High School Math, Why Add an Equal Sign?

Since both sets are open intervals, e.g. b, it can be infinitely close to 3 but cannot be equal to three. So in the case of the equal sign, the condition is true, because it can't take that value.

Concentrate. The parentheses indicate an open range, meaning that the value cannot be taken, but it can be approached infinitely.

In the case of waiting, a is equal to 0, or equal to 1, it is okay. Since, if a is equal to 0, then a is (-2,2) and b is (-2,3), and since neither set can take the maximum value, the natural satisfaction of the condition a is contained in b

Since a is included in b, there can be two cases, a=b, and a is really included in b, so an equal sign should be added.

The inclusion symbol is not typed, and this symbol shows that the relationship between A and B is that A is part of B or A is equal to B. Can this be understood?

Since a can be equal to b, it is possible that a-2 = -2In the same way, a+2 can be equal to 3

Let's consider the equal sign for now. A-2=-2, a(-2,2), b(-2,3), and a is a subset of b.

A+2=3, a(-1,3), a is also a subset of b.

Since the sets are all open sections, they must be equal signs.

For example, when a[a-2,a+2),b(-2,3), then a-2>-2, there can be no equal sign.

Inclusion in may be equal.

The basic inequality needs to satisfy a certain two-positive third, and if the equals cannot be taken, the monotonicity of the checkmark function is used to solve it.

If the two equal signs are held at the same time, the two sets are equal, and this is a sufficient condition.

Meaning 1,2 will not be equal to a-2,a.

The right endpoint does not coincide when a-2=1, and the left endpoint does not coincide when a=2.

On the left is unconditionally true, and when x=y is equal sign; If the right side is true, then xy 0 is taken, and when xy=0 is taken as an equal sign.

If a=1 2, then p:1 2 x :1 2 x 3 can push out q, then sufficient; Q is not necessary if P cannot be introduced. then P is a sufficient but not necessary condition for Q; Conversely, q is a necessary but not sufficient condition for p.

A sufficiently unnecessary condition for x b is x a

It can be understood that a is a true subset of b.

Since it is a true subset, there is no such thing as a=b.

A: -13, can't wait.

Because if a=-1, then the intersection of b= and a is an empty set, so the equal sign cannot be taken.

a= 1 then b does not become x 1, and a intersects b does not become an empty set.

It's a plus, it's written wrong, you're right, have the courage to doubt.

According to the formula, your judgment is correct, and the bottom is addition instead of subtraction.

It should indeed be plus, not subtract, and the answer is wrong.

r* * is not a plus +? If yes, it represents a positive real number.

The limbs on the orange of the book calendar are grouped and raised.

<> make a simple change and you can be old.

The most convenient way to solve this problem is to substitute a=1 to see if it meets the conditions.

In this question, the condition is met when a=1.

x > a and x >1, you can draw a number line, and when a is on point 1, the condition is also satisfied.

If the equal sign is true, there must be ac=bc

If AC is not equal to BC, the equal sign is not true.

Because the three algebras are not equal, the equations cannot be equal.

I can ask your teachers to give an example.

If the a vector is equal to zero, it is not equal. What do you see?