You are invited to Log in or Register a free Frihost Account!

# 1=.999....why? 20 frih\$

Manuel
1=.999

i am going to give 20 fri\$ points to the one who answers this.
ocalhoun
1=.999
Round
1=1

Surely that's too simple to be the correct answer, though...
Manuel
mathiaus
We did this in maths ages ago (2-3 years) when we started decimals and fractions.

1/3 = .333 recurring
2/3 = .666 recurring

1/3 + 2/3 = 3/3 = 1

.333 (R) + .666 (R) = .999 (R)

So 1 = .999 recurring (theoretically)

Of course many people say it is wrong. I disagree because I don't think 3 should be used (or any odd number for that fact). 2*(1/2 = .5) = 2/2 = 1

I think people will argue over this forever though. When you split things into three, you ALWAYS have one piece slightly larger than the other two!
Manuel
correct. i am going to pay you your 20 points
Arnie
Sorry, that is not a good mathematical proof. This one comes a bit closer. I don't care about the frih\$ though, you can keep that...

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1
Manuel
 Arnie wrote: Sorry, that is not a good mathematical proof. This one comes a bit closer. I don't care about the frih\$ though, you can keep that... x = 0.999... 10x = 9.999... 10x - x = 9.999... - 0.999... 9x = 9 x = 1

That one is cool too!

Anyway, both end in the same result and you were the second to answer it correctly.

Sorry, I would give you the points if you had been the first one.
imagefree
 mathiaus wrote: We did this in maths ages ago (2-3 years) when we started decimals and fractions. 1/3 = .333 recurring 2/3 = .666 recurring so add them together 1/3 + 2/3 = 3/3 = 1 .333 (R) + .666 (R) = .999 (R) So 1 = .999 recurring (theoretically) Of course many people say it is wrong. I disagree because I don't think 3 should be used (or any odd number for that fact). 2*(1/2 = .5) = 2/2 = 1 I think people will argue over this forever though. When you split things into three, you ALWAYS have one piece slightly larger than the other two!

It seems like the contest is over. Yet i want to say that the solution provided is not correct.

You yourself said that .333 (R) + .666 (R) = .999 (R) , but in the next step you replaced .333 (R) + .666 (R) and wrote 1 at its place.

It seems that you are trying to proof that .333 (R) + .666 (R) = 1 because you replaced the first for the second, instead of proving .999(R) = 1 .

Morever you said So 1 = .999 recurring (theoretically). Actually if you are rounding off your figures, you have to round off all the figures under consideration to maintain the quality of your results.

So, i think the solution is not correct.
rohan2kool
 imagefree wrote: You yourself said that .333 (R) + .666 (R) = .999 (R) , but in the next step you replaced .333 (R) + .666 (R) and wrote 1 at its place. It seems that you are trying to proof that .333 (R) + .666 (R) = 1 because you replaced the first for the second, instead of proving .999(R) = 1 . Morever you said So 1 = .999 recurring (theoretically). Actually if you are rounding off your figures, you have to round off all the figures under consideration to maintain the quality of your results. So, i think the solution is not correct.

he replaced .333 by 1/3 and .666 by 2/3, the sum of whose is obviously 1. Note that it is not .333 but .333... [digits continue to repeat till infinity]. He just skipped a step and replaced their sum by 1, while keeping the LHS unmodified, by which we can prove that .999 = 1.

although personally, i think that the second solution was way more cool.
mathiaus
 imagefree wrote: You yourself said that .333 (R) + .666 (R) = .999 (R) , but in the next step you replaced .333 (R) + .666 (R) and wrote 1 at its place. It seems that you are trying to proof that .333 (R) + .666 (R) = 1 because you replaced the first for the second, instead of proving .999(R) = 1 . Morever you said So 1 = .999 recurring (theoretically). Actually if you are rounding off your figures, you have to round off all the figures under consideration to maintain the quality of your results. So, i think the solution is not correct.

You misunderstood what I wrote. Please allow me to clarify...

Left 1/3 + 2/3 = 3/3 = 1 (This is where the 1 comes from on the LEFT)

Right .333 (R) + .666 (R) = .999 (R) (This is where the .999 (R) comes from on the RIGHT)

So 1 = .999 recurring (theoretically)

Arnie's way was better, I just mentioned the way I knew it (and got in first )
Guelila
 Arnie wrote: Sorry, that is not a good mathematical proof. This one comes a bit closer. I don't care about the frih\$ though, you can keep that... x = 0.999... 10x = 9.999... 10x - x = 9.999... - 0.999... 9x = 9 x = 1

I never thought about it this way
Did u figure this out by yourself or did you learn this from somewhere else?
Arnie
No, I didn't figure it out myself
hunnyhiteshseth
Well, I used to do such stuff a lot in junior classes, but now i have found an inherit flaw in all these types of proofs.

Here your assumption that 1/3 = 0.333.... is wrong. Remember they are not equal, 0.333... only gives the best approximation of 1/3
Since this basic step is flawed, rest of proof stand null and void.
Arnie
The proof I posted does not use 1/3 = 0.333... Have you read the entire topic before posting? I also already pointed out that the 1/3 proof is not mathematically correct.
hunnyhiteshseth
oh yes, about your proof. Actually I was also going to post what your faulty assumption was , but then forgot.

Anyway, your assumption that if x = 0.999... , then 10x = 9.999...... is wrong.

We assume that because, we consider that multiplying by 10 takes one 9 off from infinite 9s. So, practically, that would not make any difference. But when you are considering such type of proofs then you have to consider them.
So, 1 > 0.999... by 0.000.........(infinite times)01

Hope that makes it clear.