Mixing
Informally describe the set $A ={n : n text{ is an integer and } n = n + 1}$
$begingroup$
Informally describe the set $A ={n : n text{ is an integer and } n = n + 1}$
Just a quick set theory exercise. At first my reaction was: there is no integer that satisfies this description, i.e., $A = emptyset$. But on second thought, the set may be non null if it was
$$A = mathbb{Z}_1$$
because $n equiv k equiv 0 bmod 1 $ for any integer $n$. Maybe this question was designed to be vague (it's from a Theory of Computation text). Or maybe I'm taking too much freedom with my interpretation of equality. What do you think?
elementary-number-theory elementary-set-theory
asked Jan 26 at 18:41
ZduffZduff
1,7091020
$endgroup$
add a comment |
$begingroup$
Informally describe the set $A ={n : n text{ is an integer and } n = n + 1}$
Just a quick set theory exercise. At first my reaction was: there is no integer that satisfies this description, i.e., $A = emptyset$. But on second thought, the set may be non null if it was
$$A = mathbb{Z}_1$$
because $n equiv k equiv 0 bmod 1 $ for any integer $n$. Maybe this question was designed to be vague (it's from a Theory of Computation text). Or maybe I'm taking too much freedom with my interpretation of equality. What do you think?
elementary-number-theory elementary-set-theory
asked Jan 26 at 18:41
ZduffZduff
1,7091020
$endgroup$
$begingroup$
I agree with your first reaction : there is no integer $n$ such that $n=n+1$.
$endgroup$
– Mauro ALLEGRANZA
Jan 26 at 18:44
$begingroup$
You seem very confused about the modulo relation, do you need some clarification on that ?
$endgroup$
– J.F
Jan 26 at 18:44
$begingroup$
Yeah, could you please link me some elementary reference on that. I took an intro algebra class that covered it like 5 years ago, but haven't been exposed since. Or if it's quick to write an easily decipherable definition in the comments that would suffice. Either way, thank you.
$endgroup$
– Zduff
Jan 26 at 19:06
add a comment |
$begingroup$
Informally describe the set $A ={n : n text{ is an integer and } n = n + 1}$
Just a quick set theory exercise. At first my reaction was: there is no integer that satisfies this description, i.e., $A = emptyset$. But on second thought, the set may be non null if it was
$$A = mathbb{Z}_1$$
because $n equiv k equiv 0 bmod 1 $ for any integer $n$. Maybe this question was designed to be vague (it's from a Theory of Computation text). Or maybe I'm taking too much freedom with my interpretation of equality. What do you think?
elementary-number-theory elementary-set-theory
asked Jan 26 at 18:41
ZduffZduff
1,7091020
$endgroup$
Informally describe the set $A ={n : n text{ is an integer and } n = n + 1}$
Just a quick set theory exercise. At first my reaction was: there is no integer that satisfies this description, i.e., $A = emptyset$. But on second thought, the set may be non null if it was
$$A = mathbb{Z}_1$$
because $n equiv k equiv 0 bmod 1 $ for any integer $n$. Maybe this question was designed to be vague (it's from a Theory of Computation text). Or maybe I'm taking too much freedom with my interpretation of equality. What do you think?
elementary-number-theory elementary-set-theory
elementary-number-theory elementary-set-theory
asked Jan 26 at 18:41
ZduffZduff
1,7091020
asked Jan 26 at 18:41
ZduffZduff
1,7091020
asked Jan 26 at 18:41
ZduffZduff
1,7091020
asked Jan 26 at 18:41
ZduffZduff
1,7091020
asked Jan 26 at 18:41
ZduffZduff
1,7091020
1,7091020
$begingroup$
I agree with your first reaction : there is no integer $n$ such that $n=n+1$.
$endgroup$
– Mauro ALLEGRANZA
Jan 26 at 18:44
$begingroup$
You seem very confused about the modulo relation, do you need some clarification on that ?
$endgroup$
– J.F
Jan 26 at 18:44
$begingroup$
Yeah, could you please link me some elementary reference on that. I took an intro algebra class that covered it like 5 years ago, but haven't been exposed since. Or if it's quick to write an easily decipherable definition in the comments that would suffice. Either way, thank you.
$endgroup$
– Zduff
Jan 26 at 19:06
add a comment |
$begingroup$
I agree with your first reaction : there is no integer $n$ such that $n=n+1$.
$endgroup$
– Mauro ALLEGRANZA
Jan 26 at 18:44
$begingroup$
You seem very confused about the modulo relation, do you need some clarification on that ?
$endgroup$
– J.F
Jan 26 at 18:44
$begingroup$
Yeah, could you please link me some elementary reference on that. I took an intro algebra class that covered it like 5 years ago, but haven't been exposed since. Or if it's quick to write an easily decipherable definition in the comments that would suffice. Either way, thank you.
$endgroup$
– Zduff
Jan 26 at 19:06
$begingroup$
I agree with your first reaction : there is no integer $n$ such that $n=n+1$.
$endgroup$
– Mauro ALLEGRANZA
Jan 26 at 18:44
$begingroup$
I agree with your first reaction : there is no integer $n$ such that $n=n+1$.
$endgroup$
– Mauro ALLEGRANZA
Jan 26 at 18:44
$begingroup$
You seem very confused about the modulo relation, do you need some clarification on that ?
$endgroup$
– J.F
Jan 26 at 18:44
$begingroup$
You seem very confused about the modulo relation, do you need some clarification on that ?
$endgroup$
– J.F
Jan 26 at 18:44
$begingroup$
Yeah, could you please link me some elementary reference on that. I took an intro algebra class that covered it like 5 years ago, but haven't been exposed since. Or if it's quick to write an easily decipherable definition in the comments that would suffice. Either way, thank you.
$endgroup$
– Zduff
Jan 26 at 19:06
$begingroup$
Yeah, could you please link me some elementary reference on that. I took an intro algebra class that covered it like 5 years ago, but haven't been exposed since. Or if it's quick to write an easily decipherable definition in the comments that would suffice. Either way, thank you.
$endgroup$
– Zduff
Jan 26 at 19:06
add a comment |
1 Answer
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$begingroup$
Your first thought is correct. There are no integers $n$ with $n=n+1$.
Your second thought is interesting, but I wouldn't say that you're taking liberties with your interpretation of equality so much as with your interpretation of integer.
When we say integer, we mean an element of $Bbb{Z}$ with its usual ring structure (i.e. with its usual 0, 1, and addition, and multiplication, (and if you'd like to think of it that way, with its usual equality)), not an element of $Bbb{Z}/1Bbb{Z}$ or $Bbb{Z}/nBbb{Z}$ for any $n$.
answered Jan 26 at 18:46
jgonjgon
15.7k32143
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Your first thought is correct. There are no integers $n$ with $n=n+1$.
Your second thought is interesting, but I wouldn't say that you're taking liberties with your interpretation of equality so much as with your interpretation of integer.
When we say integer, we mean an element of $Bbb{Z}$ with its usual ring structure (i.e. with its usual 0, 1, and addition, and multiplication, (and if you'd like to think of it that way, with its usual equality)), not an element of $Bbb{Z}/1Bbb{Z}$ or $Bbb{Z}/nBbb{Z}$ for any $n$.
answered Jan 26 at 18:46
jgonjgon
15.7k32143
$endgroup$
add a comment |
$begingroup$
Your first thought is correct. There are no integers $n$ with $n=n+1$.
Your second thought is interesting, but I wouldn't say that you're taking liberties with your interpretation of equality so much as with your interpretation of integer.
When we say integer, we mean an element of $Bbb{Z}$ with its usual ring structure (i.e. with its usual 0, 1, and addition, and multiplication, (and if you'd like to think of it that way, with its usual equality)), not an element of $Bbb{Z}/1Bbb{Z}$ or $Bbb{Z}/nBbb{Z}$ for any $n$.
answered Jan 26 at 18:46
jgonjgon
15.7k32143
$endgroup$
add a comment |
$begingroup$
Your first thought is correct. There are no integers $n$ with $n=n+1$.
Your second thought is interesting, but I wouldn't say that you're taking liberties with your interpretation of equality so much as with your interpretation of integer.
When we say integer, we mean an element of $Bbb{Z}$ with its usual ring structure (i.e. with its usual 0, 1, and addition, and multiplication, (and if you'd like to think of it that way, with its usual equality)), not an element of $Bbb{Z}/1Bbb{Z}$ or $Bbb{Z}/nBbb{Z}$ for any $n$.
answered Jan 26 at 18:46
jgonjgon
15.7k32143
$endgroup$
Your first thought is correct. There are no integers $n$ with $n=n+1$.
Your second thought is interesting, but I wouldn't say that you're taking liberties with your interpretation of equality so much as with your interpretation of integer.
When we say integer, we mean an element of $Bbb{Z}$ with its usual ring structure (i.e. with its usual 0, 1, and addition, and multiplication, (and if you'd like to think of it that way, with its usual equality)), not an element of $Bbb{Z}/1Bbb{Z}$ or $Bbb{Z}/nBbb{Z}$ for any $n$.
answered Jan 26 at 18:46
jgonjgon
15.7k32143
answered Jan 26 at 18:46
jgonjgon
15.7k32143
answered Jan 26 at 18:46
jgonjgon
15.7k32143
answered Jan 26 at 18:46
jgonjgon
15.7k32143
15.7k32143
add a comment |
add a comment |
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$begingroup$
I agree with your first reaction : there is no integer $n$ such that $n=n+1$.
$endgroup$
– Mauro ALLEGRANZA
Jan 26 at 18:44
$begingroup$
You seem very confused about the modulo relation, do you need some clarification on that ?
$endgroup$
– J.F
Jan 26 at 18:44
$begingroup$
Yeah, could you please link me some elementary reference on that. I took an intro algebra class that covered it like 5 years ago, but haven't been exposed since. Or if it's quick to write an easily decipherable definition in the comments that would suffice. Either way, thank you.
$endgroup$
– Zduff
Jan 26 at 19:06