Math Problem Why 0 99 1

Do you know the limit theory? The limit of the sequence ( is for example, (Ipsilon)-Delta) exists a corresponding positive integer for an arbitrarily given real number , and when , holds. We think of is the limit of (.

That's it, actually, I don't really understand, but I think this describes a sequence of numbers or a function, but it can also be used to describe the limit of a number, because no matter what the scale you choose, that is, what the value is, it can be regarded as the limit, which is 1, and there is an idea, list an equation, let this be x, then 10x-x=9 x=1 These two second are easy to understand, the first is just for reference, because it is my own idea, I don't know if it's right, if you think it's okay, be satisfied.

Simple, that is, equal, I don't know if you went to college, if you have it, you can see the first page of mathematical analysis. Without further aside, I'm going to explain why. 1 = 3 * 1 3 you should know, then 1 3 = you know this, then 3 * 1 3 =.

Can you understand that?

In turn, add a sail all over the place.

1 2 [(slippery demeanor let liquid]

Because, this is the sum of infinite proportional sequences.

1 is not a = but a cycle of equals.

If the loop of order is x, then the loop of 10x, then the loop of + = the loop, i.e. x = 10x

So x=, so the loop is 1

If you take the whole number, it is 1,

Solution 1: Because and.

So so. Solution 2: Because and.

So so. Solution 3:

Hypothesis: 1 ≠ then 1 3 ≠

Can actually be 1 3=

Leads to contradictions. So 1=

Solution: i.e., an infinitesimal quantity, but in higher mathematics, 0 represents an infinitesimal quantity.

So that is: 1=

This is a truth, and it may be strange. Just like i,a 0=+ (a>0) a 0=- (a<0) 0 0=nan 0* =nan

These are very similar and more difficult to accept.

It's not equal, it's about equal, because this number is infinitely close to 1. Similar to a regular polygon, when it has an infinite number of sides, it is infinitely close to a circle.

So far on the Internet, the answers are all about equality, this and that, this is obviously a false proposition, what is the limit, do you understand? Today, the old man will tell you that 1= is a false proposition.

It doesn't matter if you're using 1 or 3 or whatever, I'll give you an example here, and the others are similar.

Evidence 2: Let x=; --1) then: 10x=.

--2)(2)-(1),get:9x=9? What math is this Nima?

2) Equation is one less 9 than (1), okay? Do you understand? Well, (2) there are infinite 9s after the decimal point, and (1) the decimal point is (infinite - 1) 9s!

So at all, (2)-(1) does not get 9, for the limit, look at the definition of the limit.

It's like and not first of all the definition of the limits of people online is confused. The limit of = 1 means that infinity tends to 1 and does not mean that it is equal to 1.

The question is this:

a＝10a=

10a=9+

9a=9a=1 but there is a problem with this, since a=. Then the third line. So 9a≠9.

So a≠1.

Because wireless looping = wireless looping. If it is equal, then this is no longer a wireless loop, which is already limited. Therefore, in the concept of wireless looping, this 1 will never appear, and the appearance will be limited.

So. The essence of this question is the wireless loop, and the 1 of the wireless loop just can't get out.

Let x=then 10x=

Subtract the two formulas.

9x=9, so x=1.

's infinite loop is not equal to 1, because any number divided by 0 is meaningless, but the infinite loop of a number other than 0 divided by it can be infinite, which is meaningful. The so-called infinite loop where 1 3 is equal to, and the infinite loop multiplied by 3 is of course equal to 1, and the infinite loop that is not equal to cannot be understood by the multiplication rule of finite digits, and other proofs are similar. In addition, if it exists, then necessarily, and so on, then all real numbers are equal.

So the above two numbers are not equal.

Infinite problems are not to be solved with finite thoughts.

I guess your question is that they're a little bit off going **. In fact, if they are finite, they are indeed a little bit worse, but they are infinite, and infinite is not bad. Because you can't mathematically represent that difference.

Because that 1 can never happen, if this 1 is not possible, it is no different from 0.

Let x=then 10x=

Subtract the two formulas.

9x=9, so x=1

Therefore, this solution is correct, some people will ask whether 9x is equal to 9, in fact, the number of 9s after x is the same as the number of 9s after 10x, not one difference. If you can understand that there are as many even numbers as there are integers, this question is not difficult.

Not equal, don't be misled by 3 (1 3) 1 This kind of thing resembling a series of terms is infinitesimally inexplicable according to ordinary mathematical analysis theories.

……The latter is an infinite ratio of which virtual decreasing sequence, which is equal to a1 (1-q), where a1 is the first term, q is the common ratio, and the common ratio of the slow digging sequence is, so or because it is multiplied by 3 on both sides of the equation, it can be known by the virtual core.

Preface: This question gave me a headache for a while, and I couldn't come up with a definite answer at all, but it made me think of the phrase "learning to be confused as the side", which I can accurately say, I seem to have run to the side.

I now have a variety of views on both sides, starting with the view that it is not equal to one: 1They are not an exact position on the number line, which proves that they are not a number at all.

2.No matter how small the difference between it and one is, it is still a little bit worse, that is, it is always worse than cycle 1.

Then there is the view that it is equal to one: 1A simple fraction can show that 1 3 3 cannot be regarded as an infinite number of decimals, it is not within the range of real numbers, so it is one.

Then there is the counter-argument that it is not equal to one: 1The structure of the fraction simply ignores an infinitesimal number. 2.He may still be in the range of real numbers, because it is a number with a size.

Then there is the positive rebuttal: 1He's not an infinitely cyclic decimal number, so there's no such thing as infinitesimal at all.

From the perspective of fractions alone, a a equals one is absolutely correct. 2.But he can't be classified into any of the categories of rational numbers, and he can't exist in any of the categories of rational numbers.

That's all for my point, but if I can refute it, I might find something new.