Is a number written in the square root/fraction form called a non-integer even if it can be simplified to an...
This is very simple question, but I cannot get the ansewer from the internet.
Is a number written in the square root/fraction form called a non-integer even if it can be simplified to an integer.
For example 4/2, 12/4, sqr4, sqr64 etc... do these need to be simplified before we can call them integers.
Too make this easer to understand are sqr64 and 12/4 non-integers while 8 and 3 are integers.
square-numbers
closed as unclear what you're asking by Lord Shark the Unknown, KReiser, Cesareo, Leucippus, Rebellos Nov 28 '18 at 9:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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This is very simple question, but I cannot get the ansewer from the internet.
Is a number written in the square root/fraction form called a non-integer even if it can be simplified to an integer.
For example 4/2, 12/4, sqr4, sqr64 etc... do these need to be simplified before we can call them integers.
Too make this easer to understand are sqr64 and 12/4 non-integers while 8 and 3 are integers.
square-numbers
closed as unclear what you're asking by Lord Shark the Unknown, KReiser, Cesareo, Leucippus, Rebellos Nov 28 '18 at 9:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
– lulu
Nov 27 '18 at 21:05
Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
– mathnoob
Nov 27 '18 at 21:09
Thank you to all.
– Asan Ramzan
Nov 27 '18 at 21:26
I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
– Asan Ramzan
Nov 28 '18 at 12:03
@AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
– Théophile
Nov 28 '18 at 16:45
|
show 1 more comment
This is very simple question, but I cannot get the ansewer from the internet.
Is a number written in the square root/fraction form called a non-integer even if it can be simplified to an integer.
For example 4/2, 12/4, sqr4, sqr64 etc... do these need to be simplified before we can call them integers.
Too make this easer to understand are sqr64 and 12/4 non-integers while 8 and 3 are integers.
square-numbers
This is very simple question, but I cannot get the ansewer from the internet.
Is a number written in the square root/fraction form called a non-integer even if it can be simplified to an integer.
For example 4/2, 12/4, sqr4, sqr64 etc... do these need to be simplified before we can call them integers.
Too make this easer to understand are sqr64 and 12/4 non-integers while 8 and 3 are integers.
square-numbers
square-numbers
edited Nov 28 '18 at 16:49
Asan Ramzan
asked Nov 27 '18 at 21:03
Asan RamzanAsan Ramzan
162
162
closed as unclear what you're asking by Lord Shark the Unknown, KReiser, Cesareo, Leucippus, Rebellos Nov 28 '18 at 9:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Lord Shark the Unknown, KReiser, Cesareo, Leucippus, Rebellos Nov 28 '18 at 9:27
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
– lulu
Nov 27 '18 at 21:05
Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
– mathnoob
Nov 27 '18 at 21:09
Thank you to all.
– Asan Ramzan
Nov 27 '18 at 21:26
I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
– Asan Ramzan
Nov 28 '18 at 12:03
@AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
– Théophile
Nov 28 '18 at 16:45
|
show 1 more comment
1
$frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
– lulu
Nov 27 '18 at 21:05
Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
– mathnoob
Nov 27 '18 at 21:09
Thank you to all.
– Asan Ramzan
Nov 27 '18 at 21:26
I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
– Asan Ramzan
Nov 28 '18 at 12:03
@AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
– Théophile
Nov 28 '18 at 16:45
1
1
$frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
– lulu
Nov 27 '18 at 21:05
$frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
– lulu
Nov 27 '18 at 21:05
Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
– mathnoob
Nov 27 '18 at 21:09
Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
– mathnoob
Nov 27 '18 at 21:09
Thank you to all.
– Asan Ramzan
Nov 27 '18 at 21:26
Thank you to all.
– Asan Ramzan
Nov 27 '18 at 21:26
I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
– Asan Ramzan
Nov 28 '18 at 12:03
I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
– Asan Ramzan
Nov 28 '18 at 12:03
@AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
– Théophile
Nov 28 '18 at 16:45
@AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
– Théophile
Nov 28 '18 at 16:45
|
show 1 more comment
4 Answers
4
active
oldest
votes
No. Numbers are what they are. It doesn't matter how they are represented.
$7$ is an integer. Period.
It doesn't matter if is written as $5 + 2$ or $sqrt{49}$ or $sqrt{25} + frac{sqrt[3]{16}}{2^{frac 13}}$ or $ln (e^7)$.
Those are all equal to $7$ and $7$ is an integer. Period.
====
That said, it might not be easy (or even possible) to tell if a number is or is not an integer. It's obvious that $7$ is an integer and $7.0000012142650469991421281354411.....$ isn't. But it isn't clear whether $sqrt[7]{823543}$ or $sqrt[7]{823544}$ are integers. (It turns out that those are the same numbers.)
But it doesn't matter whether we know if a number is an integer or not. It either is or isn't.
add a comment |
It is helpful to have vocabulary to distinguish between what an object fundamentally is and how it is represented.
A fraction $a/b$ is said to be irreducible or in simplest form if $a$ and $b$ are integers with no common factors, i.e., $gcd(a,b) = 1$. So, for example, $2/3$ is irreducible, while $200/300$ is not.
An integer can be written down in many ways. Not all of those ways are simple, but fundamentally the number is the same. Take $2$, for instance. We could write it as $10/5$: this is not the simplest form of the fraction, but the number is still an integer. Or we could express it as
$$1 + 1/2 + 1/4 + 1/8 + cdots;$$
again, this is not as straightforward as simply writing "$2$", but the expression still represents an integer.
add a comment |
$2$ is an integer. $4/2$ is equal to $2$, and therefore has all the properties that the number $2$ has, including being an integer. The square root of $4$ is also equal to $2$, so it's an integer as well. In some cases, you'll probably need to simplify to recognize that it is indeed an integer, but that doesn't change its properties no matter how you write it.
For example, is $sqrt{14883}$ an integer? How about $sqrt{14884}$? It might be tough to tell unless you do the simplification, but one is an integer and one isn't.
add a comment |
If it can be simplified to an integer,
it can be called an integer
after the simplification.
Until the simplification is done,
I would just call the expression
"an expression"
when it is not clear if it could be
simplified to an integer.
Considering how expressions
involving nested radicals
can be sometimes
amazingly simplified,
I think that there would be cases
where the fact that
an expression simplifies
to an integer
is a surprise.
I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
– fleablood
Nov 27 '18 at 21:22
Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
– marty cohen
Nov 27 '18 at 23:38
1
" Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
– fleablood
Nov 28 '18 at 0:04
No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
– marty cohen
Nov 28 '18 at 3:47
1
Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
– fleablood
Nov 28 '18 at 5:38
|
show 1 more comment
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
No. Numbers are what they are. It doesn't matter how they are represented.
$7$ is an integer. Period.
It doesn't matter if is written as $5 + 2$ or $sqrt{49}$ or $sqrt{25} + frac{sqrt[3]{16}}{2^{frac 13}}$ or $ln (e^7)$.
Those are all equal to $7$ and $7$ is an integer. Period.
====
That said, it might not be easy (or even possible) to tell if a number is or is not an integer. It's obvious that $7$ is an integer and $7.0000012142650469991421281354411.....$ isn't. But it isn't clear whether $sqrt[7]{823543}$ or $sqrt[7]{823544}$ are integers. (It turns out that those are the same numbers.)
But it doesn't matter whether we know if a number is an integer or not. It either is or isn't.
add a comment |
No. Numbers are what they are. It doesn't matter how they are represented.
$7$ is an integer. Period.
It doesn't matter if is written as $5 + 2$ or $sqrt{49}$ or $sqrt{25} + frac{sqrt[3]{16}}{2^{frac 13}}$ or $ln (e^7)$.
Those are all equal to $7$ and $7$ is an integer. Period.
====
That said, it might not be easy (or even possible) to tell if a number is or is not an integer. It's obvious that $7$ is an integer and $7.0000012142650469991421281354411.....$ isn't. But it isn't clear whether $sqrt[7]{823543}$ or $sqrt[7]{823544}$ are integers. (It turns out that those are the same numbers.)
But it doesn't matter whether we know if a number is an integer or not. It either is or isn't.
add a comment |
No. Numbers are what they are. It doesn't matter how they are represented.
$7$ is an integer. Period.
It doesn't matter if is written as $5 + 2$ or $sqrt{49}$ or $sqrt{25} + frac{sqrt[3]{16}}{2^{frac 13}}$ or $ln (e^7)$.
Those are all equal to $7$ and $7$ is an integer. Period.
====
That said, it might not be easy (or even possible) to tell if a number is or is not an integer. It's obvious that $7$ is an integer and $7.0000012142650469991421281354411.....$ isn't. But it isn't clear whether $sqrt[7]{823543}$ or $sqrt[7]{823544}$ are integers. (It turns out that those are the same numbers.)
But it doesn't matter whether we know if a number is an integer or not. It either is or isn't.
No. Numbers are what they are. It doesn't matter how they are represented.
$7$ is an integer. Period.
It doesn't matter if is written as $5 + 2$ or $sqrt{49}$ or $sqrt{25} + frac{sqrt[3]{16}}{2^{frac 13}}$ or $ln (e^7)$.
Those are all equal to $7$ and $7$ is an integer. Period.
====
That said, it might not be easy (or even possible) to tell if a number is or is not an integer. It's obvious that $7$ is an integer and $7.0000012142650469991421281354411.....$ isn't. But it isn't clear whether $sqrt[7]{823543}$ or $sqrt[7]{823544}$ are integers. (It turns out that those are the same numbers.)
But it doesn't matter whether we know if a number is an integer or not. It either is or isn't.
edited Nov 27 '18 at 21:27
answered Nov 27 '18 at 21:20
fleabloodfleablood
68.7k22685
68.7k22685
add a comment |
add a comment |
It is helpful to have vocabulary to distinguish between what an object fundamentally is and how it is represented.
A fraction $a/b$ is said to be irreducible or in simplest form if $a$ and $b$ are integers with no common factors, i.e., $gcd(a,b) = 1$. So, for example, $2/3$ is irreducible, while $200/300$ is not.
An integer can be written down in many ways. Not all of those ways are simple, but fundamentally the number is the same. Take $2$, for instance. We could write it as $10/5$: this is not the simplest form of the fraction, but the number is still an integer. Or we could express it as
$$1 + 1/2 + 1/4 + 1/8 + cdots;$$
again, this is not as straightforward as simply writing "$2$", but the expression still represents an integer.
add a comment |
It is helpful to have vocabulary to distinguish between what an object fundamentally is and how it is represented.
A fraction $a/b$ is said to be irreducible or in simplest form if $a$ and $b$ are integers with no common factors, i.e., $gcd(a,b) = 1$. So, for example, $2/3$ is irreducible, while $200/300$ is not.
An integer can be written down in many ways. Not all of those ways are simple, but fundamentally the number is the same. Take $2$, for instance. We could write it as $10/5$: this is not the simplest form of the fraction, but the number is still an integer. Or we could express it as
$$1 + 1/2 + 1/4 + 1/8 + cdots;$$
again, this is not as straightforward as simply writing "$2$", but the expression still represents an integer.
add a comment |
It is helpful to have vocabulary to distinguish between what an object fundamentally is and how it is represented.
A fraction $a/b$ is said to be irreducible or in simplest form if $a$ and $b$ are integers with no common factors, i.e., $gcd(a,b) = 1$. So, for example, $2/3$ is irreducible, while $200/300$ is not.
An integer can be written down in many ways. Not all of those ways are simple, but fundamentally the number is the same. Take $2$, for instance. We could write it as $10/5$: this is not the simplest form of the fraction, but the number is still an integer. Or we could express it as
$$1 + 1/2 + 1/4 + 1/8 + cdots;$$
again, this is not as straightforward as simply writing "$2$", but the expression still represents an integer.
It is helpful to have vocabulary to distinguish between what an object fundamentally is and how it is represented.
A fraction $a/b$ is said to be irreducible or in simplest form if $a$ and $b$ are integers with no common factors, i.e., $gcd(a,b) = 1$. So, for example, $2/3$ is irreducible, while $200/300$ is not.
An integer can be written down in many ways. Not all of those ways are simple, but fundamentally the number is the same. Take $2$, for instance. We could write it as $10/5$: this is not the simplest form of the fraction, but the number is still an integer. Or we could express it as
$$1 + 1/2 + 1/4 + 1/8 + cdots;$$
again, this is not as straightforward as simply writing "$2$", but the expression still represents an integer.
answered Nov 27 '18 at 21:16
ThéophileThéophile
19.5k12946
19.5k12946
add a comment |
add a comment |
$2$ is an integer. $4/2$ is equal to $2$, and therefore has all the properties that the number $2$ has, including being an integer. The square root of $4$ is also equal to $2$, so it's an integer as well. In some cases, you'll probably need to simplify to recognize that it is indeed an integer, but that doesn't change its properties no matter how you write it.
For example, is $sqrt{14883}$ an integer? How about $sqrt{14884}$? It might be tough to tell unless you do the simplification, but one is an integer and one isn't.
add a comment |
$2$ is an integer. $4/2$ is equal to $2$, and therefore has all the properties that the number $2$ has, including being an integer. The square root of $4$ is also equal to $2$, so it's an integer as well. In some cases, you'll probably need to simplify to recognize that it is indeed an integer, but that doesn't change its properties no matter how you write it.
For example, is $sqrt{14883}$ an integer? How about $sqrt{14884}$? It might be tough to tell unless you do the simplification, but one is an integer and one isn't.
add a comment |
$2$ is an integer. $4/2$ is equal to $2$, and therefore has all the properties that the number $2$ has, including being an integer. The square root of $4$ is also equal to $2$, so it's an integer as well. In some cases, you'll probably need to simplify to recognize that it is indeed an integer, but that doesn't change its properties no matter how you write it.
For example, is $sqrt{14883}$ an integer? How about $sqrt{14884}$? It might be tough to tell unless you do the simplification, but one is an integer and one isn't.
$2$ is an integer. $4/2$ is equal to $2$, and therefore has all the properties that the number $2$ has, including being an integer. The square root of $4$ is also equal to $2$, so it's an integer as well. In some cases, you'll probably need to simplify to recognize that it is indeed an integer, but that doesn't change its properties no matter how you write it.
For example, is $sqrt{14883}$ an integer? How about $sqrt{14884}$? It might be tough to tell unless you do the simplification, but one is an integer and one isn't.
edited Nov 27 '18 at 21:17
Théophile
19.5k12946
19.5k12946
answered Nov 27 '18 at 21:12
Nuclear WangNuclear Wang
41127
41127
add a comment |
add a comment |
If it can be simplified to an integer,
it can be called an integer
after the simplification.
Until the simplification is done,
I would just call the expression
"an expression"
when it is not clear if it could be
simplified to an integer.
Considering how expressions
involving nested radicals
can be sometimes
amazingly simplified,
I think that there would be cases
where the fact that
an expression simplifies
to an integer
is a surprise.
I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
– fleablood
Nov 27 '18 at 21:22
Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
– marty cohen
Nov 27 '18 at 23:38
1
" Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
– fleablood
Nov 28 '18 at 0:04
No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
– marty cohen
Nov 28 '18 at 3:47
1
Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
– fleablood
Nov 28 '18 at 5:38
|
show 1 more comment
If it can be simplified to an integer,
it can be called an integer
after the simplification.
Until the simplification is done,
I would just call the expression
"an expression"
when it is not clear if it could be
simplified to an integer.
Considering how expressions
involving nested radicals
can be sometimes
amazingly simplified,
I think that there would be cases
where the fact that
an expression simplifies
to an integer
is a surprise.
I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
– fleablood
Nov 27 '18 at 21:22
Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
– marty cohen
Nov 27 '18 at 23:38
1
" Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
– fleablood
Nov 28 '18 at 0:04
No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
– marty cohen
Nov 28 '18 at 3:47
1
Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
– fleablood
Nov 28 '18 at 5:38
|
show 1 more comment
If it can be simplified to an integer,
it can be called an integer
after the simplification.
Until the simplification is done,
I would just call the expression
"an expression"
when it is not clear if it could be
simplified to an integer.
Considering how expressions
involving nested radicals
can be sometimes
amazingly simplified,
I think that there would be cases
where the fact that
an expression simplifies
to an integer
is a surprise.
If it can be simplified to an integer,
it can be called an integer
after the simplification.
Until the simplification is done,
I would just call the expression
"an expression"
when it is not clear if it could be
simplified to an integer.
Considering how expressions
involving nested radicals
can be sometimes
amazingly simplified,
I think that there would be cases
where the fact that
an expression simplifies
to an integer
is a surprise.
answered Nov 27 '18 at 21:10
marty cohenmarty cohen
72.8k549128
72.8k549128
I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
– fleablood
Nov 27 '18 at 21:22
Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
– marty cohen
Nov 27 '18 at 23:38
1
" Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
– fleablood
Nov 28 '18 at 0:04
No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
– marty cohen
Nov 28 '18 at 3:47
1
Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
– fleablood
Nov 28 '18 at 5:38
|
show 1 more comment
I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
– fleablood
Nov 27 '18 at 21:22
Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
– marty cohen
Nov 27 '18 at 23:38
1
" Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
– fleablood
Nov 28 '18 at 0:04
No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
– marty cohen
Nov 28 '18 at 3:47
1
Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
– fleablood
Nov 28 '18 at 5:38
I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
– fleablood
Nov 27 '18 at 21:22
I disagree. The number either is or is not an integer whether we know whether it is or is not. So it IS an integer before we simplify it. It's just that before we simplify it we don't KNOW if it is an integer. But that doesn't mean it isn't an integer. (because ... it is.)
– fleablood
Nov 27 '18 at 21:22
Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
– marty cohen
Nov 27 '18 at 23:38
Suppose the expression encodes something we don't know the answer for - for example, 1 if $pi$ is a normal number, 1/2 if it is not. Is the value of the expression an integer?
– marty cohen
Nov 27 '18 at 23:38
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1
" Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
– fleablood
Nov 28 '18 at 0:04
" Is the value of the expression an integer?" If $pi$ is normal then it is an integer. If $pi$ is not normal it is not an integer. The fact that we don't know if it's an integer or not doesn't prevent it from being an integer or a non-integer.
– fleablood
Nov 28 '18 at 0:04
No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
– marty cohen
Nov 28 '18 at 3:47
No. A normal number is a real number: en.wikipedia.org/wiki/Normal_number
– marty cohen
Nov 28 '18 at 3:47
1
1
Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
– fleablood
Nov 28 '18 at 5:38
Um.... What are you talking about??? Of course normal numbers are real numbers. Everything that isnt complex is a real number. You asked if given the function $f(x) = 1$ if $x$ is normal; $f(x)$ if $x$ is not normal. You asked whether $f(pi)$ is an integer (which it is if it is $1$) or if it was not an integer (which it will not be if it is $frac 12$). And the answer is: We don't know. But EITHER $f(pi)$ is an integer OR it is not. We can NOT say $f(pi)$ is a non-integer, just because we don't know that it is an integer. We don't know it is not.
– fleablood
Nov 28 '18 at 5:38
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$frac 42=2$ so that's an integer, however you have written it. Similarly for the others.
– lulu
Nov 27 '18 at 21:05
Integers is a subset of Rational numbers, Of course an integer is also a rational number. So the point is no!
– mathnoob
Nov 27 '18 at 21:09
Thank you to all.
– Asan Ramzan
Nov 27 '18 at 21:26
I know this put on hold because its unclear, but everyone who has answered this question seems to understand what I am saying, so I don't see where the problem is. Could you elaborate
– Asan Ramzan
Nov 28 '18 at 12:03
@AsanRamzan I'm surprised the question was put on hold; I've voted to reopen it. Do you feel you have a better understanding after reading the answers?
– Théophile
Nov 28 '18 at 16:45