Simple maths but i don't understand why (about recurring decimals or repeating)

**donaldcat** · Aug17-06, 01:12 PM

When calculating recurring decimals, we let X to be that number to calculate it for example:

0.4* = x
4.4* = 10x

10x - x = 4.4* - 4*
9x = 4
x = 4/9

Therefore 0.4* = 4/9

But I when I calculate 0.9* this, i get
0.9* = X
9.9* = 10X

10X - X = 9.9* - 0.9*
9X = 9
X = 1

Therefore 0.9* = 1

but we know that 0.9* = 0.999999999............................. is close to 1 and not equal to 1
but how can we prove 0.9* is not equal to 1?

**d_leet** · Aug17-06, 01:56 PM

We can't because it IS equal to 1. Search for other threads like this on this forum, there have been a ton of them.

**Werg22** · Aug17-06, 05:21 PM

Originally Posted by donaldcat

When calculating recurring decimals, we let X to be that number to calculate it for example:

0.4* = x
4.4* = 10x

10x - x = 4.4* - 4*
9x = 4
x = 4/9

Therefore 0.4* = 4/9

But I when I calculate 0.9* this, i get
0.9* = X
9.9* = 10X

10X - X = 9.9* - 0.9*
9X = 9
X = 1

Therefore 0.9* = 1

but we know that 0.9* = 0.999999999............................. is close to 1 and not equal to 1
but how can we prove 0.9* is not equal to 1?

The value 0.9* does not have any real meaning. Infinite numbers are irrelevant to mathematics as we can only deal with rational numbers (that are certain time approximation of irrational numbers). Letting this aside, 0.9* can be interpreted as the limit of a the geometrical sum:

9*(1/10) + 9*(1/10)^2 ...9*(1/10)^n

As n increases beyond all bounds (thus creating an "infinite" number). The limit is in fact 1.

**d_leet** · Aug17-06, 05:23 PM

Originally Posted by Werg22

The value 0.9* does not have any real meaning. Infinite numbers are irrelevant to mathematics as we can only deal with rational numbers (that are certain time approximation of irrational numbers). Letting this aside, 0.9* can be interpreted as the limit of a the geometrical sum:

What? Infinite numbers seem, to me at least, to definitely be relavent to mathematics, and are you seriously saying that we only deal with rational numbers? Are pi and e rational?

**Werg22** · Aug17-06, 05:26 PM

Originally Posted by d_leet

What? Infinite numbers seem, to me at least, to definitely be relavent to mathematics, and are you seriously saying that we only deal with rational numbers? Are pi and e rational?

PI and e aren't rational, but we can only work with rational approximations. But the concept of "infinity" has no real meaning in mathematics. 0.9* dosen't mean anything if you don't take it as the previously mentionned limit.

**d_leet** · Aug17-06, 05:28 PM

Originally Posted by Werg22

PI and e aren't rational, but we can only work with rational approximations.

Yeah, but how often when working a calculus problem let's say do you really work with 3.14, or 2.718 instead of just calling it pi or e and having an exact answer?

**d_leet** · Aug17-06, 05:29 PM

Originally Posted by Werg22

But the concept of "infinity" has no real meaning in mathematics. 0.9* dosen't mean anything if you don't take it as the previously mentionned limit.

Would you care to explain this a little more in depth?

**Werg22** · Aug17-06, 05:33 PM

There is no single awnser. For example, the function 1/x has irrational values of rational x, but its integral from 1 to x will give an exact awnser for x = e^a.

**d_leet** · Aug17-06, 05:36 PM

Originally Posted by Werg22

There is no single awnser. For example, the function 1/x has irrational values of rational x, but its integral from 1 to x will give an exact awnser for x = e^a.

Umm.. Ok. That didn't really answer the question I asked though.

**Werg22** · Aug17-06, 05:39 PM

Originally Posted by d_leet

Would you care to explain this a little more in depth?

How to explain? Mathematics has roots in the study of nature. From that point of view, we can only deal with rational approximations. 0.9* is simply another notation for:

C = 9(1/10) + 9(1/10)^2... 9(1/10)^n

Where C is always closer to reality as n grows. Thus the notation 0.9* for a value that C is never equal to, but always closer to it. It can also be written;

C = B + a

Where B = 1, and a can be as close to 0 as we wish, as long as n is big enough. Finally, the value 0.9* is thus equal to 1, since 1 is the value that "C is never equal to, but always closer to it".

**Werg22** · Aug17-06, 05:44 PM

Originally Posted by d_leet

Umm.. Ok. That didn't really answer the question I asked though.

How often cannot be awnsered in mathematics

An exact awnser will only occur if two inverse operations, involving e or pi, are made. For logs, trig and roots operations such as addition and multiplication can give a rational number (example: 3^1/2 * 3^1/2).

**d_leet** · Aug17-06, 05:47 PM

Originally Posted by Werg22

How to explain? Mathematics has roots in the study of nature. From that point of view, we can only deal with rational approximations. 0.9* is simply another notation for:

Ok. But that's a very experimentalist view in my opinion, and I will say again that we often deal with pi and e in problems and proofs as just that and take those to represent the irrational number.

Originally Posted by Werg22

C = 9(1/10) + 9(1/10)^2... 9(1/10)^n

Where C is always closer to reality as n grows. Thus the notation 0.9* for a value that C is never equal to, but always closer to it. It can also be written;

C = B + a

Where B = 1, and a can be as close to 0 as we wish, as long as n is big enough. Finally, the value 0.9* is thus equal to 1, since 1 is the value that "C is never equal to, but always closer to it".

I understand exactly why .999... is equal to 1 so you don't need to explain this to me or convince me of this fact, but the work you have here makes little sense and is certainly not rigorous. And the way you set this up makes it seem that you should have

C + a = B where B is 1 and a is as close to zero as you would like depending on the value of n using the above.

**d_leet** · Aug17-06, 05:49 PM

Originally Posted by Werg22

How often cannot be awnsered in mathematics

An exact awnser will only occur if two inverse operations, involving e or pi, are made. For logs, trig and roots operations such as addition and multiplication can give a rational number (example: 3^1/2 * 3^1/2).

Give me more credit than that

I know how we can obtain exact answers using operations similar to those above, my question was asking you to justify your claim that infinite numbers are irrelevant to mathematics.

**Werg22** · Aug17-06, 06:02 PM

For your comment on how e an pi are used in proofs and problems, I would never go against that. But my point is e an pi are merely notations, when dealing with real data we can only use rational numbers. For the

C = 1 + a

What I meant to say is that, if a = a(x)

0<a(n+1) < a(n)

Also for how infinite numbers are irrelevant: they're not irrelevant as if they don't exist, they're irrelevant as in they are only notations of a limit. The dealing of infinity in mathematics only has a meaning when we look at the limit of the expressions.

**d_leet** · Aug17-06, 06:07 PM

Originally Posted by Werg22

For your comment on how e an pi are used in proofs and problems, I would never go against that. But my point is e an pi are merely notations, when dealing with real data we can only use rational numbers.

Again I agree with this, but I would think that in most cases when working specifically in pure mathematics the data is unlikely to be so experimental in nature that it is approximated by rational numbers.

Originally Posted by Werg22

C = 1 + a

What I meant to say is that, if a = a(x)

0<a(n+1) < a(n)

I understand what you mean, but the way you defined C makes it trivially less than 1 for all n and so I still think it should be

C + a = 1.

Originally Posted by Werg22

Also for how infinite numbers are irrelevant: their not irrelevant as if they don't exist, they're irrelevant as in they are only notations of a limit. The dealing of infinity in mathematics only has a meaning when we look at the limit of the expressions.

Ok that makes sense and I will agree with you on that for the most part.

**Werg22** · Aug17-06, 06:14 PM

Pardon my mistake on a, it should be

0 < |a(n+1)| < |a(n)|

Just consider my a to be negative

Sorry for the confusion.