A simple limit problem.
If you haven't studied advanced mathematics, please see the following proof:
Proof method 1 0.9 period =0.3 period +0.6 period.
0.3 cycles equals one third. 0.6 period equals two thirds+two thirds equals 1.
Proof 2: Let A=0.9 period, then 10A=9.999 period. The two formulas are subtracted: 9A=9, so A= 1.
Proof of proposition
Supplementary note: Any infinite circulating decimal can be converted into decimal form.
0. 1 period = 1/9
0.2 period =2/9
0.3 period =3/9
……
0.8 period =8/9
Then 0.9 period =9/9= 1.
Note that all the above are equal signs, not equal! ! !
In fact, the above proof is not rigorous or even wrong, but the reasons for the mistakes are generally wrong.
I'm here to point out which is wrong. I tried to explain it in common language:
First of all, it must be clear that the concept of infinity will always be an unreachable place.
We can get 0.9=0.3+0.6.
That is, one-time verification 9=3+6.
0.99=0.33+0.66, that is, one-time verification 9=3+6.
0.9999 (K9s) = 0.3333 … 3+0.666 … 6 (K3 and 6s) Verification k times 9=3+6.
0.9999 (any number of 9) = 0.3333+0.666 (any number of 3 and 6)
But we can't get 0.9 period =0.3 period +0.6 period for a simple reason.
We can't verify them one by one! ! Because we can only repeat 9=3+6 any time, but we can't repeat 9=3+6 indefinitely! Although every decimal place is 9=3+6! But mathematics is rigorous, and there can be no loopholes! So this proof 1 is incorrect. Similarly, Proof 2 is not rigorous, because we can't get 9.9 cycles-0.9 cycles =9!
Note that 9.9 cycle -0.9 cycle =9 and 0.9 cycle =0.3 cycle +0.6 cycle are true propositions, but our proof is worked out bit by bit, so there are loopholes in the process.
Generally speaking, the calculation can't be finished because the decimal places are infinite!
Here is a general proof.
For series {an}0.9
0.99 0.999 ...0.9 period
We can get the n power of the general term1-110.
0.9 cycle is the value of an when n approaches infinity.
When n approaches infinity, Lim an= 1! Note that it is an equal sign, not approximately equal to.
So a period of 0.9 is equal to 1.
When n tends to infinity, the n power of lim110 = 0 is used here!
If we have studied number theory and real variable function, then we know that rational numbers and irrational numbers are dense on real numbers, in other words, there must be rational numbers between two irrational numbers, and there must be irrational numbers between two rational numbers! If your theory that 1 is greater than 0.9 weeks holds, then 1 and 0.9 weeks must be two numbers. 1 is a rational number. 0.9 The cycle is infinite and the decimal is rational. Please give an irrational number between these two numbers! Obviously, you can't even give a number with a period greater than 0.9 and less than 1, so 0.9 period equals 1.
Prove 3 If the period of 1 is greater than 0.9, then there must be a number A that makes the period 0.9.
It is proved that if the period of 1 is greater than 0.9, then there must be a number A that makes the period a= 1-0.9 nonexistent, so the assumption is not valid.
Here's what some friends will say: 1-0.9 cycle =0.000 (infinite zeros) 1. First of all, I will say that infinity is endless, and there can't be 1 after infinity, that is, there won't be any numbers after the cyclic symbol. This is the definition of cycle. If you insist on being serious, I will prove it to you below.
Assuming that there can be a number after the cycle number, choose 0.9 cycle. If there is a number after the 0.9 cycle, assuming it is 1, it is called 0.9 cycle 1 (because I can't type the cycle number, there is a dot on it).
0.9 period 1 must be greater than 0.9 period.
0.9 period 2 must be greater than 0.9 period 1.
…
0.9 period 8 must be greater than 0.9 period 7.
0.9 cycle 9 must be greater than 0.9 cycle 8!
That is, 0.9 period 9 must be greater than 0.9 period 1.
But 0.9 cycle 9 is 0.9 cycle! According to the first article, it is less than 0.9 period 1.
Contradictions, so there won't be any numbers after the periodic symbol.
The following refutes some wrong views.
A 1/3 is about 0.3 cycles, not equal to.
I suggest you go back and read a good book. It should be said in the primary school book that if it is about 0.3, what is the number of recycled components?
Second, according to the limit theory of calculus, the period of 0.9 is only an infinite approximation to 1, which is not equal.
If the function f(n) = 0.99...9 (n 9s), then F (n) approaches 1 at infinity, but when n approaches positive infinity, f(n)= 1.
Suppose the function f(n) = 0.99.9 (n 9s) and the image, then the image of f (n) can never coincide with y= 1, but it is an infinite approximation, so the cycle of 0.999 is not equal to 1.
The error is like this in the face of any large number, but we can never reach infinity. For example, when x tends to infinity, 1/x=0, but the image of this function can never coincide with y=0.
(Quoted from a post of bar friend conny572)