Is it possible to represent $e=2.71ldots$ and $pi=3.14ldots$ algebraicially, without using infinity or...
$begingroup$
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
edited Jan 4 at 23:50
Blue
49.5k870157
asked Jan 4 at 23:20
Math LoverMath Lover
17410
$endgroup$
closed as off-topic by jgon, Cesareo, Leucippus, user91500, Saad Jan 5 at 8:55
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.
|
show 9 more comments
$begingroup$
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
edited Jan 4 at 23:50
Blue
49.5k870157
asked Jan 4 at 23:20
Math LoverMath Lover
17410
$endgroup$
closed as off-topic by jgon, Cesareo, Leucippus, user91500, Saad Jan 5 at 8:55
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
$begingroup$
Yes, use the symbols “$e$” and “$pi$” ;)
$endgroup$
– MPW
Jan 4 at 23:23
5
$begingroup$
@Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
$endgroup$
– Cheerful Parsnip
Jan 4 at 23:25
2
$begingroup$
Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
$endgroup$
– SmileyCraft
Jan 4 at 23:25
1
$begingroup$
This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
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– Noble Mushtak
Jan 4 at 23:29
3
$begingroup$
@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 $pi$.
$endgroup$
– JMoravitz
Jan 4 at 23:41
|
show 9 more comments
$begingroup$
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
edited Jan 4 at 23:50
Blue
49.5k870157
asked Jan 4 at 23:20
Math LoverMath Lover
17410
$endgroup$
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
number-theory
edited Jan 4 at 23:50
Blue
49.5k870157
asked Jan 4 at 23:20
Math LoverMath Lover
17410
edited Jan 4 at 23:50
Blue
49.5k870157
asked Jan 4 at 23:20
Math LoverMath Lover
17410
edited Jan 4 at 23:50
Blue
49.5k870157
edited Jan 4 at 23:50
Blue
49.5k870157
edited Jan 4 at 23:50
Blue
49.5k870157
49.5k870157
asked Jan 4 at 23:20
Math LoverMath Lover
17410
asked Jan 4 at 23:20
Math LoverMath Lover
17410
asked Jan 4 at 23:20
Math LoverMath Lover
17410
17410
closed as off-topic by jgon, Cesareo, Leucippus, user91500, Saad Jan 5 at 8:55
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.
closed as off-topic by jgon, Cesareo, Leucippus, user91500, Saad Jan 5 at 8:55
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
$begingroup$
Yes, use the symbols “$e$” and “$pi$” ;)
$endgroup$
– MPW
Jan 4 at 23:23
5
$begingroup$
@Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
$endgroup$
– Cheerful Parsnip
Jan 4 at 23:25
2
$begingroup$
Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
$endgroup$
– SmileyCraft
Jan 4 at 23:25
1
$begingroup$
This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
$endgroup$
– Noble Mushtak
Jan 4 at 23:29
3
$begingroup$
@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 $pi$.
$endgroup$
– JMoravitz
Jan 4 at 23:41
|
show 9 more comments
6
$begingroup$
Yes, use the symbols “$e$” and “$pi$” ;)
$endgroup$
– MPW
Jan 4 at 23:23
5
$begingroup$
@Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
$endgroup$
– Cheerful Parsnip
Jan 4 at 23:25
2
$begingroup$
Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
$endgroup$
– SmileyCraft
Jan 4 at 23:25
1
$begingroup$
This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
$endgroup$
– Noble Mushtak
Jan 4 at 23:29
3
$begingroup$
@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 $pi$.
$endgroup$
– JMoravitz
Jan 4 at 23:41
6
6
$begingroup$
Yes, use the symbols “$e$” and “$pi$” ;)
$endgroup$
– MPW
Jan 4 at 23:23
$begingroup$
Yes, use the symbols “$e$” and “$pi$” ;)
$endgroup$
– MPW
Jan 4 at 23:23
5
5
$begingroup$
@Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
$endgroup$
– Cheerful Parsnip
Jan 4 at 23:25
$begingroup$
@Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
$endgroup$
– Cheerful Parsnip
Jan 4 at 23:25
2
2
$begingroup$
Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
$endgroup$
– SmileyCraft
Jan 4 at 23:25
$begingroup$
Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
$endgroup$
– SmileyCraft
Jan 4 at 23:25
1
1
$begingroup$
This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
$endgroup$
– Noble Mushtak
Jan 4 at 23:29
$begingroup$
This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
$endgroup$
– Noble Mushtak
Jan 4 at 23:29
3
3
$begingroup$
@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 $pi$.
$endgroup$
– JMoravitz
Jan 4 at 23:41
$begingroup$
@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 $pi$.
$endgroup$
– JMoravitz
Jan 4 at 23:41
|
show 9 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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."
$endgroup$
$begingroup$
+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 $e$.
$endgroup$
– fleablood
Jan 5 at 0:04
$begingroup$
Never mind. I see the OP has editted the post to mean "written algebraically"
$endgroup$
– fleablood
Jan 5 at 0:07
add a comment |
$begingroup$
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 "$pi$"
answered Jan 4 at 23:49
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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."
$endgroup$
$begingroup$
+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 $e$.
$endgroup$
– fleablood
Jan 5 at 0:04
$begingroup$
Never mind. I see the OP has editted the post to mean "written algebraically"
$endgroup$
– fleablood
Jan 5 at 0:07
add a comment |
$begingroup$
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."
$endgroup$
$begingroup$
+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 $e$.
$endgroup$
– fleablood
Jan 5 at 0:04
$begingroup$
Never mind. I see the OP has editted the post to mean "written algebraically"
$endgroup$
– fleablood
Jan 5 at 0:07
add a comment |
$begingroup$
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."
$endgroup$
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."
answered Jan 4 at 23:58
community wiki
David K
$begingroup$
+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 $e$.
$endgroup$
– fleablood
Jan 5 at 0:04
$begingroup$
Never mind. I see the OP has editted the post to mean "written algebraically"
$endgroup$
– fleablood
Jan 5 at 0:07
add a comment |
$begingroup$
+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 $e$.
$endgroup$
– fleablood
Jan 5 at 0:04
$begingroup$
Never mind. I see the OP has editted the post to mean "written algebraically"
$endgroup$
– fleablood
Jan 5 at 0:07
$begingroup$
+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 $e$.
$endgroup$
– fleablood
Jan 5 at 0:04
$begingroup$
+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 $e$.
$endgroup$
– fleablood
Jan 5 at 0:04
$begingroup$
Never mind. I see the OP has editted the post to mean "written algebraically"
$endgroup$
– fleablood
Jan 5 at 0:07
$begingroup$
Never mind. I see the OP has editted the post to mean "written algebraically"
$endgroup$
– fleablood
Jan 5 at 0:07
add a comment |
$begingroup$
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 "$pi$"
answered Jan 4 at 23:49
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
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 "$pi$"
answered Jan 4 at 23:49
fleabloodfleablood
73.8k22891
$endgroup$
add a comment |
$begingroup$
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 "$pi$"
answered Jan 4 at 23:49
fleabloodfleablood
73.8k22891
$endgroup$
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 "$pi$"
answered Jan 4 at 23:49
fleabloodfleablood
73.8k22891
answered Jan 4 at 23:49
fleabloodfleablood
73.8k22891
answered Jan 4 at 23:49
fleabloodfleablood
73.8k22891
answered Jan 4 at 23:49
fleabloodfleablood
73.8k22891
73.8k22891
add a comment |
add a comment |
6
$begingroup$
Yes, use the symbols “$e$” and “$pi$” ;)
$endgroup$
– MPW
Jan 4 at 23:23
5
$begingroup$
@Math Lover: You may be interested in learning about transcendental numbers. Neither $e$ nor $pi$ can be written using radicals.
$endgroup$
– Cheerful Parsnip
Jan 4 at 23:25
2
$begingroup$
Not really, since $pi$ and $e$ are transcendental en.wikipedia.org/wiki/Transcendental_number
$endgroup$
– SmileyCraft
Jan 4 at 23:25
1
$begingroup$
This article proves both $e$ and $pi$ are transcendental: sixthform.info/maths/files/pitrans.pdf
$endgroup$
– Noble Mushtak
Jan 4 at 23:29
3
$begingroup$
@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 $pi$.
$endgroup$
– JMoravitz
Jan 4 at 23:41