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1 = 0.99... (RossGr)
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JSSheridan
Golden Member
I would say that 0.33... + 0.66... = 1;
But as an engineer, 0.99... is close enough to be 1 at appropiate times, and sometimes it is not. Engineering is the art of approximation. Peace.
But as an engineer, 0.99... is close enough to be 1 at appropiate times, and sometimes it is not. Engineering is the art of approximation. Peace.
x=.9(repeating)
10x=9.9(repeating)
10x
- x
-------
9x
9.9(repeating)
- .9(repeating)
--------------------
9.0
9x=9.0
x=1
For the non believers out there.
.9(repeating) is not an irrational number.
That like saying that .3, .6, or anything for that matter(repeating) is an irrational number.
I will use the above proof to prove that any repeating decimal is NOT an irrational number.
Using the above 1/3+2/3 proof I will clarify using the proof I used.
1/3 = .3(repeating) correct, use long division to prove this
2/3 = .6(repeating) correct, use long division to prove this
These are, now, both proven to be rational numbers.
Basic arithmetic rules apply to rational numbers.
Given x=.3 repeating, and 10x=3.3 repeating; this is only moving the decimal place by multiplying by 10. Whoever said that 9x = 8.9 repeating is wrong, go back to school; flame on.
10x
- x
-------
9x
3.3 repeating
- .3 repeating
-----------------
3.0 Arithmetic rules of subtraction apply to rational number; by defenition, no repeating decimal is irrational.
9x = 3
x = 3/9 = 1/3
Given y=.6 repeating, and 10y=6.6 repeating
10y
-y
-----
9y
6.6repeating
-.6 repeating
-----------------
6.0
9y=6
y=6/9=2/3
x=1/3=.3(repeating)
y=2/3=.6(repeating)
x+y=1
1/3+2/3=1 if you try to say this is false, you're an idiot.
Stop trying to disprove this.
10x=9.9(repeating)
10x
- x
-------
9x
9.9(repeating)
- .9(repeating)
--------------------
9.0
9x=9.0
x=1
For the non believers out there.
.9(repeating) is not an irrational number.
That like saying that .3, .6, or anything for that matter(repeating) is an irrational number.
I will use the above proof to prove that any repeating decimal is NOT an irrational number.
Using the above 1/3+2/3 proof I will clarify using the proof I used.
1/3 = .3(repeating) correct, use long division to prove this
2/3 = .6(repeating) correct, use long division to prove this
These are, now, both proven to be rational numbers.
Basic arithmetic rules apply to rational numbers.
Given x=.3 repeating, and 10x=3.3 repeating; this is only moving the decimal place by multiplying by 10. Whoever said that 9x = 8.9 repeating is wrong, go back to school; flame on.
10x
- x
-------
9x
3.3 repeating
- .3 repeating
-----------------
3.0 Arithmetic rules of subtraction apply to rational number; by defenition, no repeating decimal is irrational.
9x = 3
x = 3/9 = 1/3
Given y=.6 repeating, and 10y=6.6 repeating
10y
-y
-----
9y
6.6repeating
-.6 repeating
-----------------
6.0
9y=6
y=6/9=2/3
x=1/3=.3(repeating)
y=2/3=.6(repeating)
x+y=1
1/3+2/3=1 if you try to say this is false, you're an idiot.
Stop trying to disprove this.
Mathematically, the trouble with this computation is that it cannot be completed. You are adding an infinite number of digits, this is impossible. This computation falls into the catogory of demonstration, it is not a proof. The fact that the result is valid does not justify the questionable methods used to obtain it.
My proof is a valid mathematical approach to the problem. I essentially show that .999... exists in every neighborhood on the real line that contains 1. That is you cannot find an interval around 1 which does not contain .999.... No matter how small an interval you choose. That is the defintion of equality on the real line. So mathematically 1= .999...
I feel that my proof contain exellect concepts for see why they are equal. It shows clearly that adding any small incremet to .999... results in 1 + . {a bunch of zeros}999.....
you will always have 9s left over, on any number you add to .999... results in 1 + some more and that is the definition of 1.
That is the defintion of equality on the real line.
Careful there boss. Because for every number that you pick there is still a neighborhood that does not contain the chosen number........
The flaw is when it is stated that x = .999...
.999... is not defined. In you define it as the limit... then you are already defining x as 1.
jeremy806...
Careful there boss. Because for every number that you pick there is still a neighborhood that does not contain the chosen number........
The flaw is when it is stated that x = .999...
.999... is not defined. In you define it as the limit... then you are already defining x as 1.
jeremy806...
Dinominant
Member
It is because we are using a base 10 math system. It is the same as saying in binary, 1.1... equals 10. If we used a base 9 math system, then yes 0.9... in base 10 equals 1 in base 9. Using algebra in a fancy way doesn't explain anything when you use irrational numbers that can be explained rationally in another system.
BTW
x=0.9?
9x=8.9... not 9.0
0.9? does not equal 1
BTW
x=0.9?
9x=8.9... not 9.0
0.9? does not equal 1
Shalmanese
Platinum Member
What about the interval (0.9999+1)/2 to 1?
If you can play funny tricks with numbers, so can I.
Since when is .999... undefined? The ellipsis is accepted notation for an infinite repetition of the preceding pattern when it appears at the end of a number. It represents the existance of a unspecified but finite number when it appears inside a number. So there are no limits involved. .999.... is by definition short hand for an infinite number of 9s there is no need to take a limit. This notaion indicates that the limit has been taken.
RossGr
So you're saying that fractions cannot be proven as infinitely repeating decimals and vice versa.
Given that statement is true, I am wrong.
But I don't believe that statement to be true. It's been a while since I went to math class, so I could be wrong. I doubt it, though.
So you're saying that fractions cannot be proven as infinitely repeating decimals and vice versa.
Given that statement is true, I am wrong.
But I don't believe that statement to be true. It's been a while since I went to math class, so I could be wrong. I doubt it, though.
Although the proof, 1/3 +2/3 =1, is tantalizing---it bothers me that .9repeating=1.
Is it possible that simple addition of irrational numbers is not clearly defined? I would be more comfortable with the addition of 2 infinite series-can each of the irrational numbers above, be expressed as the sum of an infinite series?
Is it possible that simple addition of irrational numbers is not clearly defined? I would be more comfortable with the addition of 2 infinite series-can each of the irrational numbers above, be expressed as the sum of an infinite series?
PowerEngineer
Diamond Member
Not another thread on this....
.999999.... is definitely equal to 1.0
Moderator, do us all a favor and lock this thread!!!
:beer:
illusion88
Lifer
come on!
It will be fun!
/me sticks some popcorn in the airpopper. You wanna get down on this?
Oh dear... how can I begin to share knowledge when someone makes this sort of comment, Must I teach even the most basic elements. Why do you call .999... irrational? is .333.... irrational? I think it is about the 5th grade where you learn that infinitly repeating and terminating decimals represent rational numbers. If you do not even know that much about the number system I do not think you should contribute to this thread.
BTW an irrational number is one which there is NO repeating patterns, clearly .999.... is not of this type.
And what about the number 1 - 0.(9), by (9) I mean reapeting 9.
It is 0.0.....................1, but what kind of a pattern is that?? No pattern. Is it a rational number then? Can you go outside rational numbers just by using the operator "-" and rational numbers?
Rational numbers are such that can be represented as fractions: p/q
1/3 = 0.(3) etc. where p, q are integers.
but 0.(9)?
that would be:
999.../1000... - what is that? Infinity / Infinity?
OK, let's assume 0.(9) is a rational number, 0.000........................1 must be a rational number as well (look above).
Now you said that because there is no number between 0.(9) and 1 - they are equal. (Something about neighbourhood).
The same holds for 1.000................1 and 1.
By transitivity we quickly end up with the following:
0.(9) = 1 = 1.000..................1 that implies that:
0.(9) = 1.000................1
But is it equal?
To show that it is not equal I can try to find a number that is between 0.(9) and 1.000.......1, why wouldn't I pick the number 1 then?
Or can't you see that this can be easly extended to all rational numbers? (All of them are equal?). Well no.
I know this discussion is pointless so I won't go on. I belive that 0.(9) has a limit of 1, but is not equal to 1 as a number. I see no trouble I could run into this way.
EDIT:
Right, I gave it more thought and I think it all depends on what set on numbers we are working on:
If it is RATIONAL numbers, then:
If we consider 0.(9) to be a rational number (because there is a pattern - 9 is always followed by 9), then:
Consider this operation:
1 - 0.(9) = ?
it is equal to 0, that implies that 1 = 0.(9), there I convinced myself 🙂, why not 0.000...............1 ? Because that is not a rational number - no repeating pattern!
Now we end up with this lack of symmetricy (sp?), 0.(9) is a rational number, but 0.000..........1 isn't.
If we assume that 0.000................1 is a rational number, 0.(9) != 1.
If it is REAL numbers, then:
0.000...............1 is a real number and it is the solution to
1 - 0.(9) = ?
and therefore 1 != 0.(9).
Now I have to think if 0.000...........1 is a number at all 😀 (if it ain't, then 0.(9)=1 both under rational and real numbers!).
It is 0.0.....................1, but what kind of a pattern is that?? No pattern. Is it a rational number then? Can you go outside rational numbers just by using the operator "-" and rational numbers?
Rational numbers are such that can be represented as fractions: p/q
1/3 = 0.(3) etc. where p, q are integers.
but 0.(9)?
that would be:
999.../1000... - what is that? Infinity / Infinity?
OK, let's assume 0.(9) is a rational number, 0.000........................1 must be a rational number as well (look above).
Now you said that because there is no number between 0.(9) and 1 - they are equal. (Something about neighbourhood).
The same holds for 1.000................1 and 1.
By transitivity we quickly end up with the following:
0.(9) = 1 = 1.000..................1 that implies that:
0.(9) = 1.000................1
But is it equal?
To show that it is not equal I can try to find a number that is between 0.(9) and 1.000.......1, why wouldn't I pick the number 1 then?
Or can't you see that this can be easly extended to all rational numbers? (All of them are equal?). Well no.
I know this discussion is pointless so I won't go on. I belive that 0.(9) has a limit of 1, but is not equal to 1 as a number. I see no trouble I could run into this way.
EDIT:
Right, I gave it more thought and I think it all depends on what set on numbers we are working on:
If it is RATIONAL numbers, then:
If we consider 0.(9) to be a rational number (because there is a pattern - 9 is always followed by 9), then:
Consider this operation:
1 - 0.(9) = ?
it is equal to 0, that implies that 1 = 0.(9), there I convinced myself 🙂, why not 0.000...............1 ? Because that is not a rational number - no repeating pattern!
Now we end up with this lack of symmetricy (sp?), 0.(9) is a rational number, but 0.000..........1 isn't.
If we assume that 0.000................1 is a rational number, 0.(9) != 1.
If it is REAL numbers, then:
0.000...............1 is a real number and it is the solution to
1 - 0.(9) = ?
and therefore 1 != 0.(9).
Now I have to think if 0.000...........1 is a number at all 😀 (if it ain't, then 0.(9)=1 both under rational and real numbers!).
Jeff7
Lifer
How can you accurately plot 0.99repeating on a number line? It would need to be an infinitessimally small distance in front of the 1.0 point. How would this be accomplished? I just wonder that current math has any good way (other than fractions) of dealing with an infinite string of numbers. Conventional math seems to rely on having a set amount of digits to work with.
It seems like what was proven is that the limit of 0.999... = 1 as the number of repeated digits goes to infinity. I have no problem believing that 1 is the limit of the convergent series 0.999..., however just because it is the limit does not mean it is equal. Because infinity isn't a number, merely a concept, you can't claim that 0.999... with infinite digits actually equals 1, only that it approaches it as the limit with more and more 9's.
In plenty of cases, the limit of a convergent series is never actually reached by that series.
In plenty of cases, the limit of a convergent series is never actually reached by that series.
Did you bother to read my rules for the use of the ellipsis? .000...1 implies that a finite but unspeficed number of zeros has been left out.
There is NO NUMBER with an infinite number of zeros followed by a 1. This statement is self contradictory. If the string of zeros is followed by a 1 (or anything other then a zero!) it is not infinite, it terminates at the 1 and can be counted.
in general your .000........1 = 10^-n where you have not specified n, this is a rational number.
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