Mixing
Rules for factors of a composite number
0
Is there any rule that the factors of a composite number must be of the form $n^m$, where $n$ is a real number and $m ge 1$?
Example 1: the factors of $4$ are : $2^2, 1^1, 4^1$.
Example 2: The factors of $-4$ are $-2^2, -4^1, -1^1$, and so on.
real-numbers prime-factorization
edited Nov 22 '18 at 13:38
Klangen
1,65711334
asked Jun 12 '18 at 6:44
pro neonpro neon
1305
|
show 2 more comments
0
Is there any rule that the factors of a composite number must be of the form $n^m$, where $n$ is a real number and $m ge 1$?
Example 1: the factors of $4$ are : $2^2, 1^1, 4^1$.
Example 2: The factors of $-4$ are $-2^2, -4^1, -1^1$, and so on.
real-numbers prime-factorization
edited Nov 22 '18 at 13:38
Klangen
1,65711334
asked Jun 12 '18 at 6:44
pro neonpro neon
1305
I'm not sure I completely understand your question. Can you give us a couple of examples of what you mean?
– Billy
Jun 12 '18 at 6:45
@Billy I have edited!
– pro neon
Jun 12 '18 at 6:53
Well, first of all, since we're talking about factors, $n$ should be an integer, not a real number. Second, when you say the factors of $4$ are $2^2,1^1,4^1$, what does that really mean? The prime factorisation of $4$ is $2cdot2$, which can also be written as $2^2$ if you want (but it's not necessary). The factors of $4$ are $1,2$ and $4$. I don't know what else you could mean.
– Arthur
Jun 12 '18 at 7:15
If $n$ is a factor, you can always write it in the form $n^m$ by setting $m=1$. This doesn't seem to get you anywhere, though.
– saulspatz
Jun 12 '18 at 8:06
1
You could say that $4$ has a factor of $5^0$, but it's kindof pointless. So we don't do that. So yes, we usually require the exponent to be at least $1$.
– Arthur
Jun 12 '18 at 11:56
|
show 2 more comments
0
0
0
Is there any rule that the factors of a composite number must be of the form $n^m$, where $n$ is a real number and $m ge 1$?
Example 1: the factors of $4$ are : $2^2, 1^1, 4^1$.
Example 2: The factors of $-4$ are $-2^2, -4^1, -1^1$, and so on.
real-numbers prime-factorization
edited Nov 22 '18 at 13:38
Klangen
1,65711334
asked Jun 12 '18 at 6:44
pro neonpro neon
1305
Is there any rule that the factors of a composite number must be of the form $n^m$, where $n$ is a real number and $m ge 1$?
Example 1: the factors of $4$ are : $2^2, 1^1, 4^1$.
Example 2: The factors of $-4$ are $-2^2, -4^1, -1^1$, and so on.
real-numbers prime-factorization
real-numbers prime-factorization
edited Nov 22 '18 at 13:38
Klangen
1,65711334
asked Jun 12 '18 at 6:44
pro neonpro neon
1305
edited Nov 22 '18 at 13:38
Klangen
1,65711334
asked Jun 12 '18 at 6:44
pro neonpro neon
1305
edited Nov 22 '18 at 13:38
Klangen
1,65711334
edited Nov 22 '18 at 13:38
Klangen
1,65711334
edited Nov 22 '18 at 13:38
Klangen
1,65711334
1,65711334
asked Jun 12 '18 at 6:44
pro neonpro neon
1305
asked Jun 12 '18 at 6:44
pro neonpro neon
1305
asked Jun 12 '18 at 6:44
pro neonpro neon
1305
1305
I'm not sure I completely understand your question. Can you give us a couple of examples of what you mean?
– Billy
Jun 12 '18 at 6:45
@Billy I have edited!
– pro neon
Jun 12 '18 at 6:53
Well, first of all, since we're talking about factors, $n$ should be an integer, not a real number. Second, when you say the factors of $4$ are $2^2,1^1,4^1$, what does that really mean? The prime factorisation of $4$ is $2cdot2$, which can also be written as $2^2$ if you want (but it's not necessary). The factors of $4$ are $1,2$ and $4$. I don't know what else you could mean.
– Arthur
Jun 12 '18 at 7:15
If $n$ is a factor, you can always write it in the form $n^m$ by setting $m=1$. This doesn't seem to get you anywhere, though.
– saulspatz
Jun 12 '18 at 8:06
1
You could say that $4$ has a factor of $5^0$, but it's kindof pointless. So we don't do that. So yes, we usually require the exponent to be at least $1$.
– Arthur
Jun 12 '18 at 11:56
|
show 2 more comments
I'm not sure I completely understand your question. Can you give us a couple of examples of what you mean?
– Billy
Jun 12 '18 at 6:45
@Billy I have edited!
– pro neon
Jun 12 '18 at 6:53
Well, first of all, since we're talking about factors, $n$ should be an integer, not a real number. Second, when you say the factors of $4$ are $2^2,1^1,4^1$, what does that really mean? The prime factorisation of $4$ is $2cdot2$, which can also be written as $2^2$ if you want (but it's not necessary). The factors of $4$ are $1,2$ and $4$. I don't know what else you could mean.
– Arthur
Jun 12 '18 at 7:15
If $n$ is a factor, you can always write it in the form $n^m$ by setting $m=1$. This doesn't seem to get you anywhere, though.
– saulspatz
Jun 12 '18 at 8:06
1
You could say that $4$ has a factor of $5^0$, but it's kindof pointless. So we don't do that. So yes, we usually require the exponent to be at least $1$.
– Arthur
Jun 12 '18 at 11:56
I'm not sure I completely understand your question. Can you give us a couple of examples of what you mean?
– Billy
Jun 12 '18 at 6:45
I'm not sure I completely understand your question. Can you give us a couple of examples of what you mean?
– Billy
Jun 12 '18 at 6:45
@Billy I have edited!
– pro neon
Jun 12 '18 at 6:53
@Billy I have edited!
– pro neon
Jun 12 '18 at 6:53
Well, first of all, since we're talking about factors, $n$ should be an integer, not a real number. Second, when you say the factors of $4$ are $2^2,1^1,4^1$, what does that really mean? The prime factorisation of $4$ is $2cdot2$, which can also be written as $2^2$ if you want (but it's not necessary). The factors of $4$ are $1,2$ and $4$. I don't know what else you could mean.
– Arthur
Jun 12 '18 at 7:15
Well, first of all, since we're talking about factors, $n$ should be an integer, not a real number. Second, when you say the factors of $4$ are $2^2,1^1,4^1$, what does that really mean? The prime factorisation of $4$ is $2cdot2$, which can also be written as $2^2$ if you want (but it's not necessary). The factors of $4$ are $1,2$ and $4$. I don't know what else you could mean.
– Arthur
Jun 12 '18 at 7:15
If $n$ is a factor, you can always write it in the form $n^m$ by setting $m=1$. This doesn't seem to get you anywhere, though.
– saulspatz
Jun 12 '18 at 8:06
If $n$ is a factor, you can always write it in the form $n^m$ by setting $m=1$. This doesn't seem to get you anywhere, though.
– saulspatz
Jun 12 '18 at 8:06
1
1
You could say that $4$ has a factor of $5^0$, but it's kindof pointless. So we don't do that. So yes, we usually require the exponent to be at least $1$.
– Arthur
Jun 12 '18 at 11:56
You could say that $4$ has a factor of $5^0$, but it's kindof pointless. So we don't do that. So yes, we usually require the exponent to be at least $1$.
– Arthur
Jun 12 '18 at 11:56
|
show 2 more comments
0
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I'm not sure I completely understand your question. Can you give us a couple of examples of what you mean?
– Billy
Jun 12 '18 at 6:45
@Billy I have edited!
– pro neon
Jun 12 '18 at 6:53
Well, first of all, since we're talking about factors, $n$ should be an integer, not a real number. Second, when you say the factors of $4$ are $2^2,1^1,4^1$, what does that really mean? The prime factorisation of $4$ is $2cdot2$, which can also be written as $2^2$ if you want (but it's not necessary). The factors of $4$ are $1,2$ and $4$. I don't know what else you could mean.
– Arthur
Jun 12 '18 at 7:15
If $n$ is a factor, you can always write it in the form $n^m$ by setting $m=1$. This doesn't seem to get you anywhere, though.
– saulspatz
Jun 12 '18 at 8:06
1
You could say that $4$ has a factor of $5^0$, but it's kindof pointless. So we don't do that. So yes, we usually require the exponent to be at least $1$.
– Arthur
Jun 12 '18 at 11:56