be less than 0

That is to say, the value of the f function must be less than 0 in the defined domain where x is greater than or equal to 1.

1.If m is equal to 0 and f is constant to 0, it is not true.

2.If m is greater than 0 and the quadratic function opening is upward, between -m-3 and 2m, f is less than 0, and it is impossible to make f constant less than 0 when x is greater than 1.

3.If m is less than 0, then the quadratic function opening is downward, and the size of the two intersection points x1=2m, x2=-m-3 needs to be determined.

When m is greater than -1 and less than 0, f is greater than 0 on (-m-3, 2m), and less than 0 when it is greater than 2m. Since m is less than 0, f is constant when x is greater than 1, less than 0 is constant.

When m is less than -1, f is less than 0 when it is greater than -m-3Ream-m-3<1, resulting in m>-4

Therefore, -4 (2) according to condition 2, you can also get a value range of one m, find it yourself.

Sorry, I don't know what the landlord is trying to say. But let me tell you about my approach to the problem.

First of all, the condition (1) tells us that for all the values, we need either f(x)<0 and g(x)<0

So, let's simplify f(x) first

f(x)=m(x^2+mx+3x-2mx-6m)=m(x^2+(3-m)x-6m)

Since m is a constant, then this should be a quadratic function, and we know that this opening should either rush up or down.

For g(x), the primary function, i.e., the linear function, is g(x)>=0 when x>=1.

Then f(1) <0, and m<0, and reducing f(x) in the above equation can be obtained: f(1)=4m-m 2<0

In the second problem, it is said that there is x, so that when x is at a value less than -4, it can have f(x) and g(x) heterotypes.

First, we know that when x<=-4, g(x)<=-10

That is, when x=-4, f(x)>0

Then the simplified formula has f(-4)=8m-2m 2>0

Combine the requirements of f(1) and f(-4). You can get the range of values of m.