Existence of maps on $mathbb{N} cup {0}$ satisfying $phi(ab)=phi(a)+phi(b)$
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How many maps $phi : mathbb{N} cup {0} to mathbb{N} cup {0} $ are there with the property that $phi(ab)=phi(a)+phi(b)$, for all $a,b in mathbb{N} cup {0} $?
My Attempt is
$$phi(0)+phi(m)=phi(0) implies phi(m)=0quad text{ for all } m in mathbb{N} cup {0}$$
Hence there is only one such map.
Is it correct?
algebra-precalculus
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add a comment |
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How many maps $phi : mathbb{N} cup {0} to mathbb{N} cup {0} $ are there with the property that $phi(ab)=phi(a)+phi(b)$, for all $a,b in mathbb{N} cup {0} $?
My Attempt is
$$phi(0)+phi(m)=phi(0) implies phi(m)=0quad text{ for all } m in mathbb{N} cup {0}$$
Hence there is only one such map.
Is it correct?
algebra-precalculus
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6
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Yes, it is correct.
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– SvanN
Dec 21 '18 at 17:46
1
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Looks good to me
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– pwerth
Dec 21 '18 at 17:46
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I would also mention that conversely $phi(m) equiv 0$ is indeed a solution (what's written so far technically only proves there is at most one such map).
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– Daniel Schepler
Dec 21 '18 at 18:19
2
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Possible duplicate of How many mappings $phi:Bbb{N}cup{0}toBbb{N}cup{0}$ exist such that $phi(ab)=phi(a)+phi(b)$?
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– Empty
Dec 21 '18 at 18:27
add a comment |
$begingroup$
How many maps $phi : mathbb{N} cup {0} to mathbb{N} cup {0} $ are there with the property that $phi(ab)=phi(a)+phi(b)$, for all $a,b in mathbb{N} cup {0} $?
My Attempt is
$$phi(0)+phi(m)=phi(0) implies phi(m)=0quad text{ for all } m in mathbb{N} cup {0}$$
Hence there is only one such map.
Is it correct?
algebra-precalculus
$endgroup$
How many maps $phi : mathbb{N} cup {0} to mathbb{N} cup {0} $ are there with the property that $phi(ab)=phi(a)+phi(b)$, for all $a,b in mathbb{N} cup {0} $?
My Attempt is
$$phi(0)+phi(m)=phi(0) implies phi(m)=0quad text{ for all } m in mathbb{N} cup {0}$$
Hence there is only one such map.
Is it correct?
algebra-precalculus
algebra-precalculus
edited Dec 21 '18 at 17:45
asked Dec 21 '18 at 17:41
user408906
6
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Yes, it is correct.
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– SvanN
Dec 21 '18 at 17:46
1
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Looks good to me
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– pwerth
Dec 21 '18 at 17:46
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I would also mention that conversely $phi(m) equiv 0$ is indeed a solution (what's written so far technically only proves there is at most one such map).
$endgroup$
– Daniel Schepler
Dec 21 '18 at 18:19
2
$begingroup$
Possible duplicate of How many mappings $phi:Bbb{N}cup{0}toBbb{N}cup{0}$ exist such that $phi(ab)=phi(a)+phi(b)$?
$endgroup$
– Empty
Dec 21 '18 at 18:27
add a comment |
6
$begingroup$
Yes, it is correct.
$endgroup$
– SvanN
Dec 21 '18 at 17:46
1
$begingroup$
Looks good to me
$endgroup$
– pwerth
Dec 21 '18 at 17:46
$begingroup$
I would also mention that conversely $phi(m) equiv 0$ is indeed a solution (what's written so far technically only proves there is at most one such map).
$endgroup$
– Daniel Schepler
Dec 21 '18 at 18:19
2
$begingroup$
Possible duplicate of How many mappings $phi:Bbb{N}cup{0}toBbb{N}cup{0}$ exist such that $phi(ab)=phi(a)+phi(b)$?
$endgroup$
– Empty
Dec 21 '18 at 18:27
6
6
$begingroup$
Yes, it is correct.
$endgroup$
– SvanN
Dec 21 '18 at 17:46
$begingroup$
Yes, it is correct.
$endgroup$
– SvanN
Dec 21 '18 at 17:46
1
1
$begingroup$
Looks good to me
$endgroup$
– pwerth
Dec 21 '18 at 17:46
$begingroup$
Looks good to me
$endgroup$
– pwerth
Dec 21 '18 at 17:46
$begingroup$
I would also mention that conversely $phi(m) equiv 0$ is indeed a solution (what's written so far technically only proves there is at most one such map).
$endgroup$
– Daniel Schepler
Dec 21 '18 at 18:19
$begingroup$
I would also mention that conversely $phi(m) equiv 0$ is indeed a solution (what's written so far technically only proves there is at most one such map).
$endgroup$
– Daniel Schepler
Dec 21 '18 at 18:19
2
2
$begingroup$
Possible duplicate of How many mappings $phi:Bbb{N}cup{0}toBbb{N}cup{0}$ exist such that $phi(ab)=phi(a)+phi(b)$?
$endgroup$
– Empty
Dec 21 '18 at 18:27
$begingroup$
Possible duplicate of How many mappings $phi:Bbb{N}cup{0}toBbb{N}cup{0}$ exist such that $phi(ab)=phi(a)+phi(b)$?
$endgroup$
– Empty
Dec 21 '18 at 18:27
add a comment |
1 Answer
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You've got the correct conclusion, but it could use a tiny bit more justification. I'd express it as $$varphi(0)+varphi(m)=varphi(0m)=varphi(0),$$ just to make it perfectly clear.
answered Dec 21 '18 at 18:09
Cameron BuieCameron Buie
86.2k772161
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1 Answer
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$begingroup$
You've got the correct conclusion, but it could use a tiny bit more justification. I'd express it as $$varphi(0)+varphi(m)=varphi(0m)=varphi(0),$$ just to make it perfectly clear.
answered Dec 21 '18 at 18:09
Cameron BuieCameron Buie
86.2k772161
$endgroup$
add a comment |
$begingroup$
You've got the correct conclusion, but it could use a tiny bit more justification. I'd express it as $$varphi(0)+varphi(m)=varphi(0m)=varphi(0),$$ just to make it perfectly clear.
answered Dec 21 '18 at 18:09
Cameron BuieCameron Buie
86.2k772161
$endgroup$
add a comment |
$begingroup$
You've got the correct conclusion, but it could use a tiny bit more justification. I'd express it as $$varphi(0)+varphi(m)=varphi(0m)=varphi(0),$$ just to make it perfectly clear.
answered Dec 21 '18 at 18:09
Cameron BuieCameron Buie
86.2k772161
$endgroup$
You've got the correct conclusion, but it could use a tiny bit more justification. I'd express it as $$varphi(0)+varphi(m)=varphi(0m)=varphi(0),$$ just to make it perfectly clear.
answered Dec 21 '18 at 18:09
Cameron BuieCameron Buie
86.2k772161
answered Dec 21 '18 at 18:09
Cameron BuieCameron Buie
86.2k772161
answered Dec 21 '18 at 18:09
Cameron BuieCameron Buie
86.2k772161
answered Dec 21 '18 at 18:09
Cameron BuieCameron Buie
86.2k772161
86.2k772161
add a comment |
add a comment |
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6
$begingroup$
Yes, it is correct.
$endgroup$
– SvanN
Dec 21 '18 at 17:46
1
$begingroup$
Looks good to me
$endgroup$
– pwerth
Dec 21 '18 at 17:46
$begingroup$
I would also mention that conversely $phi(m) equiv 0$ is indeed a solution (what's written so far technically only proves there is at most one such map).
$endgroup$
– Daniel Schepler
Dec 21 '18 at 18:19
2
$begingroup$
Possible duplicate of How many mappings $phi:Bbb{N}cup{0}toBbb{N}cup{0}$ exist such that $phi(ab)=phi(a)+phi(b)$?
$endgroup$
– Empty
Dec 21 '18 at 18:27