Am I wrong?

[wolfgang]

Let X= 0.9999.............

then 10X=9.99999..........

10X- X = 9.99999......... - 0.999999.......

9X = 9

that means:

X=1

so

1= 0.999999........

the same is in case of:

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

but, 1/3 = 0.33333.........

so

0.333333......+0.33333.....+0.33333.......= 0.999999.......

which is < 1

Am I wrong?

[Guest]

Interesting puzzle....

I think that though 0.9999... stretches to infinity, it is not right to say that 9.999.... - 0.999.... is equal to 9.

It may be very close, but not equal to 9.

Edited January 8, 2012 by SMV

[Guest]

interesting...

i think the statement is correct .... as the diff. b/w 1 & 0.999..... is 0.000....(infinite 0's)1 which is actually nothing.. so there's no difference b/w 1 & 0.999....

think about it this way...

1/9 = 0.111........

0.111..... * 9 = 0.999....... but 1/9 * 9 = 1

therefore...... 0.999.... & 1 are same

[Guest]

As strange as it looks, you are definitely correct in your first example. The repeating decimal, 0.99999.... does in fact = 1. Thiis does mean you are wrong when you say 0.9999999... <1. Your first method is the technique for converting a repeating decimal into a fraction. If you use the same teqnique for 0.3333..., you'r answer will be 1/3. Likewise for 0.666666... to be 2/3.

[witzar]

0.999... = 1

0.333... = 1/3

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

[Guest]

I agree with the comments already posted. The infinite decimal 0-99999.... is equal to 1.

However, while on the subject of fractions, figure this out:

10/10=1; 9/9-=1; 8/8 =1 .......1/1=1, so 0/0 = 1

Now 0/10 =0; 0/9=0; ......0/1=0 and so 0/0=0 !

Further, 10/0 = infinity; 9/0 = infinity, ... 1/0 = infinity, so 0/0 = infinity !

Show the flaw in the logic.

[Guest]

.9999999999... is equal to 1.

consider the geometric series 9*(1/10)^n from n=1 to n=infinity

the series can be written .9+.09+.009+.0009+.00009+.000009...

the sum of the series is 9*(1/(1-1/10)-1)=9*1/9=1

As for donjar, x/0 is not infinity. It is undefined unless it is taken as a limit, but I will proceed under the assumption that that is what you mean. regardless, those three functions are not continuous at x=0. In fact, the value of 0/0, strange as it may sound, is dependent on context.

[mewminator]

the number of 9's after the decimal point should decrease by 1, I know infinity-1 is still infinity, but in this case it would be a different value,9.99999 x 10 would be 99.9999,so the 9's after the decimal should decrease, but stay infinity, though I don't know how to achieve that

[Guest]

For those of you who aren't convinced that 0.999... = 1, here's something to think about.

We all know that pi is irrational. However, we've come up with a numerical representation of pi: 3.1415926.... My question to you: what do we mean by that sequence of digits? Can you define it in terms of rational numbers? Then, how would that apply to 0.9999?

[Guest]

I agree with the comments already posted. The infinite decimal 0-99999.... is equal to 1.

However, while on the subject of fractions, figure this out:

10/10=1; 9/9-=1; 8/8 =1 .......1/1=1, so 0/0 = 1

Now 0/10 =0; 0/9=0; ......0/1=0 and so 0/0=0 !

Further, 10/0 = infinity; 9/0 = infinity, ... 1/0 = infinity, so 0/0 = infinity !

Show the flaw in the logic.

Division by zero does not equal infinity or zero. Dividing by zero causes the line to diverge. If you take the limit* of 1/x as x approaches 0 from the positive side, it goes towards positive infinity. If you take the limit of 1/x as x approaches 0 from the negative side, it goes towards negative infinity.

A few problems that will illustrate this point:

1/1 = 1; 1/-1 = -1

1/.1 = 10; 1/-.5 = -10

1/.01 = 100; 1/-.01 = -100

1/.000001 = 1000000; 1/-.000001 = -1000000

etc.

*Taking limits is something you will learn in Calculus 1 if you aren't already aware of the concept.

[Guest]

Division by zero does not equal infinity or zero. Dividing by zero causes the line to diverge. If you take the limit* of 1/x as x approaches 0 from the positive side, it goes towards positive infinity. If you take the limit of 1/x as x approaches 0 from the negative side, it goes towards negative infinity.

A few problems that will illustrate this point:

1/1 = 1; 1/-1 = -1

1/.1 = 10; 1/-.5 = -10

1/.01 = 100; 1/-.01 = -100

1/.000001 = 1000000; 1/-.000001 = -1000000

etc.

*Taking limits is something you will learn in Calculus 1 if you aren't already aware of the concept.

It depends on the system you're using; however, most of the time, division by 0 is undefined. Division by a number is equivalent to multiplication by that number's multiplicative inverse, and 0 does not have a multiplicative inverse in any ring. If your domain is, say, the Riemann sphere, division by zero is meaningful, but one assumes that we're using the reals here.

[Guest]

Interesting!

yes you are quite right, but there are no difference between 1 and .999999..........(to infinite).

But you mention that, 0.999999.........<1, it's incurrect. That way 1/3=0.33333333......

and, 1/3+1/3+1/3=0.3333....+0.3333....+0.3333....=1 (or 0.99999.......)

[bhramarraj]

Since many of the friends have not used spoiler so i am also not using it.

I think unless we do not ascertain a fixed value to an unknown quantity, say X, we can not apply mathematics to this quantity with respect to other quantities.

For Example X could be treated as 0.999 and then 10X = 9.99,

So 10X - X = 9.99 - 0.999 = 8.991 or 9X = 8.991;

Now we may see that X can not be equal to 1.

Without ascertaining a fixed value to a certain parameter we may fall into infinity.....!!!!!!

Am I right...?