Is it possible to represent $e=2.71ldots$ and $pi=3.14ldots$ algebraicially, without using infinity or...












-1














Is it possible to represent $e=2.71ldots$ and $pi=3.14ldots$ without using the infinity sign nor using infinitely-many numbers, and algebraically, like representing the golden ratio, which is $1:dfrac{sqrt{5}+1}{2}$?










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put on hold as off-topic by jgon, Cesareo, Leucippus, user91500, Saad yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jgon, Cesareo, Leucippus, user91500, Saad

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 6




    Yes, use the symbols “$e$” and “$pi$” ;)
    – MPW
    yesterday






  • 5




    @Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
    – Cheerful Parsnip
    yesterday






  • 2




    Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
    – SmileyCraft
    yesterday






  • 1




    This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
    – Noble Mushtak
    yesterday






  • 3




    @MathLover in your most recent comment, referring to $arccos(-1)$ as possibly also being $3pi$, not only $pi$, that is incorrect. The way $arccos$ is defined is as a function and functions only have one output at a time, not several. $arccos(-1)$ is equal to $pi$ and only $pi$.
    – JMoravitz
    yesterday


















-1














Is it possible to represent $e=2.71ldots$ and $pi=3.14ldots$ without using the infinity sign nor using infinitely-many numbers, and algebraically, like representing the golden ratio, which is $1:dfrac{sqrt{5}+1}{2}$?










share|cite|improve this question









New contributor




Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as off-topic by jgon, Cesareo, Leucippus, user91500, Saad yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jgon, Cesareo, Leucippus, user91500, Saad

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 6




    Yes, use the symbols “$e$” and “$pi$” ;)
    – MPW
    yesterday






  • 5




    @Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
    – Cheerful Parsnip
    yesterday






  • 2




    Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
    – SmileyCraft
    yesterday






  • 1




    This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
    – Noble Mushtak
    yesterday






  • 3




    @MathLover in your most recent comment, referring to $arccos(-1)$ as possibly also being $3pi$, not only $pi$, that is incorrect. The way $arccos$ is defined is as a function and functions only have one output at a time, not several. $arccos(-1)$ is equal to $pi$ and only $pi$.
    – JMoravitz
    yesterday
















-1












-1








-1







Is it possible to represent $e=2.71ldots$ and $pi=3.14ldots$ without using the infinity sign nor using infinitely-many numbers, and algebraically, like representing the golden ratio, which is $1:dfrac{sqrt{5}+1}{2}$?










share|cite|improve this question









New contributor




Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Is it possible to represent $e=2.71ldots$ and $pi=3.14ldots$ without using the infinity sign nor using infinitely-many numbers, and algebraically, like representing the golden ratio, which is $1:dfrac{sqrt{5}+1}{2}$?







number-theory






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Check out our Code of Conduct.











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Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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edited yesterday









Blue

47.7k870151




47.7k870151






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asked yesterday









Math Lover

546




546




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Math Lover is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.




put on hold as off-topic by jgon, Cesareo, Leucippus, user91500, Saad yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jgon, Cesareo, Leucippus, user91500, Saad

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by jgon, Cesareo, Leucippus, user91500, Saad yesterday


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – jgon, Cesareo, Leucippus, user91500, Saad

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 6




    Yes, use the symbols “$e$” and “$pi$” ;)
    – MPW
    yesterday






  • 5




    @Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
    – Cheerful Parsnip
    yesterday






  • 2




    Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
    – SmileyCraft
    yesterday






  • 1




    This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
    – Noble Mushtak
    yesterday






  • 3




    @MathLover in your most recent comment, referring to $arccos(-1)$ as possibly also being $3pi$, not only $pi$, that is incorrect. The way $arccos$ is defined is as a function and functions only have one output at a time, not several. $arccos(-1)$ is equal to $pi$ and only $pi$.
    – JMoravitz
    yesterday
















  • 6




    Yes, use the symbols “$e$” and “$pi$” ;)
    – MPW
    yesterday






  • 5




    @Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
    – Cheerful Parsnip
    yesterday






  • 2




    Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
    – SmileyCraft
    yesterday






  • 1




    This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
    – Noble Mushtak
    yesterday






  • 3




    @MathLover in your most recent comment, referring to $arccos(-1)$ as possibly also being $3pi$, not only $pi$, that is incorrect. The way $arccos$ is defined is as a function and functions only have one output at a time, not several. $arccos(-1)$ is equal to $pi$ and only $pi$.
    – JMoravitz
    yesterday










6




6




Yes, use the symbols “$e$” and “$pi$” ;)
– MPW
yesterday




Yes, use the symbols “$e$” and “$pi$” ;)
– MPW
yesterday




5




5




@Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
– Cheerful Parsnip
yesterday




@Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
– Cheerful Parsnip
yesterday




2




2




Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
– SmileyCraft
yesterday




Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
– SmileyCraft
yesterday




1




1




This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
– Noble Mushtak
yesterday




This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
– Noble Mushtak
yesterday




3




3




@MathLover in your most recent comment, referring to $arccos(-1)$ as possibly also being $3pi$, not only $pi$, that is incorrect. The way $arccos$ is defined is as a function and functions only have one output at a time, not several. $arccos(-1)$ is equal to $pi$ and only $pi$.
– JMoravitz
yesterday






@MathLover in your most recent comment, referring to $arccos(-1)$ as possibly also being $3pi$, not only $pi$, that is incorrect. The way $arccos$ is defined is as a function and functions only have one output at a time, not several. $arccos(-1)$ is equal to $pi$ and only $pi$.
– JMoravitz
yesterday












2 Answers
2






active

oldest

votes


















2














To summarize (some) of the comments:



The numbers that you can write algebraically, like the golden ratio--using any finite sequence of operations consisting of addition, subtraction, multiplication, division, square roots, cube roots, and/or higher-order roots applied to integers--is a subset of the algebraic numbers.



Transcendental numbers are real numbers that are not algebraic numbers.
It can be shown that $pi$ and $e$ are transcendental;
for example, see How hard is the proof of $pi$ or $e$ being transcendental?.



Hence neither $pi$ nor $e$ can be written algebraically.
So the answer is "no."






share|cite|improve this answer























  • +1 for bringing up the concept of "written algebraically". However maybe we should define what "written algebraically" and explain to the OP that this is actually what s/he was asking. After all. We can always represent $pi$ and $e$ as.... $pi$ and $e$.
    – fleablood
    yesterday










  • Never mind. I see the OP has editted the post to mean "written algebraically"
    – fleablood
    yesterday



















2














Yes. You can represent $e$ in one symbol "$e$" and you can represent $pi$ in one symbol by using the single symbol "$pi$".



However you can't express them as integers because.... they are not integers. You can't express $frac 35$ as an integer either.



You can't express them as a terminating decimal because they are not an integer scaled to a power of $10$.



Nor can you express them as a ratio between two finite integers because the are not rational.



Only difference between them and numbers such as $sqrt 2$ or the golden ratio which are also not rational, is that they are not roots to any polynomial with integer coefficients. $sqrt 2$ is the one of the roots to $x^2 -2 = 0$ and the golden ratio is solution to $x^2 - 2x -4= 0$. Neither $pi$ nor $e$ are a solution to any polynomials with integer coefficients.



Numbers that are solutions to polynomials are called algebraic numbers and the irrational algebraic numbers can't be expressed as finite decimals. But if we invent the notation, and we are mathematicians--- we can invent anything we like as like as it makes sense---, $sqrt[k]{m}$ to mean a $k$-th root of $m$, then we can express any algebraic number as some combination of roots.



Numbers, such as $e$ and $pi$ which are not algebraic are called transcendental. And, no, they can not be expressed via a digital decimal system. Nor can they be expressed as a series of root signs.



But so what? Using digital decimals to represent numbers is completely arbitrary and does not in any way make a number more "real" than any other. (The main reason we use digital decimals is because the are convergent and thus every thing can be approximate and accurately expressed through and infinite number of decimals if necessary.)



But to "represent" a number means nothing more than having a symbol for it. And if a number can by defined --- in a way that makes sense so that we know such a number exists--- we can define any symbol we want for them.



So I stand by my answer. You can represent $e$ and $pi$ with one symbol each: "$e$" and "$pi$"






share|cite|improve this answer




























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    To summarize (some) of the comments:



    The numbers that you can write algebraically, like the golden ratio--using any finite sequence of operations consisting of addition, subtraction, multiplication, division, square roots, cube roots, and/or higher-order roots applied to integers--is a subset of the algebraic numbers.



    Transcendental numbers are real numbers that are not algebraic numbers.
    It can be shown that $pi$ and $e$ are transcendental;
    for example, see How hard is the proof of $pi$ or $e$ being transcendental?.



    Hence neither $pi$ nor $e$ can be written algebraically.
    So the answer is "no."






    share|cite|improve this answer























    • +1 for bringing up the concept of "written algebraically". However maybe we should define what "written algebraically" and explain to the OP that this is actually what s/he was asking. After all. We can always represent $pi$ and $e$ as.... $pi$ and $e$.
      – fleablood
      yesterday










    • Never mind. I see the OP has editted the post to mean "written algebraically"
      – fleablood
      yesterday
















    2














    To summarize (some) of the comments:



    The numbers that you can write algebraically, like the golden ratio--using any finite sequence of operations consisting of addition, subtraction, multiplication, division, square roots, cube roots, and/or higher-order roots applied to integers--is a subset of the algebraic numbers.



    Transcendental numbers are real numbers that are not algebraic numbers.
    It can be shown that $pi$ and $e$ are transcendental;
    for example, see How hard is the proof of $pi$ or $e$ being transcendental?.



    Hence neither $pi$ nor $e$ can be written algebraically.
    So the answer is "no."






    share|cite|improve this answer























    • +1 for bringing up the concept of "written algebraically". However maybe we should define what "written algebraically" and explain to the OP that this is actually what s/he was asking. After all. We can always represent $pi$ and $e$ as.... $pi$ and $e$.
      – fleablood
      yesterday










    • Never mind. I see the OP has editted the post to mean "written algebraically"
      – fleablood
      yesterday














    2












    2








    2






    To summarize (some) of the comments:



    The numbers that you can write algebraically, like the golden ratio--using any finite sequence of operations consisting of addition, subtraction, multiplication, division, square roots, cube roots, and/or higher-order roots applied to integers--is a subset of the algebraic numbers.



    Transcendental numbers are real numbers that are not algebraic numbers.
    It can be shown that $pi$ and $e$ are transcendental;
    for example, see How hard is the proof of $pi$ or $e$ being transcendental?.



    Hence neither $pi$ nor $e$ can be written algebraically.
    So the answer is "no."






    share|cite|improve this answer














    To summarize (some) of the comments:



    The numbers that you can write algebraically, like the golden ratio--using any finite sequence of operations consisting of addition, subtraction, multiplication, division, square roots, cube roots, and/or higher-order roots applied to integers--is a subset of the algebraic numbers.



    Transcendental numbers are real numbers that are not algebraic numbers.
    It can be shown that $pi$ and $e$ are transcendental;
    for example, see How hard is the proof of $pi$ or $e$ being transcendental?.



    Hence neither $pi$ nor $e$ can be written algebraically.
    So the answer is "no."







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    answered yesterday


























    community wiki





    David K













    • +1 for bringing up the concept of "written algebraically". However maybe we should define what "written algebraically" and explain to the OP that this is actually what s/he was asking. After all. We can always represent $pi$ and $e$ as.... $pi$ and $e$.
      – fleablood
      yesterday










    • Never mind. I see the OP has editted the post to mean "written algebraically"
      – fleablood
      yesterday


















    • +1 for bringing up the concept of "written algebraically". However maybe we should define what "written algebraically" and explain to the OP that this is actually what s/he was asking. After all. We can always represent $pi$ and $e$ as.... $pi$ and $e$.
      – fleablood
      yesterday










    • Never mind. I see the OP has editted the post to mean "written algebraically"
      – fleablood
      yesterday
















    +1 for bringing up the concept of "written algebraically". However maybe we should define what "written algebraically" and explain to the OP that this is actually what s/he was asking. After all. We can always represent $pi$ and $e$ as.... $pi$ and $e$.
    – fleablood
    yesterday




    +1 for bringing up the concept of "written algebraically". However maybe we should define what "written algebraically" and explain to the OP that this is actually what s/he was asking. After all. We can always represent $pi$ and $e$ as.... $pi$ and $e$.
    – fleablood
    yesterday












    Never mind. I see the OP has editted the post to mean "written algebraically"
    – fleablood
    yesterday




    Never mind. I see the OP has editted the post to mean "written algebraically"
    – fleablood
    yesterday











    2














    Yes. You can represent $e$ in one symbol "$e$" and you can represent $pi$ in one symbol by using the single symbol "$pi$".



    However you can't express them as integers because.... they are not integers. You can't express $frac 35$ as an integer either.



    You can't express them as a terminating decimal because they are not an integer scaled to a power of $10$.



    Nor can you express them as a ratio between two finite integers because the are not rational.



    Only difference between them and numbers such as $sqrt 2$ or the golden ratio which are also not rational, is that they are not roots to any polynomial with integer coefficients. $sqrt 2$ is the one of the roots to $x^2 -2 = 0$ and the golden ratio is solution to $x^2 - 2x -4= 0$. Neither $pi$ nor $e$ are a solution to any polynomials with integer coefficients.



    Numbers that are solutions to polynomials are called algebraic numbers and the irrational algebraic numbers can't be expressed as finite decimals. But if we invent the notation, and we are mathematicians--- we can invent anything we like as like as it makes sense---, $sqrt[k]{m}$ to mean a $k$-th root of $m$, then we can express any algebraic number as some combination of roots.



    Numbers, such as $e$ and $pi$ which are not algebraic are called transcendental. And, no, they can not be expressed via a digital decimal system. Nor can they be expressed as a series of root signs.



    But so what? Using digital decimals to represent numbers is completely arbitrary and does not in any way make a number more "real" than any other. (The main reason we use digital decimals is because the are convergent and thus every thing can be approximate and accurately expressed through and infinite number of decimals if necessary.)



    But to "represent" a number means nothing more than having a symbol for it. And if a number can by defined --- in a way that makes sense so that we know such a number exists--- we can define any symbol we want for them.



    So I stand by my answer. You can represent $e$ and $pi$ with one symbol each: "$e$" and "$pi$"






    share|cite|improve this answer


























      2














      Yes. You can represent $e$ in one symbol "$e$" and you can represent $pi$ in one symbol by using the single symbol "$pi$".



      However you can't express them as integers because.... they are not integers. You can't express $frac 35$ as an integer either.



      You can't express them as a terminating decimal because they are not an integer scaled to a power of $10$.



      Nor can you express them as a ratio between two finite integers because the are not rational.



      Only difference between them and numbers such as $sqrt 2$ or the golden ratio which are also not rational, is that they are not roots to any polynomial with integer coefficients. $sqrt 2$ is the one of the roots to $x^2 -2 = 0$ and the golden ratio is solution to $x^2 - 2x -4= 0$. Neither $pi$ nor $e$ are a solution to any polynomials with integer coefficients.



      Numbers that are solutions to polynomials are called algebraic numbers and the irrational algebraic numbers can't be expressed as finite decimals. But if we invent the notation, and we are mathematicians--- we can invent anything we like as like as it makes sense---, $sqrt[k]{m}$ to mean a $k$-th root of $m$, then we can express any algebraic number as some combination of roots.



      Numbers, such as $e$ and $pi$ which are not algebraic are called transcendental. And, no, they can not be expressed via a digital decimal system. Nor can they be expressed as a series of root signs.



      But so what? Using digital decimals to represent numbers is completely arbitrary and does not in any way make a number more "real" than any other. (The main reason we use digital decimals is because the are convergent and thus every thing can be approximate and accurately expressed through and infinite number of decimals if necessary.)



      But to "represent" a number means nothing more than having a symbol for it. And if a number can by defined --- in a way that makes sense so that we know such a number exists--- we can define any symbol we want for them.



      So I stand by my answer. You can represent $e$ and $pi$ with one symbol each: "$e$" and "$pi$"






      share|cite|improve this answer
























        2












        2








        2






        Yes. You can represent $e$ in one symbol "$e$" and you can represent $pi$ in one symbol by using the single symbol "$pi$".



        However you can't express them as integers because.... they are not integers. You can't express $frac 35$ as an integer either.



        You can't express them as a terminating decimal because they are not an integer scaled to a power of $10$.



        Nor can you express them as a ratio between two finite integers because the are not rational.



        Only difference between them and numbers such as $sqrt 2$ or the golden ratio which are also not rational, is that they are not roots to any polynomial with integer coefficients. $sqrt 2$ is the one of the roots to $x^2 -2 = 0$ and the golden ratio is solution to $x^2 - 2x -4= 0$. Neither $pi$ nor $e$ are a solution to any polynomials with integer coefficients.



        Numbers that are solutions to polynomials are called algebraic numbers and the irrational algebraic numbers can't be expressed as finite decimals. But if we invent the notation, and we are mathematicians--- we can invent anything we like as like as it makes sense---, $sqrt[k]{m}$ to mean a $k$-th root of $m$, then we can express any algebraic number as some combination of roots.



        Numbers, such as $e$ and $pi$ which are not algebraic are called transcendental. And, no, they can not be expressed via a digital decimal system. Nor can they be expressed as a series of root signs.



        But so what? Using digital decimals to represent numbers is completely arbitrary and does not in any way make a number more "real" than any other. (The main reason we use digital decimals is because the are convergent and thus every thing can be approximate and accurately expressed through and infinite number of decimals if necessary.)



        But to "represent" a number means nothing more than having a symbol for it. And if a number can by defined --- in a way that makes sense so that we know such a number exists--- we can define any symbol we want for them.



        So I stand by my answer. You can represent $e$ and $pi$ with one symbol each: "$e$" and "$pi$"






        share|cite|improve this answer












        Yes. You can represent $e$ in one symbol "$e$" and you can represent $pi$ in one symbol by using the single symbol "$pi$".



        However you can't express them as integers because.... they are not integers. You can't express $frac 35$ as an integer either.



        You can't express them as a terminating decimal because they are not an integer scaled to a power of $10$.



        Nor can you express them as a ratio between two finite integers because the are not rational.



        Only difference between them and numbers such as $sqrt 2$ or the golden ratio which are also not rational, is that they are not roots to any polynomial with integer coefficients. $sqrt 2$ is the one of the roots to $x^2 -2 = 0$ and the golden ratio is solution to $x^2 - 2x -4= 0$. Neither $pi$ nor $e$ are a solution to any polynomials with integer coefficients.



        Numbers that are solutions to polynomials are called algebraic numbers and the irrational algebraic numbers can't be expressed as finite decimals. But if we invent the notation, and we are mathematicians--- we can invent anything we like as like as it makes sense---, $sqrt[k]{m}$ to mean a $k$-th root of $m$, then we can express any algebraic number as some combination of roots.



        Numbers, such as $e$ and $pi$ which are not algebraic are called transcendental. And, no, they can not be expressed via a digital decimal system. Nor can they be expressed as a series of root signs.



        But so what? Using digital decimals to represent numbers is completely arbitrary and does not in any way make a number more "real" than any other. (The main reason we use digital decimals is because the are convergent and thus every thing can be approximate and accurately expressed through and infinite number of decimals if necessary.)



        But to "represent" a number means nothing more than having a symbol for it. And if a number can by defined --- in a way that makes sense so that we know such a number exists--- we can define any symbol we want for them.



        So I stand by my answer. You can represent $e$ and $pi$ with one symbol each: "$e$" and "$pi$"







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        fleablood

        68.4k22685




        68.4k22685















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