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Forums - General - 0.9999.... = 1.0

 

Are you convinced?

Yes 34 58.62%
 
No 20 34.48%
 
not sure 1 1.72%
 
Total:55
Jay520 on 30 April 2013
SuperMarioWorld said:

the step 'subtract x' is a trick. you are treating x as a constant at the start of the proof and a variable on the 'subract x' step. so when you say subtract x it means subtract 0.99999 therefore your proof doesn't really hold up mathematically. its a trick


I don't care what you call it x. I made X equal to 0.99... 

At that point, everything I did was done to both sides, meaning I didn't change anything. Look, I can do the same for any other number you can think of. Here, I'll do some examples just for you.

x = 5 given
10x = 50 multiply by 10
9x = 45 subtract x
x = 5 divide by 9
5=5 substitution
x = 12 given
10x =120 multiply by 10
9x = 108 subtract x
x = 12 divide by 9
12 = 12 substitution
x = (1/9) given
10x = 10/9 multiply by 10
9x = 1 subtract x
x = 1/9 divide by 9
1/9 = 1/9 substitution

etc, etc.

So please tell me why this wouldn't work for 0.999...



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Jay520 on 30 April 2013
dsgrue3 said:

You went from 10x = 9.9999... and by subtracting x somehow arrived on 9x = 9; this is wrong. You should have arrived on 9.000.....1x = 9.

Series will converge to 1, but never arrive at it. It's like saying that you can divide 1 in half and eventually get to 0. You can't. You can approach 0, but never reach it. 

As to 1+1 =3 for sufficiently large values of 1, it's just a joke. 1.49 +1.49 = 2.98.


No. X = 0.999.... So when, I subtract x, I can subtract either X or 0.999.... It wouldn't make a difference.

Btw, 9.000....1x does not make sense. You can't just have a 1 there after an infinite amount of zeros.



dsgrue3 on 30 April 2013
Jay520 said:
dsgrue3 said:

You went from 10x = 9.9999... and by subtracting x somehow arrived on 9x = 9; this is wrong. You should have arrived on 9.000.....1x = 9.

Series will converge to 1, but never arrive at it. It's like saying that you can divide 1 in half and eventually get to 0. You can't. You can approach 0, but never reach it. 

As to 1+1 =3 for sufficiently large values of 1, it's just a joke. 1.49 +1.49 = 2.98.


No. X = 0.999.... So when, I subtract x, I can subtract either X or 0.999.... It wouldn't make a difference.

Btw, 9.000....1x does not make sense. You can't just have a 1 there after an infinite amount of zeros.

You must have a 1 there. Otherwise you're using circular logic.

I'm not disputing 1 = 0.999999 though, Pezus example made it very clear.



Jay520 on 30 April 2013
dsgrue3 said:
Jay520 said:
dsgrue3 said:

You went from 10x = 9.9999... and by subtracting x somehow arrived on 9x = 9; this is wrong. You should have arrived on 9.000.....1x = 9.

Series will converge to 1, but never arrive at it. It's like saying that you can divide 1 in half and eventually get to 0. You can't. You can approach 0, but never reach it. 

As to 1+1 =3 for sufficiently large values of 1, it's just a joke. 1.49 +1.49 = 2.98.


No. X = 0.999.... So when, I subtract x, I can subtract either X or 0.999.... It wouldn't make a difference.

Btw, 9.000....1x does not make sense. You can't just have a 1 there after an infinite amount of zeros.

You must have a 1 there. Otherwise you're using circular logic.

I'm not disputing 1 = 0.999999 though, Pezus example made it very clear.

uh...that's not circular logic. It's an infinite amount of zeros. To put a 1 at the end would mean that it's a finite amount of zeros, which is wrong.

You seem to think I'm subtracting 0.9999x from 10x. I'm not. On the left, I'm subtracting X from 10x, which is 9x. On the right side I'm subtracting 0.9999 from 9.0000 which is 9. In both cases, I'm subtracting the same value (since X = 0.999...) so no rule is broken.



Netyaroze on 30 April 2013

The difference between 0.9999..... and 1 is infinitely small but infinites can never end so 0.9999..... can never really equal 1 its always just almost 1 in theory.



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dsgrue3 on 30 April 2013
Jay520 said:

uh...that's not circular logic. It's an infinite amount of zeros. To put a 1 at the end would mean that it's a finite amount of zeros, which is wrong.

You seem to think I'm subtracting 0.9999x from 10x. I'm not. On the left, I'm subtracting X from 10x, which is 9x. On the right side I'm subtracting 0.9999 from 9.0000 which is 9. In both cases, I'm subtracting the same value (since X = 0.999...) so no rule is broken.

I see it now.



drkohler on 30 April 2013
Netyaroze said:

The difference between 0.9999..... and 1 is infinitely small but infinites can never end so 0.9999..... can never really equal 1 its always just almost 1 in theory.

 No. you don't get it. 0.(period)9 is EXACTLY 1

0.(period)9 can be written as a convergent geometrical series which has the mathematical value of 1.



dsgrue3 on 30 April 2013

x = 2
x^2 = 4
x = +/- 2

Fuzzy math is fuzzy. This is why it doesn't sit well with me.

I understand it through Pezus' example with fractions, but this isn't an equation. It's just an assertion.

It isn't a proof at all.

Surely you admit -2 != 2?



dsgrue3 on 30 April 2013
pezus said:
dsgrue3 said:
x = 2
x^2 = 4
x = +/- 2

Fuzzy math is fuzzy. This is why it doesn't sit well with me.

I understand it through Pezus' example with fractions, but this isn't an equation. It's just an assertion.

It isn't a proof at all.

Surely you admit -2 != 2?

Where is this equation and who are you replying to? o.O

I'm showing the OP and others that you can manipulate values willy nilly and it isn't a proof.

-

x = 0.9999… given
10x = 9.9999…. multiply by 10
9x = 9 subtract x
x = 1 divide by 9
0.999... = 1 substitution

-

This was his  "proof"

-

x = 2 given
x^2 = 4 square both sides
x = +/- 2 take sqr root


-2 = 2 substitution

- 

It is invalid. I'm pretty sure in order for you to do this you need an equation which resolves to 0.



Soleron on 30 April 2013
dsgrue3 said:
x = 2
x^2 = 4
x = +/- 2

Fuzzy math is fuzzy. This is why it doesn't sit well with me.

I understand it through Pezus' example with fractions, but this isn't an equation. It's just an assertion.

It isn't a proof at all.

Surely you admit -2 != 2?

you're not allowed to multiply or divide equations by x without handling the fact you just added/removed a solution. In this case the second negative solution is an artifact because you multiplied by x. This happens all the time in real physics problems and you can just ignore the other solution. There is a more formal way to deal with it in pure maths but I'm not practiced with it.