Math information representations

In summary: We cannot use the assumption that we have a complete list of real numbers, because in this case we cannot find any number in the diagonal, therefore no list, no information, no result (or proof if you like).Please read this again and again, until you understand it.Thank you very much.In summary, the conversation discusses the concept
  • #1
Organic
1,224
0
Any two noticeable R numbers a and b can be an open interval (a,b) of infinitely many R numbers, that cannot be separated form each other by any representation, and each R number can be represented only by aleph0 different representations.

Is it right ?

If what i wrote holds, than please tell me how can Cantor use some aleph0 representation to prove that there are more numbers then this aleph0 representation?

What i clime is: Cantor cannot start by using the assumption saying that he produces its new diagonal number (which is not in the list) from a complete list of R numbers, because a complete list of R numbers has no representation of any kind, therefore no information of any kind, and therefore there is no basis for any conclusion like: R numbers are not countable | R numbers are countable .

An example:

0 . ? ? ? ? ? ...
0 . ? ? ? ? ? ...
0 . ? ? ? ? ? ...
0 . ? ? ? ? ? ...
0 . ? ? ? ? ? ...

We cannot use the assetion that we have a complete list of R numbers, because
in this case we cannot find any number in the diagonal, therefore no list, no information, no result (or proof if you like).



Please show by Math language, how can we come to some result by using no information as a meaningful input?
 
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  • #2


Originally posted by Organic
Any two distinguished R numbers a and b ... that cannot be separated form each other by any representation...

It can be my english, but that doesn't make sense...
 
  • #3
Thank you Guybrush Threepwood,

I have changed 'distinguished' to 'noticeable' in my first post.
 
  • #4
well actually my comnfusion is:
is (a != b) as in "2 distinguished numbers from R"
or (a = b) as in "cannot be separated form each other by any representation"
 
  • #5
By Modern math a complete list of all R numbers is uncountable, according to Cantor's second diagonal proof: http://home.ican.net/~arandall/abelard/math12/Cantor.html [Broken]

I clime that by using the diagonal method we cannot come to this conclusion.
 
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  • #6
Any two noticeable R numbers a and b can be an open interval (a,b)

What do you mean by "noticable"?

Anyways, the real numbers [itex]a[/itex] and [itex]b[/itex] cannt be an open interval; they're real numbers, not intervals. What is correct to state is:

Given any two real numbers [itex]a[/itex] and [itex]b[/itex], we can form an open interval, which we call [itex](a, b)[/itex], defined by

[tex]
(a, b) := \{x | a < x < b\}
[/tex]

that is, [itex](a, b)[/itex] is defined to be the set of all real numbers greater than [itex]a[/itex] and less than [itex]b[/itex].


an open interval (a,b) of infinitely many R numbers

[itex](a, b)[/itex] contains infinitely many real numbers iff [itex]a<b[/itex]; if [itex]b<=a[/itex], then [itex](a, b)[/itex] is the empty set.


that cannot be separated form each other by any representation

What do you mean by "representation" and what do you mean by representations seperating real numbers?

For [itex]a<b[/itex], I can always decompose the open interval [itex](a, b)[/itex] into, for instance, three disjoint pieces, though I'm not sure if this has any bearing on your statement because I don't know what you mean (iow I'm purely guessing at what you could mean):

[tex]
(a, b) = (a, \frac{a+b}{2}) \, \dot{\cup} \, \{\frac{a+b}{2}\} \, \dot{\cup} \, (\frac{a+b}{2}, b)
[/tex]


and each R number can be represented only by aleph0 different representations.

Again, what is a representation?


For reference:

General form of a proof by contradiction:

P is the statement in which we are interested.

e.g. P := "There is a 'list' of all real numbers"

Assume P is true, and use this assumption to prove P is false.

Therefore, we can logically conclude that P is false. (Because if P is true, we can derive a contradiction)

In symbolic logic:

[tex](P \Rightarrow \neg P) \Rightarrow \neg P[/tex]
 
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  • #7
Hi Hurkyl,

Thank you very much for your reply.


By "noticable" i mean that a < b.

'representation' for some R number can be by base^power method with finite or infinitely many digits like 0.46385... or by using equivalence classes of increasing sequences of rational numbers like 1/2 - 3/45 + 7/88 ... and so on.

You wrote:
General form of a proof by contradiction:

P is the statement in which we are interested.

e.g. P := "There is a 'list' of all real numbers"

Assume P is true, and use this assumption to prove P is false.

What i clime is: Cantor cannot start by using the P statement saying that he produces its new diagonal number (which is not in the list) from a complete list of R numbers, because a complete list of R numbers has no representation of any kind, therefore no information of any kind, and therefore there is no basis for any conclusion like: R numbers are not countable | R numbers are countable .

An example:

? . ? ? ? ? ? ...
? . ? ? ? ? ? ...
? . ? ? ? ? ? ...
? . ? ? ? ? ? ...
? . ? ? ? ? ? ...

We cannot use the assumption that we have a complete list of R numbers, because in this case we cannot find any number in the diagonal, therefore no list, no information, no result (or proof if you like).


Please show by Math language, how can we come to some result by using no information as a meaningful input?
 
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  • #8
Originally posted by Organic
'representation' for some R number can be by base^power method with finite or infinitely many digits

is there some other posibility?


also i think you don't understand the meaning of a proof by contradiction. What Cantor did was the following:

1) assume that the statement "real numbers are countable" is true. Thus there exists a list of all the real numbers.

2) Show that the list is not complete (i.e there eis at least one real number that is not in the list)

3) Since the list is not complete, it means it does not contain all the real numbers.

see, just like Hurkyl said...
 
  • #9
Hi Guybrush Threepwood,

Please read what i write very carefully, because i am talking about a very fine idea that it is hard to understand from first look.

The main idea is this:

For any mathematical conclusion you need some input (some information).

When Cantor finds some irrational number, which is not in the list, all he can say that the list is not complete, and if he add this number to the list then there is another number which is not in the list, and so on.

If the list of R numbers was really a complete list, then the diagonal number must be in the list.

But then we have no list at all and we can't conclude that R is uncountable or R is countable, because we have no digits (the input information) for Cantor's function that define the new number, which is not the list.

So the basic law here is: No (input) information, no proof.

All we can get is a closed loop, based on base^power representation method, which gives us a list of 2^aleph0-1 numbers, where the missing one is some diagonal number, which is not in the list, and this is because we technically still did not find the way to represent 2^aleph0 numbers.
 
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  • #10
Originally posted by Organic
All we can get is a closed loop, based on base^power representation method, which gives us a list of 2^aleph0-1 numbers, where the missing one is some diagonal number, which is not in the list.

and I ask again, is there another posibility for representing numbers?

the Cantor's proof doesn't care in which base you choose to represent the real numbers. It can be base 2, base 10, or base 16. He does however use the proof by contradiction.
 
  • #11
Yes you can use instead of base^power representation method (and by writing this i mean any base value) the equivalence classes of increasing sequences of rational numbers.

For example:

1 <--> 1/2 3/45 7/88 ...
2 <--> 3/4 2/67 9/44 ...
3 <--> 1/9 8/17 3/37 ...
...

Then if our rule is to add 1 to the numerator then in this case Cantor's function result is: 2/2 3/67 4/37 ... which is a new sequence not in the list.

But it really doesn't matter, because still we can clime that we have some sequence as Cantor's function result, only because we deal with 2^aleph0-1 sequences, where the missing one is Cantor's function result, and this missing sequence depends on some abitrary order and the rule which we use to define Cantor's function result.

When we have 2^aleph0 sequences, Cantor's function result is unknown because we have no input at all (no rational form, no base^power form, no what so ever).

Because we are "runnig after our own tail" in a closed system, we cannot find a proof by contradiction.

But again these is because we did not find yet a way to represent 2^aleph0 numbers.

For more detailed information, please read this:

http://www.geocities.com/complementarytheory/NewDiagonalView.pdf
 
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  • #12
Originally posted by Organic
Yes you can use instead of base^power representation method (and by writing this i mean any base value) the equivalence classes of increasing sequences of rational numbers.

For example:

1 <--> 1/2 3/45 7/88 ...
2 <--> 3/4 2/67 9/44 ...
3 <--> 1/9 8/17 3/37 ...
...

When we have 2^aleph0 sequences, Cantor's function result is unknown because we have no input at all (no rational form, no base^power form, no what so ever).

how can you say that? You wrote the input yourself...
I can't quite understand you.
Are you saying that you can't write real numbers in Cantor's proof??
Also the equivalence classes (whatever...) you used are equivalent with the usual representation of real numbers (waht you call base^power)

But again these is because we did not find yet a way to represent 2^aleph0 numbers.

how did you come up with that number? you're using base 2 or something? and 2aleph0 is still aleph0...
 
  • #13
Guybrush Threepwood,

Cantor's diagonal proof says that aleph0 < 2^aleph0 = |R|.

Please tell me if you know this.
 
  • #16
And my argument is:

Cantor cannot prove that 2^aleph0 > aleph0, by using its
diagonal method, because of a very simple reason, which is:

When we have the complete list of 2^aleph0 numbers, then we don't have any known method that can represent its numbers, therefore without any input (which is some R number for Cantor's function, which its output is some R number not in the list) we HAVE NO WAY to conclude one of the two options, which are:

1) All R numbers are uncountable.

2) ALL R numbers are countable.

We can have some R number as an output of Cantor's function (which is some R number not in the list) if and only if we have some R number as its input.

When we have 2^aleph0 R numbers, then we don't know how to represent the input, therefore NO INFORMATION can only lead to NO CONCLUSION.
 
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  • #17
Originally posted by Organic
When we have the complete list of 2^aleph0 numbers, then we don't have any known method that can represent it, therefore without any input (which is some irrational number for Cantor's function, which its output is some irrational number not in the list) we are not able to conclude one of the two options, which are:

I don't understand what you mean here

Cantor said "let's assume that all real numbers are countable". That means we will be able to make a list of all the real numbers. The list has [itex]\aleph_0[/itex] entries (because we assumed the real numbers are countable). I dont' know what list of [itex]2^{\aleph_0}[/itex] numbers you are reffering to.
 
  • #18
And how do we check if ALL R numbers are countable XOR uncountable?

We try to create a bijection (1-1 and onto) among all N numbers and all R numbers, for example:

1 <--> 0.14321...
2 <--> 0.37130...
3 <--> 0.24031...
4 <--> 0.28850...
5 <--> 0.99210...
...

Now let us use Cantor's diagonal method on this list.


Cantor's method:

Let us use some function that get infinitely many digits as input,
and define some number as an output, according to some rule.

Let us say that the rule is:

A) Every 0 in the original diagonal number is turned to 1 in Cantor's new number.

B) Every non-zero in the original diagonal number is turned to 0 in Cantor's new number.

(we can define any rule that we like)

Now, the input for Cantor's function is:

The first digit of the first number, which in the list,
The second digit of the second number, which in the list,
and so on.

1 <--> 0.14321...
2 <--> 0.37130...
3 <--> 0.24031...
4 <--> 0.28850...
5 <--> 0.99210...
...

In this example the input R number is 0.17050...
and the output R number is 0.00101... (which depends on the arbitrary R list, and the rule that have been used by Cantor's function).

The output number is not in the list because it is different from any R number (which is in the list) by at least one digit.

If we add this new number to the list, then we can find a new input
of infinitely many digits for Cantor's function, and create another R number, which is not in the list, for example:

1 <--> 0.143211...
2 <--> 0.371301...
3 <--> 0.240311...
4 <--> 0.288501...
5 <--> 0.992101...
6 <--> 0.001011...
...

In this example the input R number is 0.170501...
and the output R number (which is not in the list) is 0.001010...

So, no matter how infinitely many Cantor's functions outputs we add
to the list, we still can find at list one R number which is not in the list.

Therefore we can conclude that there is no bijection between ALL N numbers and ALL R numbers and we can conclude that |N|<|R| or aleph0 < 2^aleph0.

The information's point of view on this proof:

I clime that Cantor cannot conclude that aleph0 < 2^aleph0 because
of the simple reason, which is:

No matter what representation we use to define some input for Cantor's function, first we have to define if there are an invariant signatures of R numbers that stays in any representation's form.

And by today's math language we can recognize two signatures, which are (in this case we shall use the decimal form, but again, it does not matter, we can use another base instead of 10, or the equivalence classes of increasing sequences of rational numbers*):

a) periodic repetitions for rational numbers (i.e. 5.000..., 0.101010...)

b) non-periodic repetitions for irrational numbers (i.e. .4589203...)

Now we come to the heart of my argument:

If we still can find some rational(periodic repetitions) or irrational(non-periodic repetitions)number which is not in the list (Q or R list), WE CAN CONCLUDE that our Q(all rational numbers) or R(all rational + all irrational numbers) list IS NOT COMPLETE.

In the case of a complete Q list, no rational number can be found as an input for Cantor's function.

In the case of a complete R list, no rational(periodic repetitions) or irrational(non-periodic repetitions) number can be found as an input for Cantor's function.

If R is not complete then we cannot conclude anything about the bijection (1-1 and onto) between N numbers and some NOT COMPLETE list of R numbers.







*
1 <--> 1/2 3/45 7/88 ...
2 <--> 3/4 2/67 9/44 ...
3 <--> 1/9 8/17 3/37 ...
...
 
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  • #19
Originally posted by Organic
And how do we check if ALL R numbers are countable XOR uncountable?

We try to create a bijection (1-1 and onto) among all N numbers and all R numbers, for example:

so do you think that Cantor wrote a list with all the natural numbers and their real corespondents?
sorry, but you clearly don't understand the proof by contradiction
the ideea of the proof is:
1) a countable list of all real numbers exist (doesn't matter how we found it...)
2) based on 1) show that there is a real number who's not in the list
3) thus we conclude that the countable list of real numbers does not exist an furthermore that there are more real numbers than the list can handle.

No matter what representation we use to define some input for Cantor's function, first we have to define if there are an invariant signatures of R numbers that stays in any representation's form.

please explain what do you mean by invariant signature of a number

(in this case we shall use the decimal form, but again, it does not matter, we can use another base instead of 10, or the equivalence classes of increasing sequences of rational numbers*)

what you call equivalence class of increasing sequences of rational numbers is still a rational number. You can multiply all the rational numbers you want the result is still a rational number. So this representation does not cover the irrational numbers
 
  • #20
so do you think that Cantor wrote a list with all the natural numbers and their real corespondents?

No, you missed the point.

Let us say that we want to mark any R number by any N number.

For doing this we have to represent each N number and each R number
by some common representation technique (let us say that we are using the base^power representation technique, where the base is 10).

We have no problem to represent all N numbers (aleph0 numbers), but we don't have any representation technique to represent R numbers (2^aleph0 numbers).

Through this point of view we find this:

aleph0 = (2^aleph0)-1, which means that we can mark (2^aleph0)-1 R numbers by (2^aleph0)-1 N numbers.

And where is the missing R number?

The missing R number is some result of Cantor's function, which depends on our ability to give some input to this function.

When we deal with 2^aleph0 elements then we find that all of our representations techniques does not have the power to produce the information for some meaningful conclusion.

We cannot prove anything without information.

All we can say that 2^aleph0 objects still cannot be represented by any representations techniques that we developed until today.

But this is a technical problem and not some deep mathematical true.

Any information system that constructed by rules is limited by its rules, therefore any extrapolation that does not change the rules is nothing but an information system that closed on itself, that recycling old information as if it is a new information.
------------------------------------------------------------------------------

Let me give you a concrete example by using Cantor's diagonal method
on decimal representation of the rational numbers.

The name 'rational numbers' is because each rational number is the ratio between two integers, for example:

1/2 6/1 567/3452 ... and so on.

"By using decimal representation method we can represent a list of rational numbers (repetitions over scales, for example: 0.123123123...) where the missing rational number is based on the diagonal rational number used as an input for Cantor's function (the function that defines the rational number, which is not in the list), where the result (the rational number, which is not in the list) depends on some arbitrary order of the list and some rule, which is used by Cantor's function.

For example:

0 . 1 7 1 1 3 1 7 1 1 3 1 7 ...
1 . 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 . 4 2 1 3 4 2 1 3 4 2 1 3 ...
0 . 1 0 1 0 1 0 1 0 1 0 1 0 ...
0 . 3 3 3 3 3 3 3 3 3 3 3 3 ...
2 . 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 . 3 5 4 9 5 5 1 3 5 4 9 5 ...
3 . 0 0 0 0 0 0 0 0 0 0 0 0 ...
0 . 6 4 1 6 4 1 6 4 1 6 4 1 ...
0 . 3 0 2 0 3 0 2 0 3 0 2 0 ...
0 . 6 1 3 6 1 3 6 1 3 6 1 3 ...
0 . 2 7 1 0 2 7 1 0 2 7 1 0 ...
...

In this case Cantor's function result is 0.0101010101010101... which is not in the list.


The rule is:

a) Every 0 in the original diagonal number is turned to 1 in Cantor's new number.

b) Every non-zero in the original diagonal number is turned to 0 in Cantor's new number.

We can add 0.0101010101010101... to the list, and then rearrenge the list in such a way that give us another rational number as cantor's function result, which means that our list is still not complete.


my conjecture is:

When we have the Aleph0 complete list of rational numbers, represented by decimal representation method, then and only then we can find only some irrational diagonal number (no repetitions over scales, for example: 0.123005497762...) as an input for Cantor's function."

I call this conjecture "Aleph0-1 conjecture" because i clime that if even 1 rational number is not in the list, we shall find it as the result of Cantor's function, but when the list is the complete list of all rational numbers (an aleph0 list), we can find only some irrational number as Contor function result.

------------------------------------------------------------------------------

Now take what we have found on Q numbers (the difference between the inputs of aleph0(irrational input) and aleph0-1(rational or irrational inputs)) and translate it to R numbers.

when we have (2^aleph0)-1 numbers the input for Cantor's function is some irrational number.

when we have 2^aleph0 numbers the input for Cantor's function cannot be represented, therefore we have no input For Cantor's function,
and without input (some information) we cannot conclude any meaningful conclusion on the countability of R numbers.
 
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  • #21
Why do you keep saying that we can't "represent" all of the numbers in R when we clearly can using the standard decimal representation (the thing you keep calling the base^power technique).

Just because we can't write all of the numbers in R as a list doesn't mean we don't have a representation for all of the numbers. We just can't write them as a list.
 
  • #22
Hi master_coda,

I added more information to my previous post.

Please read it again.

Thank you.

Organic
 
  • #23
If we have a listing of [itex]2^{\aleph_0}[/itex] numbers, what makes you think that we can't represent the result of "Cantor's function"?
 
  • #24
Because when we have the 2^aleph0 list, then there is no number that left out of the list, and can be found as Cantor's function result, which is always out of the list.

If we still can represent the number of Cantor's function result, it means that we don't have the complete 2^aleph0 list, and we can't start from the assumption that we have such a list (as Cantor did).
 
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  • #25
That's exactly the point. Given a complete listing of the real numbers using "Cantor's function" we can construct another valid real number that is not in the list. Thus the listing is not complete. CONTRADICTION

Because we have a contradiction, we must have made a false assumption somewhere. Where could that assumption be? Well, perhaps the number we constructed with Cantor's function is not in fact a real number. But Cantor's function produces an infinite string of digits. And we can prove from the fundamental properties of the real numbers that any real number can be represented by an infinite string of digits, and any infinite string of digits represents a real number.

Thus the number generated by Cantor's function is in fact a real number that is not in the list of all real numbers. There is only one other assumption that is could be false...the assumption that a complete listing of the real number exists. Thus we know that such a listing does not exist.

Therefore [itex]\lvert\mathbb{R}\rvert>\lvert\mathbb{N}\rvert[/itex].
 
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  • #26
Q numbers represented by decimal form also have infinitely meny digits, but |N|=|Q|.

You wrote:
Given a complete listing of the real numbers using "Cantor's function" we can construct another valid real number that is not in the list.

If you write this then you missed my argument, because what i clime is this:

Ginen a complete list of real numbers and you have no input for Cantor's function, to find some R number, which is not in the list.
 
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  • #27
Originally posted by Organic
Q numbers represented by decimal form also have infinitely meny digits, but |N|=|Q|.

But not every infinite string of digits is a rational number. Finding a real number not in our list of Q don't prove anything.

Originally posted by Organic
If you write this then you missed my arument, because what i clime is this:

Ginen a complete list of real numbers and you have no input for Cantor's function, to find some R number which is not in the list.

But the input of the function isn't the number not in the list. The input is the list we already have.

What the proof is saying is that:
"Given a list of all the real numbers, there are real numbers not in the list"
 
  • #28
How can you have an output (Cantor's function result, which is not the list) if you have no input ?

When the R list is complete then today's Math does not have the way to represent it, therefore no input --> no output --> no proof.

And this is a thechnical problem of and not deep mathematical true.

Please read this:

http://www.geocities.com/complementarytheory/NewDiagonalView.pdf
 
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  • #29
Originally posted by Organic
How can you have an output (Cantor's function result, which is not the list) if you have no input ?
[/url]

But you do have an input...it comes from the list.
 
  • #30
No dear master_coda,

If the list is complete, then any R number must be in the list, therefore we cannot find any R number (by using Cantor's function) which is not in the list.

And we cannot find any output result (by using Cantor's function) which is not in the list, if and only if we have no input.

Question: And when we have no input?

Answer: When we have no way to represent the list.

Therefore when the R list is complete, we cannot represent it, and cannot conclude if it is uncountable XOR countable.

Without information we cannot prove anything about R numbers.
 
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  • #31
Originally posted by Organic
No dear master_coda,

If the list is complete, then any R number must be in the list, therefore we cannot find any R number (by using Cantor's function) which is not in the list.

And we cannot find any output result (by using Cantor's function) which is not in the list, if and only if we have no input.

Question: And when we have no input?

Answer: When we have no way to represent the list.

Therefore when the R list is complete, we cannot represent it, and cannot conclude if it is uncountable XOR countable.

Without information we cannot prove anything about R numbers.

You don't even have a basic understanding of a proof by contradiction. The whole point of the a proof is that even though you can't find a real number not in the list, you can still find a real number not in the list.

When a proposition and it's negation are both true, you have a contradiction. That means that you are reasoning from a false premise. In this case, the premise is that it is possible to construct a complete list. You can't write a complete list of the reals because if you could, you could show that something is both true and false.
 
  • #33
... Question: And when we have no input?

Answer: When we have no way to represent the list.

The correct answer is "When no such list exists."
 

1. What are the different types of math information representations?

There are several types of math information representations, including numerical, graphical, symbolic, and verbal representations. Numerical representations involve numbers and calculations, while graphical representations use graphs, charts, and diagrams. Symbolic representations use mathematical symbols and equations, and verbal representations involve written or spoken descriptions of mathematical concepts.

2. How do math information representations help in problem solving?

Math information representations provide a visual or written representation of a problem, making it easier to understand and solve. They allow us to see patterns and relationships between numbers and concepts, and can help us identify the most efficient method for solving a problem.

3. What are some common mistakes people make when using math information representations?

One common mistake is misinterpreting or misreading the information represented. This can lead to incorrect calculations or solutions. Another mistake is relying too heavily on one type of representation and not considering other perspectives or methods for solving a problem.

4. How can math information representations be used in real life?

Math information representations are used in various fields such as science, engineering, finance, and statistics. They can help us understand and analyze data, make predictions, and solve real-world problems. For example, graphs and charts are commonly used to represent data in scientific research, and financial analysts use mathematical models to make predictions about the stock market.

5. What are some ways to improve our ability to use math information representations?

Practice is key to improving our ability to use math information representations. It is important to familiarize ourselves with different types of representations and practice interpreting and creating them. Additionally, seeking help from teachers or peers and actively seeking out opportunities to use math representations in real-life situations can also improve our skills.

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