The thing is 0.333… And 1/3 represent the same thing. Base 10 struggles to represent the thirds in decimal form. You get other decimal issues like this in other base formats too
In this context, yes, because of the cancellation on the fractions when you recover.
1/3 x 3 = 1
I would say without the context, there is an infinitesimal difference. The approximation solution above essentially ignores the problem which is more of a functional flaw in base 10 than a real number theory issue
Base 10 simply is not capable of displaying it in a concise format. We could say that this is a notation issue. No notation is perfect. Base 10 has some confusing implications
This seems to be conflating 0.333...3 with 0.333... One is infinitesimally close to 1/3, the other is a decimal representation of 1/3. Indeed, if 1-0.999... resulted in anything other than 0, that would necessarily be a number with more significant digits than 0.999... which would mean that the ... failed to be an infinite repetition.
While I agree that my proof is blunt, yours doesn’t prove that .999… is equal to -1. With your assumption, the infinite 9’s behave like they’re finite, adding the 0 to the end, and you forgot to move the decimal point in the beginning of the number when you multiplied by 10.
x=0.999…999
10x=9.999…990 assuming infinite decimals behave like finite ones.
Now x - 10x = 0.999…999 - 9.999…990
-9x = -9.000…009
x = 1.000…001
Thus, adding or subtracting the infinitesimal makes no difference, meaning it behaves like 0.
Edit: Having written all this I realised that you probably meant the infinitely large number consisting of only 9’s, but with infinity you can’t really prove anything like this. You can’t have one infinite number being 10 times larger than another. It’s like assuming division by 0 is well defined.
0a=0b, thus
a=b, meaning of course your …999 can equal -1.
Edit again: what my proof shows is that even if you assume that .000…001≠0, doing regular algebra makes it behave like 0 anyway. Your proof shows that you can’t to regular maths with infinite numbers, which wasn’t in question. Infinity exists, the infinitesimal does not.
The algebraic proof that 0.999… = 1 must first prove why you can assign 0.999… to x.
My “proof” abuses algebraic notation like this - you cannot assign infinity to a variable. After that, regular algebraic rules become meaningless.
The proper proof would use the definition that the value of a limit approaching another value is exactly that value. For any epsilon > 0, 0.999… will be within the epsilon environment of 1 (= the interval 1 ± epsilon), therefore 0.999… is 1.
The explanation I’ve seen is that … is notation for something that can be otherwise represented as sums of infinite series.
In the case of 0.999…, it can be shown to converge toward 1 with the convergence rule for geometric series.
If |r| < 1, then:
ar + ar² + ar³ + … = ar / (1 - r)
Thus:
0.999… = 9(1/10) + 9(1/10)² + 9(1/10)³ + …
= 9(1/10) / (1 - 1/10)
= (9/10) / (9/10)
= 1
Just for fun, let’s try 0.424242…
0.424242… = 42(1/100) + 42(1/100)² + 42(1/100)³
= 42(1/100) / (1 - 1/100)
= (42/100) / (99/100)
= 42/99
= 0.424242…
So there you go, nothing gained from that other than seeing that 0.999… is distinct from other known patterns of repeating numbers after the decimal point.
x=.9999…
10x=9.9999…
Subtract x from both sides
9x=9
x=1
There it is, folks.
Somehow I have the feeling that this is not going to convince people who think that 0.9999… /= 1, but only make them madder.
Personally I like to point to the difference, or rather non-difference, between 0.333… and ⅓, then ask them what multiplying each by 3 is.
Oh shit, don’t think I saw that before. That makes it intuitive as hell.
The thing is 0.333… And 1/3 represent the same thing. Base 10 struggles to represent the thirds in decimal form. You get other decimal issues like this in other base formats too
(I think, if I remember correctly. Lol)
Cut a banana into thirds and you lose material from cutting it hence .9999
That’s not how fractions and math work though.
I was taught that if 0.9999… didn’t equal 1 there would have to be a number that exists between the two. Since there isn’t, then 0.9999…=1
Not even a number between, but there is no distance between the two. There is no value X for 1-x = 0.9~
We can’t notate 0.0~ …01 in any way.
Divide 1 by 3: 1÷3=0.3333…
Multiply the result by 3 reverting the operation: 0.3333… x 3 = 0.9999… or just 1
0.9999… = 1
In this context, yes, because of the cancellation on the fractions when you recover.
1/3 x 3 = 1
I would say without the context, there is an infinitesimal difference. The approximation solution above essentially ignores the problem which is more of a functional flaw in base 10 than a real number theory issue
The context doesn’t make a difference
In base 10 --> 1/3 is 0.333…
In base 12 --> 1/3 is 0.4
But they’re both the same number.
Base 10 simply is not capable of displaying it in a concise format. We could say that this is a notation issue. No notation is perfect. Base 10 has some confusing implications
This seems to be conflating
0.333...3with0.333...One is infinitesimally close to 1/3, the other is a decimal representation of 1/3. Indeed, if1-0.999...resulted in anything other than 0, that would necessarily be a number with more significant digits than0.999...which would mean that the...failed to be an infinite repetition.Unfortunately not an ideal proof.
It makes certain assumptions:
Similarly, I could prove that the number which consists of infinite 9’s to the left of the decimal separator is equal to -1:
...999.0 = x ...990.0 = 10x Calculate x - 10x: x - 10x = ...999.0 - ...990.0 -9x = 9 x = -1And while this is true for 10-adic numbers, it is certainly not true for the real numbers.
While I agree that my proof is blunt, yours doesn’t prove that .999… is equal to -1. With your assumption, the infinite 9’s behave like they’re finite, adding the 0 to the end, and you forgot to move the decimal point in the beginning of the number when you multiplied by 10.
x=0.999…999
10x=9.999…990 assuming infinite decimals behave like finite ones.
Now x - 10x = 0.999…999 - 9.999…990
-9x = -9.000…009
x = 1.000…001
Thus, adding or subtracting the infinitesimal makes no difference, meaning it behaves like 0.
Edit: Having written all this I realised that you probably meant the infinitely large number consisting of only 9’s, but with infinity you can’t really prove anything like this. You can’t have one infinite number being 10 times larger than another. It’s like assuming division by 0 is well defined.
0a=0b, thus
a=b, meaning of course your …999 can equal -1.
Edit again: what my proof shows is that even if you assume that .000…001≠0, doing regular algebra makes it behave like 0 anyway. Your proof shows that you can’t to regular maths with infinite numbers, which wasn’t in question. Infinity exists, the infinitesimal does not.
Yes, but similar flaws exist for your proof.
The algebraic proof that 0.999… = 1 must first prove why you can assign 0.999… to x.
My “proof” abuses algebraic notation like this - you cannot assign infinity to a variable. After that, regular algebraic rules become meaningless.
The proper proof would use the definition that the value of a limit approaching another value is exactly that value. For any epsilon > 0, 0.999… will be within the epsilon environment of 1 (= the interval 1 ± epsilon), therefore 0.999… is 1.
The explanation I’ve seen is that … is notation for something that can be otherwise represented as sums of infinite series.
In the case of 0.999…, it can be shown to converge toward 1 with the convergence rule for geometric series.
If |r| < 1, then:
ar + ar² + ar³ + … = ar / (1 - r)
Thus:
0.999… = 9(1/10) + 9(1/10)² + 9(1/10)³ + …
= 9(1/10) / (1 - 1/10)
= (9/10) / (9/10)
= 1
Just for fun, let’s try 0.424242…
0.424242… = 42(1/100) + 42(1/100)² + 42(1/100)³
= 42(1/100) / (1 - 1/100)
= (42/100) / (99/100)
= 42/99
= 0.424242…
So there you go, nothing gained from that other than seeing that 0.999… is distinct from other known patterns of repeating numbers after the decimal point.
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