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

[donaldcat]

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

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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.

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]

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]

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]

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.

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.

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]

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.

 

[d_leet]

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.

Ok now that makes a lot more sense, sorry about that, and thanks for clearing that up.

 

[uart]

OK Werg22, let's assume that irrational numbers don't exist and then look at the problem of finding the zeros of the equation f(x) = x^2 + x - 1.

Assume that x can be written as p/q, where p and q have no common factors. This gives,

p^2/q^2 + p/q = 1

Multiply throughout by q and rearrange to get,

p^2/q = q - p

Notice that the RHS is an integer. This implies that q must be +/- 1, other wise LHS cannot be an integer.


Now take the original equation and multiply throughout by q^2/p. This gives,

q^2/p = p + q

Again the RHS is clearly an integer and so p must be equal to +/- 1, otherwist the LHS cannot be integer.

So what is our conclusion, if the original equation has any solution then it can only be +/-1. Sadly however you can easily verify that neither +1 or -1 is a solution to f(x)=0, so we'd have to conclude that the equation has no solutions! But f(0)=-1 and f(1)=+1, hence the function has to pass though zero somewhere in between. So we have a clear contradiction! You can't just ignore irrational numbers or you'll end up with all sorts of contradictions like this.

 

[Hurkyl]

Sigh, I miss PF for a couple days, and my punishment is to have to clean up one of these messes.

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?

What you "know" is wrong. The decimal number system is defined so that 0.9* = 1, and calculations like the one you performed are the reasons such a definition was chosen.

The value 0.9* does not have any real meaning.
...
0.9* can be interpreted as the limit ... The limit is in fact 1.

Unless your arguing that 1 does not have any real meaning either, this contradiction sums up the primary objection I have to your posts.

0.9* is the notation for a decimal number. That decimal number has no more and no less meaning than any other decimal number. 1 just happens to be another notation for that same decimal number.


The secondary objection is that mathematics is a much, much broader subject than the study of the arithmetic of terminating decimal expansions. There is absolutely no trouble with working with numbers like pi or e in their exact form, even to a staunch constructivist.

Mathematics has roots in the study of nature.

A tertiary objection is that nature is full of irrational numbers. So, they better have some meaning, even if we can only approximate them. (As an aside, note that most rational numbers don't even have terminating decimal expansions)

Furthermore, approximating things with terminating decimals is a convenient course of action for scientific pursuits... not a necessary one. There is absolutely no reason at all why we couldn't approximate measured quantities with irrational values.

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.

And as a quaternary objection (I love that word), this is IMHO one of the most common fallacies that prevent a person from understanding decimal notation. 0.9* is a notation for one particular number... it is not some ambiguous number with an unspecified, but finite number of 9's. "0.9...9 (n 9's)" is a correct alternative notation for your C. 0.9* is not.

And, to be precise, you should write something like C(n), to express the fact that what you've written is a function of n.

 

[HallsofIvy]

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.

No, that's not at all true. A person who has actually studied some "higher" mathematics ("Calculus and Beyond") can work very easily with \pi and e- as easily as with \sqrt{2}, notapproximations. "0.9* dosen't mean anything if you don't take it as the previously mentionned limit" is very misleading since 0.999... is the "previously mentioned limit" by definition of the decimal numeral system. Would you say "2 doesn't mean anything unless you take it to be 1+ 1"?

 

[HallsofIvy]

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".

This is absolute nonsense! I think we have another physicist trying to explain mathematics.

 

[HallsofIvy]

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.

Can you give one example of a rational value of x such that 1/x is irrational?

 

[HallsofIvy]

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.

Your use of the term "real data" makes it clear that you are not talking about mathematics.

 

[robert Ihnot]

Did anybody above catch his simple error? ".4 =x, 4.4 =10X" is wrong. It's 10x=4. Thus its 9X=3.6 and X=.4, just where we started!

 

[d_leet]

Did anybody above catch his simple error? ".4 =x, 4.4 =10X" is wrong. It's 10x=4. Thus its 9X=3.6 and X=.4, just where we started!

Huh? He uses .4* representing .444444444... with fours repeating forever, so I don't see a mistake unless I'm missing your point.

 

[Robokapp]

yeah...basically 0.9999... is the notation of a number that is as close to 1 that it pretty much acts like it is how I see it. 0.999... is equal to 1 for any operation I've tried.

For example
0.333...+0.666...=0.999...
1/3+2/3=3/3=1

if a=b then 3a=3b and that makes 3(0.333...)=3(1/3) which results in 0.999...=1

 

[matt grime]

yeah...basically 0.9999... is the notation of a number that is as close to 1 that it pretty much acts like it is how I see it. 0.999... is equal to 1 for any operation I've tried.

No, no, no. There is no room for being inexact in this subject lest is encourages the cranks, not least because it appears as if you expect there to be an occasion when they are different representations of real numbers in base 10. It is 1. It is not so close we can't tell them apart. It just is 1 from the definitions.

 

[Robokapp]

Ah. I see it now. That'd explain why they behave identically in operations. Thank you Matt.