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.
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.
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.
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
pi=c/d
it’s a fraction, just not with integers, so it’s not rational, so it’s not a fraction.
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.
Cut a banana into thirds and you lose material from cutting it hence .9999
That’s not how fractions and math work though.
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:
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.
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.
x = 5/9 is not 9/9. 5/9 = .55555…
You’re proving that 0.555… equals 5/9 (which it does), not that it equals 1 (which it doesn’t).
It’s absolutely not the same result as x = 0.999… as you claim.
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.