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.
Cut a banana into thirds and you lose material from cutting it hence .9999
I’d just say that not all fractions can be broken down into a proper decimal for a whole number, just like pie never actually ends. We just stop and say it’s close enough to not be important. Need to know about a circle on your whiteboard? 3.14 is accurate enough. Need the entire observable universe measured to within a single atoms worth of accuracy? It only takes 39 digits after the 3.
pi isn’t even a fraction. like, it’s actually an important thing that it isn’t
The problem is, that’s exactly what the … is for. It is a little weird to our heads, granted, but it does allow the conversion. 0.33 is not the same thing as 0.333… The first is close to one third. The second one is one third. It’s how we express things as a decimal that don’t cleanly map to base ten. It may look funky, but it works.
Pi isn’t a fraction (in the sense of a rational fraction, an algebraic fraction where the numerator and denominator are both polynomials, like a ratio of 2 integers) – it’s an irrational number, i.e. a number with no fractional form; as opposed to rational numbers, which are defined as being able to be expressed as a fraction. Furthermore, π is a transcendental number, meaning it’s never a solution to f(x) = 0, where f(x) is a non-zero finite-degree polynomial expression with rational coefficients. That’s like, literally part of the definition. They cannot be compared to rational numbers like fractions.
Every rational number (and therefore every fraction) can be expressed using either repeating decimals or terminating decimals. Contrastly, irrational numbers only have decimal expansions which are both non-repeating and non-terminating.
Since |r|<1 → ∑[n=1, ∞] arⁿ = ar/(1-r), and 0.999... is equivalent to that sum with a = 9 and r = 1/10 (visually, 0.999... = 9(0.1) + 9(0.01) + 9(0.001) + ...), it’s easy to see after plugging in, 0.999... = ∑[n=1, ∞] 9(1/10)ⁿ = 9(1/10) / (1 - 1/10) = 0.9/0.9 = 1). This was a proof present in Euler’s Elements of Algebra.
There are a lot of concepts in mathematics which do not have good real world analogues.
i, the _imaginary number_for figuring out roots, as one example.
I am fairly certain you cannot actually do the mathematics to predict or approximate the size of an atom or subatomic particle without using complex algebra involving i.
It’s been a while since I watched the entire series Leonard Susskind has up on youtube explaining the basics of the actual math for quantum mechanics, but yeah I am fairly sure it involves complex numbers.
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
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
You’re just rounding up an irrational number. You have a non terminating, non repeating number, that will go on forever, because it can never actually get up to its whole value.
1/3 is a rational number, because it can be depicted by a ratio of two integers. You clearly don’t know what you’re talking about, you’re getting basic algebra level facts wrong. Maybe take a hint and read some real math instead of relying on your bad intuition.
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...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.
Unfortunately not an ideal proof.
It makes certain assumptions:
- That a number 0.999… exists and is well-defined
- That multiplication and subtraction for this number work as expected
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 = -1
And 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.
X=.5555…
10x=5.5555…
Subtract x from both sides.
9x=5
X=1 .5555 must equal 1.
There it isn’t. Because that math is bullshit.
Lol what? How did you conclude that if 9x = 5 then x = 1? Surely you didn’t pass algebra in high school, otherwise you could see that getting x from 9x = 5 requires dividing both sides by 9, which yields x = 5/9, i.e. 0.555... = 5/9 since x = 0.555....
Also, you shouldn’t just use uppercase X in place of lowercase x or vice versa. Case is usually significant for variable names.
