Number Theory: Problem with proofsCongruence equation problemWhat does $ a pmod b$ mean?Linear congruence fill in the missing step?Proof: if $n > 1$ then $LD(n) $ is a prime numberNumber Theory Lemma About Linear Congruence (Explanation Needed)Chinese remainder theorem, how to get a ≡ b (mod pq) from a ≡ b (mod p) and a ≡ b (mod q)?Proof verification: $a+bequiv b+a ;; (mod;;n)$ and $abequiv ba ;; (mod;;n)$.Showing two different definitions of a primitive root are the sameweird gcd problem in number theoryNumber Theory Linear Diophantine Equations
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Number Theory: Problem with proofs
Congruence equation problemWhat does $ a pmod b$ mean?Linear congruence fill in the missing step?Proof: if $n > 1$ then $LD(n) $ is a prime numberNumber Theory Lemma About Linear Congruence (Explanation Needed)Chinese remainder theorem, how to get a ≡ b (mod pq) from a ≡ b (mod p) and a ≡ b (mod q)?Proof verification: $a+bequiv b+a ;; (mod;;n)$ and $abequiv ba ;; (mod;;n)$.Showing two different definitions of a primitive root are the sameweird gcd problem in number theoryNumber Theory Linear Diophantine Equations
$begingroup$
There are two propositions in the chapter of Number Theory in my book, the proofs of which I am having trouble to understand.
For Proposition 3
I cannot understand the proof from "Therefore ..." in the third line. I was thinking that maybe "$m-l$ divides $k$" will be "$m-l$ divides $n$". Also, is $k/d$ an integer?
Also, can someone please give an example to clarify this theorem?
For Proposition 4
What does $f(x) in Z[x]$ mean? Why are the third brackets used?
I cannot understand how "By proposition 2, $a_ja^j equiv b_jb^j$ (mod $n$)...".
Can someone please explain?
number-theory modular-arithmetic congruence-relations
asked yesterday
MrAPMrAP
1,18521432
$endgroup$
add a comment |
$begingroup$
There are two propositions in the chapter of Number Theory in my book, the proofs of which I am having trouble to understand.
For Proposition 3
I cannot understand the proof from "Therefore ..." in the third line. I was thinking that maybe "$m-l$ divides $k$" will be "$m-l$ divides $n$". Also, is $k/d$ an integer?
Also, can someone please give an example to clarify this theorem?
For Proposition 4
What does $f(x) in Z[x]$ mean? Why are the third brackets used?
I cannot understand how "By proposition 2, $a_ja^j equiv b_jb^j$ (mod $n$)...".
Can someone please explain?
number-theory modular-arithmetic congruence-relations
asked yesterday
MrAPMrAP
1,18521432
$endgroup$
$begingroup$
I think the book has a typo and instead of $m-l$ divides $k$" it is supposed to be "$d$ divides $k$". That fits with $(m-l)d = knimplies m-l = frac kd n$ which assumes $frac kd$ is an integer.... on the other hand, that is one heck of a type to make!
$endgroup$
– fleablood
yesterday
add a comment |
$begingroup$
There are two propositions in the chapter of Number Theory in my book, the proofs of which I am having trouble to understand.
For Proposition 3
I cannot understand the proof from "Therefore ..." in the third line. I was thinking that maybe "$m-l$ divides $k$" will be "$m-l$ divides $n$". Also, is $k/d$ an integer?
Also, can someone please give an example to clarify this theorem?
For Proposition 4
What does $f(x) in Z[x]$ mean? Why are the third brackets used?
I cannot understand how "By proposition 2, $a_ja^j equiv b_jb^j$ (mod $n$)...".
Can someone please explain?
number-theory modular-arithmetic congruence-relations
asked yesterday
MrAPMrAP
1,18521432
$endgroup$
There are two propositions in the chapter of Number Theory in my book, the proofs of which I am having trouble to understand.
For Proposition 3
I cannot understand the proof from "Therefore ..." in the third line. I was thinking that maybe "$m-l$ divides $k$" will be "$m-l$ divides $n$". Also, is $k/d$ an integer?
Also, can someone please give an example to clarify this theorem?
For Proposition 4
What does $f(x) in Z[x]$ mean? Why are the third brackets used?
I cannot understand how "By proposition 2, $a_ja^j equiv b_jb^j$ (mod $n$)...".
Can someone please explain?
number-theory modular-arithmetic congruence-relations
number-theory modular-arithmetic congruence-relations
asked yesterday
MrAPMrAP
1,18521432
asked yesterday
MrAPMrAP
1,18521432
asked yesterday
MrAPMrAP
1,18521432
asked yesterday
MrAPMrAP
1,18521432
asked yesterday
MrAPMrAP
1,18521432
1,18521432
$begingroup$
I think the book has a typo and instead of $m-l$ divides $k$" it is supposed to be "$d$ divides $k$". That fits with $(m-l)d = knimplies m-l = frac kd n$ which assumes $frac kd$ is an integer.... on the other hand, that is one heck of a type to make!
$endgroup$
– fleablood
yesterday
add a comment |
$begingroup$
I think the book has a typo and instead of $m-l$ divides $k$" it is supposed to be "$d$ divides $k$". That fits with $(m-l)d = knimplies m-l = frac kd n$ which assumes $frac kd$ is an integer.... on the other hand, that is one heck of a type to make!
$endgroup$
– fleablood
yesterday
$begingroup$
I think the book has a typo and instead of $m-l$ divides $k$" it is supposed to be "$d$ divides $k$". That fits with $(m-l)d = knimplies m-l = frac kd n$ which assumes $frac kd$ is an integer.... on the other hand, that is one heck of a type to make!
$endgroup$
– fleablood
yesterday
$begingroup$
I think the book has a typo and instead of $m-l$ divides $k$" it is supposed to be "$d$ divides $k$". That fits with $(m-l)d = knimplies m-l = frac kd n$ which assumes $frac kd$ is an integer.... on the other hand, that is one heck of a type to make!
$endgroup$
– fleablood
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For proposition 3, I agree with fleablood's assessment in a comment above that this is probably a typo.
For proposition $4$, note that the thing you're asking about appears in a parenthesis, following the abbreviation "i.e.". That means that even without knowing what "$fin Bbb Z[x]$" means, we can infer that it's merely an alternative formulation of whatever came before, namely that $f$ is a polynomial with integral coefficients.
To actually answer the question, given a ring $R$, the notation $R[x]$ means "the ring (or set) of polynomials with coefficients in $R$". So $Bbb Z[x]$ is the ring of polynomials with integer coefficients. And $fin Bbb Z[x]$ means exactly that $f$ is an element of this ring.
As for why we use the brackets there? That's just convention. You could have $Bbb Z(x)$ as well, but that usually means the ring (or set) of rational functions with integer coefficients.
Finally, the "by proposition 2" thing, note that proposition 2 states that if we have a product of two things, and we change one of the factors to a congruent factor, the product is unchanged modulo $n$. So $$a_ja^j = a_ja^j-1cdot aequiv a_ja^j-1bpmod n$$
So we can swap one $a$ for a $b$. Now just swap the other $j-1$ $a$'s for $b$'s, one by one as proposition 2 says you're allowed to do, and finally you reach $a_jb^j$.
answered yesterday
ArthurArthur
119k7118202
$endgroup$
$begingroup$
heck of a typo to make; and heck of a way to define and introduce notation (as an aside inside a parenthesis). If anything it should go the other way: "Let $f(x)in mathbb Z[x] (i.e. $f(x)$ is a polynomial with integer coefficients)". This book isn't winning me over.
$endgroup$
– fleablood
yesterday
$begingroup$
@fleablood When I learned complex analysis, the book had enough strategically placed typos that I gave up and used wikipedia to learn the theorems instead. And the educational value of wikipedia math articles is not always the greatest, as they mostly seem like reference works. So I know how it is.
$endgroup$
– Arthur
yesterday
$begingroup$
Shouldn't you write "$f(x)in Bbb Z[x]$" instead of "$fin Bbb Z[x]$" since $f$ denotes the set just like R (relation) denotes the set?
$endgroup$
– MrAP
yesterday
$begingroup$
@MrAP it's a matter of what you want to convey. $f$ is the name of the polynomial. $f(x)$ is the expression that defines the polynomial. At least that's how I think about it. $Bbb Z$ is the ring / set of integers, $Bbb Z[x]$ is the ring / set of polynomials with integer coefficients. Also, $R$ isn't a relation here. I did say "ring". Like $Bbb Z$, or $Bbb R$. If that isn't a term you've come across yet, then I am sorry for confusing you.
$endgroup$
– Arthur
yesterday
$begingroup$
By $R$ I meant the a relation. $R:Ato B$ is a subset of $Atimes B$. So similarly $f:Ato B$ should also be a subset of $Atimes B$ since a function is a relation. So $f$ denotes a set which represents a function. And normally we write that an element belongs to a set. Here as you have written, it would mean that a set belongs to another set which is normally not the case. That is what I am trying to express.
$endgroup$
– MrAP
yesterday
|
show 3 more comments
$begingroup$
I think the book made a typo and it isn't $m-l$ that divides $k$ (I don't see that as even true) but that $d$ divides $k$.
This follow as $d$ divides $kn$ but is relatively prime to $n$ so $d$ must divide $k$.
And hence, yes, $frac kd$ is an integer, for which we conclude $m-l equiv 0 pmod n$.
Note: This will not be true if $a, b, n$ will have a common divisor (other than $1$). Consider $8 equiv 20 mod 12$ but $2equiv 5 mod 12$ is .... wrong. (Although $2 equiv 5 pmod 3$.....)
--- I empathize. For a typo that is a doozy to make and utter destroys the intent of the proof.
....
$mathbb Z[x]$ means the set of all polynomials with integer coefficients.
So $f(x) in mathbb Z[x]$ means "Let $f(x)$ be a polynomial with integer coefficients.
answered yesterday
fleabloodfleablood
72.9k22789
$endgroup$
$begingroup$
$(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
$endgroup$
– MrAP
yesterday
$begingroup$
IF $d$ and $n$ are relatively prime (i.e. $gcd(d,n)=1$) then $d$ and $n$ have no factors in common. So if $(m-l)d = kn$ then $d$ divides $kn$. But as $d$ and $n$ have nothing in common, that means $d$ divides $k$. Likewise that means $n$ divides $(m-l)d$ but $n$ and $d$ have nothing in common so $n$ divides $m-l$..... BTW "$n$ divides $A$" and "$Aequiv 0 pmod n$" mean the exact same thing. And "$n$ divides $m-l$" and "$mequiv l pmod n$" and "$m-l equiv 0 pmod n$" all mean the same thing.
$endgroup$
– fleablood
yesterday
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For proposition 3, I agree with fleablood's assessment in a comment above that this is probably a typo.
For proposition $4$, note that the thing you're asking about appears in a parenthesis, following the abbreviation "i.e.". That means that even without knowing what "$fin Bbb Z[x]$" means, we can infer that it's merely an alternative formulation of whatever came before, namely that $f$ is a polynomial with integral coefficients.
To actually answer the question, given a ring $R$, the notation $R[x]$ means "the ring (or set) of polynomials with coefficients in $R$". So $Bbb Z[x]$ is the ring of polynomials with integer coefficients. And $fin Bbb Z[x]$ means exactly that $f$ is an element of this ring.
As for why we use the brackets there? That's just convention. You could have $Bbb Z(x)$ as well, but that usually means the ring (or set) of rational functions with integer coefficients.
Finally, the "by proposition 2" thing, note that proposition 2 states that if we have a product of two things, and we change one of the factors to a congruent factor, the product is unchanged modulo $n$. So $$a_ja^j = a_ja^j-1cdot aequiv a_ja^j-1bpmod n$$
So we can swap one $a$ for a $b$. Now just swap the other $j-1$ $a$'s for $b$'s, one by one as proposition 2 says you're allowed to do, and finally you reach $a_jb^j$.
answered yesterday
ArthurArthur
119k7118202
$endgroup$
$begingroup$
heck of a typo to make; and heck of a way to define and introduce notation (as an aside inside a parenthesis). If anything it should go the other way: "Let $f(x)in mathbb Z[x] (i.e. $f(x)$ is a polynomial with integer coefficients)". This book isn't winning me over.
$endgroup$
– fleablood
yesterday
$begingroup$
@fleablood When I learned complex analysis, the book had enough strategically placed typos that I gave up and used wikipedia to learn the theorems instead. And the educational value of wikipedia math articles is not always the greatest, as they mostly seem like reference works. So I know how it is.
$endgroup$
– Arthur
yesterday
$begingroup$
Shouldn't you write "$f(x)in Bbb Z[x]$" instead of "$fin Bbb Z[x]$" since $f$ denotes the set just like R (relation) denotes the set?
$endgroup$
– MrAP
yesterday
$begingroup$
@MrAP it's a matter of what you want to convey. $f$ is the name of the polynomial. $f(x)$ is the expression that defines the polynomial. At least that's how I think about it. $Bbb Z$ is the ring / set of integers, $Bbb Z[x]$ is the ring / set of polynomials with integer coefficients. Also, $R$ isn't a relation here. I did say "ring". Like $Bbb Z$, or $Bbb R$. If that isn't a term you've come across yet, then I am sorry for confusing you.
$endgroup$
– Arthur
yesterday
$begingroup$
By $R$ I meant the a relation. $R:Ato B$ is a subset of $Atimes B$. So similarly $f:Ato B$ should also be a subset of $Atimes B$ since a function is a relation. So $f$ denotes a set which represents a function. And normally we write that an element belongs to a set. Here as you have written, it would mean that a set belongs to another set which is normally not the case. That is what I am trying to express.
$endgroup$
– MrAP
yesterday
|
show 3 more comments
$begingroup$
For proposition 3, I agree with fleablood's assessment in a comment above that this is probably a typo.
For proposition $4$, note that the thing you're asking about appears in a parenthesis, following the abbreviation "i.e.". That means that even without knowing what "$fin Bbb Z[x]$" means, we can infer that it's merely an alternative formulation of whatever came before, namely that $f$ is a polynomial with integral coefficients.
To actually answer the question, given a ring $R$, the notation $R[x]$ means "the ring (or set) of polynomials with coefficients in $R$". So $Bbb Z[x]$ is the ring of polynomials with integer coefficients. And $fin Bbb Z[x]$ means exactly that $f$ is an element of this ring.
As for why we use the brackets there? That's just convention. You could have $Bbb Z(x)$ as well, but that usually means the ring (or set) of rational functions with integer coefficients.
Finally, the "by proposition 2" thing, note that proposition 2 states that if we have a product of two things, and we change one of the factors to a congruent factor, the product is unchanged modulo $n$. So $$a_ja^j = a_ja^j-1cdot aequiv a_ja^j-1bpmod n$$
So we can swap one $a$ for a $b$. Now just swap the other $j-1$ $a$'s for $b$'s, one by one as proposition 2 says you're allowed to do, and finally you reach $a_jb^j$.
answered yesterday
ArthurArthur
119k7118202
$endgroup$
$begingroup$
heck of a typo to make; and heck of a way to define and introduce notation (as an aside inside a parenthesis). If anything it should go the other way: "Let $f(x)in mathbb Z[x] (i.e. $f(x)$ is a polynomial with integer coefficients)". This book isn't winning me over.
$endgroup$
– fleablood
yesterday
$begingroup$
@fleablood When I learned complex analysis, the book had enough strategically placed typos that I gave up and used wikipedia to learn the theorems instead. And the educational value of wikipedia math articles is not always the greatest, as they mostly seem like reference works. So I know how it is.
$endgroup$
– Arthur
yesterday
$begingroup$
Shouldn't you write "$f(x)in Bbb Z[x]$" instead of "$fin Bbb Z[x]$" since $f$ denotes the set just like R (relation) denotes the set?
$endgroup$
– MrAP
yesterday
$begingroup$
@MrAP it's a matter of what you want to convey. $f$ is the name of the polynomial. $f(x)$ is the expression that defines the polynomial. At least that's how I think about it. $Bbb Z$ is the ring / set of integers, $Bbb Z[x]$ is the ring / set of polynomials with integer coefficients. Also, $R$ isn't a relation here. I did say "ring". Like $Bbb Z$, or $Bbb R$. If that isn't a term you've come across yet, then I am sorry for confusing you.
$endgroup$
– Arthur
yesterday
$begingroup$
By $R$ I meant the a relation. $R:Ato B$ is a subset of $Atimes B$. So similarly $f:Ato B$ should also be a subset of $Atimes B$ since a function is a relation. So $f$ denotes a set which represents a function. And normally we write that an element belongs to a set. Here as you have written, it would mean that a set belongs to another set which is normally not the case. That is what I am trying to express.
$endgroup$
– MrAP
yesterday
|
show 3 more comments
$begingroup$
For proposition 3, I agree with fleablood's assessment in a comment above that this is probably a typo.
For proposition $4$, note that the thing you're asking about appears in a parenthesis, following the abbreviation "i.e.". That means that even without knowing what "$fin Bbb Z[x]$" means, we can infer that it's merely an alternative formulation of whatever came before, namely that $f$ is a polynomial with integral coefficients.
To actually answer the question, given a ring $R$, the notation $R[x]$ means "the ring (or set) of polynomials with coefficients in $R$". So $Bbb Z[x]$ is the ring of polynomials with integer coefficients. And $fin Bbb Z[x]$ means exactly that $f$ is an element of this ring.
As for why we use the brackets there? That's just convention. You could have $Bbb Z(x)$ as well, but that usually means the ring (or set) of rational functions with integer coefficients.
Finally, the "by proposition 2" thing, note that proposition 2 states that if we have a product of two things, and we change one of the factors to a congruent factor, the product is unchanged modulo $n$. So $$a_ja^j = a_ja^j-1cdot aequiv a_ja^j-1bpmod n$$
So we can swap one $a$ for a $b$. Now just swap the other $j-1$ $a$'s for $b$'s, one by one as proposition 2 says you're allowed to do, and finally you reach $a_jb^j$.
answered yesterday
ArthurArthur
119k7118202
$endgroup$
For proposition 3, I agree with fleablood's assessment in a comment above that this is probably a typo.
For proposition $4$, note that the thing you're asking about appears in a parenthesis, following the abbreviation "i.e.". That means that even without knowing what "$fin Bbb Z[x]$" means, we can infer that it's merely an alternative formulation of whatever came before, namely that $f$ is a polynomial with integral coefficients.
To actually answer the question, given a ring $R$, the notation $R[x]$ means "the ring (or set) of polynomials with coefficients in $R$". So $Bbb Z[x]$ is the ring of polynomials with integer coefficients. And $fin Bbb Z[x]$ means exactly that $f$ is an element of this ring.
As for why we use the brackets there? That's just convention. You could have $Bbb Z(x)$ as well, but that usually means the ring (or set) of rational functions with integer coefficients.
Finally, the "by proposition 2" thing, note that proposition 2 states that if we have a product of two things, and we change one of the factors to a congruent factor, the product is unchanged modulo $n$. So $$a_ja^j = a_ja^j-1cdot aequiv a_ja^j-1bpmod n$$
So we can swap one $a$ for a $b$. Now just swap the other $j-1$ $a$'s for $b$'s, one by one as proposition 2 says you're allowed to do, and finally you reach $a_jb^j$.
answered yesterday
ArthurArthur
119k7118202
edited yesterday
answered yesterday
ArthurArthur
119k7118202
answered yesterday
ArthurArthur
119k7118202
answered yesterday
ArthurArthur
119k7118202
119k7118202
$begingroup$
heck of a typo to make; and heck of a way to define and introduce notation (as an aside inside a parenthesis). If anything it should go the other way: "Let $f(x)in mathbb Z[x] (i.e. $f(x)$ is a polynomial with integer coefficients)". This book isn't winning me over.
$endgroup$
– fleablood
yesterday
$begingroup$
@fleablood When I learned complex analysis, the book had enough strategically placed typos that I gave up and used wikipedia to learn the theorems instead. And the educational value of wikipedia math articles is not always the greatest, as they mostly seem like reference works. So I know how it is.
$endgroup$
– Arthur
yesterday
$begingroup$
Shouldn't you write "$f(x)in Bbb Z[x]$" instead of "$fin Bbb Z[x]$" since $f$ denotes the set just like R (relation) denotes the set?
$endgroup$
– MrAP
yesterday
$begingroup$
@MrAP it's a matter of what you want to convey. $f$ is the name of the polynomial. $f(x)$ is the expression that defines the polynomial. At least that's how I think about it. $Bbb Z$ is the ring / set of integers, $Bbb Z[x]$ is the ring / set of polynomials with integer coefficients. Also, $R$ isn't a relation here. I did say "ring". Like $Bbb Z$, or $Bbb R$. If that isn't a term you've come across yet, then I am sorry for confusing you.
$endgroup$
– Arthur
yesterday
$begingroup$
By $R$ I meant the a relation. $R:Ato B$ is a subset of $Atimes B$. So similarly $f:Ato B$ should also be a subset of $Atimes B$ since a function is a relation. So $f$ denotes a set which represents a function. And normally we write that an element belongs to a set. Here as you have written, it would mean that a set belongs to another set which is normally not the case. That is what I am trying to express.
$endgroup$
– MrAP
yesterday
|
show 3 more comments
$begingroup$
heck of a typo to make; and heck of a way to define and introduce notation (as an aside inside a parenthesis). If anything it should go the other way: "Let $f(x)in mathbb Z[x] (i.e. $f(x)$ is a polynomial with integer coefficients)". This book isn't winning me over.
$endgroup$
– fleablood
yesterday
$begingroup$
@fleablood When I learned complex analysis, the book had enough strategically placed typos that I gave up and used wikipedia to learn the theorems instead. And the educational value of wikipedia math articles is not always the greatest, as they mostly seem like reference works. So I know how it is.
$endgroup$
– Arthur
yesterday
$begingroup$
Shouldn't you write "$f(x)in Bbb Z[x]$" instead of "$fin Bbb Z[x]$" since $f$ denotes the set just like R (relation) denotes the set?
$endgroup$
– MrAP
yesterday
$begingroup$
@MrAP it's a matter of what you want to convey. $f$ is the name of the polynomial. $f(x)$ is the expression that defines the polynomial. At least that's how I think about it. $Bbb Z$ is the ring / set of integers, $Bbb Z[x]$ is the ring / set of polynomials with integer coefficients. Also, $R$ isn't a relation here. I did say "ring". Like $Bbb Z$, or $Bbb R$. If that isn't a term you've come across yet, then I am sorry for confusing you.
$endgroup$
– Arthur
yesterday
$begingroup$
By $R$ I meant the a relation. $R:Ato B$ is a subset of $Atimes B$. So similarly $f:Ato B$ should also be a subset of $Atimes B$ since a function is a relation. So $f$ denotes a set which represents a function. And normally we write that an element belongs to a set. Here as you have written, it would mean that a set belongs to another set which is normally not the case. That is what I am trying to express.
$endgroup$
– MrAP
yesterday
$begingroup$
heck of a typo to make; and heck of a way to define and introduce notation (as an aside inside a parenthesis). If anything it should go the other way: "Let $f(x)in mathbb Z[x] (i.e. $f(x)$ is a polynomial with integer coefficients)". This book isn't winning me over.
$endgroup$
– fleablood
yesterday
$begingroup$
heck of a typo to make; and heck of a way to define and introduce notation (as an aside inside a parenthesis). If anything it should go the other way: "Let $f(x)in mathbb Z[x] (i.e. $f(x)$ is a polynomial with integer coefficients)". This book isn't winning me over.
$endgroup$
– fleablood
yesterday
$begingroup$
@fleablood When I learned complex analysis, the book had enough strategically placed typos that I gave up and used wikipedia to learn the theorems instead. And the educational value of wikipedia math articles is not always the greatest, as they mostly seem like reference works. So I know how it is.
$endgroup$
– Arthur
yesterday
$begingroup$
@fleablood When I learned complex analysis, the book had enough strategically placed typos that I gave up and used wikipedia to learn the theorems instead. And the educational value of wikipedia math articles is not always the greatest, as they mostly seem like reference works. So I know how it is.
$endgroup$
– Arthur
yesterday
$begingroup$
Shouldn't you write "$f(x)in Bbb Z[x]$" instead of "$fin Bbb Z[x]$" since $f$ denotes the set just like R (relation) denotes the set?
$endgroup$
– MrAP
yesterday
$begingroup$
Shouldn't you write "$f(x)in Bbb Z[x]$" instead of "$fin Bbb Z[x]$" since $f$ denotes the set just like R (relation) denotes the set?
$endgroup$
– MrAP
yesterday
$begingroup$
@MrAP it's a matter of what you want to convey. $f$ is the name of the polynomial. $f(x)$ is the expression that defines the polynomial. At least that's how I think about it. $Bbb Z$ is the ring / set of integers, $Bbb Z[x]$ is the ring / set of polynomials with integer coefficients. Also, $R$ isn't a relation here. I did say "ring". Like $Bbb Z$, or $Bbb R$. If that isn't a term you've come across yet, then I am sorry for confusing you.
$endgroup$
– Arthur
yesterday
$begingroup$
@MrAP it's a matter of what you want to convey. $f$ is the name of the polynomial. $f(x)$ is the expression that defines the polynomial. At least that's how I think about it. $Bbb Z$ is the ring / set of integers, $Bbb Z[x]$ is the ring / set of polynomials with integer coefficients. Also, $R$ isn't a relation here. I did say "ring". Like $Bbb Z$, or $Bbb R$. If that isn't a term you've come across yet, then I am sorry for confusing you.
$endgroup$
– Arthur
yesterday
$begingroup$
By $R$ I meant the a relation. $R:Ato B$ is a subset of $Atimes B$. So similarly $f:Ato B$ should also be a subset of $Atimes B$ since a function is a relation. So $f$ denotes a set which represents a function. And normally we write that an element belongs to a set. Here as you have written, it would mean that a set belongs to another set which is normally not the case. That is what I am trying to express.
$endgroup$
– MrAP
yesterday
$begingroup$
By $R$ I meant the a relation. $R:Ato B$ is a subset of $Atimes B$. So similarly $f:Ato B$ should also be a subset of $Atimes B$ since a function is a relation. So $f$ denotes a set which represents a function. And normally we write that an element belongs to a set. Here as you have written, it would mean that a set belongs to another set which is normally not the case. That is what I am trying to express.
$endgroup$
– MrAP
yesterday
|
show 3 more comments
$begingroup$
I think the book made a typo and it isn't $m-l$ that divides $k$ (I don't see that as even true) but that $d$ divides $k$.
This follow as $d$ divides $kn$ but is relatively prime to $n$ so $d$ must divide $k$.
And hence, yes, $frac kd$ is an integer, for which we conclude $m-l equiv 0 pmod n$.
Note: This will not be true if $a, b, n$ will have a common divisor (other than $1$). Consider $8 equiv 20 mod 12$ but $2equiv 5 mod 12$ is .... wrong. (Although $2 equiv 5 pmod 3$.....)
--- I empathize. For a typo that is a doozy to make and utter destroys the intent of the proof.
....
$mathbb Z[x]$ means the set of all polynomials with integer coefficients.
So $f(x) in mathbb Z[x]$ means "Let $f(x)$ be a polynomial with integer coefficients.
answered yesterday
fleabloodfleablood
72.9k22789
$endgroup$
$begingroup$
$(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
$endgroup$
– MrAP
yesterday
$begingroup$
IF $d$ and $n$ are relatively prime (i.e. $gcd(d,n)=1$) then $d$ and $n$ have no factors in common. So if $(m-l)d = kn$ then $d$ divides $kn$. But as $d$ and $n$ have nothing in common, that means $d$ divides $k$. Likewise that means $n$ divides $(m-l)d$ but $n$ and $d$ have nothing in common so $n$ divides $m-l$..... BTW "$n$ divides $A$" and "$Aequiv 0 pmod n$" mean the exact same thing. And "$n$ divides $m-l$" and "$mequiv l pmod n$" and "$m-l equiv 0 pmod n$" all mean the same thing.
$endgroup$
– fleablood
yesterday
add a comment |
$begingroup$
I think the book made a typo and it isn't $m-l$ that divides $k$ (I don't see that as even true) but that $d$ divides $k$.
This follow as $d$ divides $kn$ but is relatively prime to $n$ so $d$ must divide $k$.
And hence, yes, $frac kd$ is an integer, for which we conclude $m-l equiv 0 pmod n$.
Note: This will not be true if $a, b, n$ will have a common divisor (other than $1$). Consider $8 equiv 20 mod 12$ but $2equiv 5 mod 12$ is .... wrong. (Although $2 equiv 5 pmod 3$.....)
--- I empathize. For a typo that is a doozy to make and utter destroys the intent of the proof.
....
$mathbb Z[x]$ means the set of all polynomials with integer coefficients.
So $f(x) in mathbb Z[x]$ means "Let $f(x)$ be a polynomial with integer coefficients.
answered yesterday
fleabloodfleablood
72.9k22789
$endgroup$
$begingroup$
$(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
$endgroup$
– MrAP
yesterday
$begingroup$
IF $d$ and $n$ are relatively prime (i.e. $gcd(d,n)=1$) then $d$ and $n$ have no factors in common. So if $(m-l)d = kn$ then $d$ divides $kn$. But as $d$ and $n$ have nothing in common, that means $d$ divides $k$. Likewise that means $n$ divides $(m-l)d$ but $n$ and $d$ have nothing in common so $n$ divides $m-l$..... BTW "$n$ divides $A$" and "$Aequiv 0 pmod n$" mean the exact same thing. And "$n$ divides $m-l$" and "$mequiv l pmod n$" and "$m-l equiv 0 pmod n$" all mean the same thing.
$endgroup$
– fleablood
yesterday
add a comment |
$begingroup$
I think the book made a typo and it isn't $m-l$ that divides $k$ (I don't see that as even true) but that $d$ divides $k$.
This follow as $d$ divides $kn$ but is relatively prime to $n$ so $d$ must divide $k$.
And hence, yes, $frac kd$ is an integer, for which we conclude $m-l equiv 0 pmod n$.
Note: This will not be true if $a, b, n$ will have a common divisor (other than $1$). Consider $8 equiv 20 mod 12$ but $2equiv 5 mod 12$ is .... wrong. (Although $2 equiv 5 pmod 3$.....)
--- I empathize. For a typo that is a doozy to make and utter destroys the intent of the proof.
....
$mathbb Z[x]$ means the set of all polynomials with integer coefficients.
So $f(x) in mathbb Z[x]$ means "Let $f(x)$ be a polynomial with integer coefficients.
answered yesterday
fleabloodfleablood
72.9k22789
$endgroup$
I think the book made a typo and it isn't $m-l$ that divides $k$ (I don't see that as even true) but that $d$ divides $k$.
This follow as $d$ divides $kn$ but is relatively prime to $n$ so $d$ must divide $k$.
And hence, yes, $frac kd$ is an integer, for which we conclude $m-l equiv 0 pmod n$.
Note: This will not be true if $a, b, n$ will have a common divisor (other than $1$). Consider $8 equiv 20 mod 12$ but $2equiv 5 mod 12$ is .... wrong. (Although $2 equiv 5 pmod 3$.....)
--- I empathize. For a typo that is a doozy to make and utter destroys the intent of the proof.
....
$mathbb Z[x]$ means the set of all polynomials with integer coefficients.
So $f(x) in mathbb Z[x]$ means "Let $f(x)$ be a polynomial with integer coefficients.
answered yesterday
fleabloodfleablood
72.9k22789
edited yesterday
answered yesterday
fleabloodfleablood
72.9k22789
answered yesterday
fleabloodfleablood
72.9k22789
answered yesterday
fleabloodfleablood
72.9k22789
72.9k22789
$begingroup$
$(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
$endgroup$
– MrAP
yesterday
$begingroup$
IF $d$ and $n$ are relatively prime (i.e. $gcd(d,n)=1$) then $d$ and $n$ have no factors in common. So if $(m-l)d = kn$ then $d$ divides $kn$. But as $d$ and $n$ have nothing in common, that means $d$ divides $k$. Likewise that means $n$ divides $(m-l)d$ but $n$ and $d$ have nothing in common so $n$ divides $m-l$..... BTW "$n$ divides $A$" and "$Aequiv 0 pmod n$" mean the exact same thing. And "$n$ divides $m-l$" and "$mequiv l pmod n$" and "$m-l equiv 0 pmod n$" all mean the same thing.
$endgroup$
– fleablood
yesterday
add a comment |
$begingroup$
$(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
$endgroup$
– MrAP
yesterday
$begingroup$
IF $d$ and $n$ are relatively prime (i.e. $gcd(d,n)=1$) then $d$ and $n$ have no factors in common. So if $(m-l)d = kn$ then $d$ divides $kn$. But as $d$ and $n$ have nothing in common, that means $d$ divides $k$. Likewise that means $n$ divides $(m-l)d$ but $n$ and $d$ have nothing in common so $n$ divides $m-l$..... BTW "$n$ divides $A$" and "$Aequiv 0 pmod n$" mean the exact same thing. And "$n$ divides $m-l$" and "$mequiv l pmod n$" and "$m-l equiv 0 pmod n$" all mean the same thing.
$endgroup$
– fleablood
yesterday
$begingroup$
$(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
$endgroup$
– MrAP
yesterday
$begingroup$
$(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
$endgroup$
– MrAP
yesterday
$begingroup$
IF $d$ and $n$ are relatively prime (i.e. $gcd(d,n)=1$) then $d$ and $n$ have no factors in common. So if $(m-l)d = kn$ then $d$ divides $kn$. But as $d$ and $n$ have nothing in common, that means $d$ divides $k$. Likewise that means $n$ divides $(m-l)d$ but $n$ and $d$ have nothing in common so $n$ divides $m-l$..... BTW "$n$ divides $A$" and "$Aequiv 0 pmod n$" mean the exact same thing. And "$n$ divides $m-l$" and "$mequiv l pmod n$" and "$m-l equiv 0 pmod n$" all mean the same thing.
$endgroup$
– fleablood
yesterday
$begingroup$
IF $d$ and $n$ are relatively prime (i.e. $gcd(d,n)=1$) then $d$ and $n$ have no factors in common. So if $(m-l)d = kn$ then $d$ divides $kn$. But as $d$ and $n$ have nothing in common, that means $d$ divides $k$. Likewise that means $n$ divides $(m-l)d$ but $n$ and $d$ have nothing in common so $n$ divides $m-l$..... BTW "$n$ divides $A$" and "$Aequiv 0 pmod n$" mean the exact same thing. And "$n$ divides $m-l$" and "$mequiv l pmod n$" and "$m-l equiv 0 pmod n$" all mean the same thing.
$endgroup$
– fleablood
yesterday
add a comment |
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$begingroup$
I think the book has a typo and instead of $m-l$ divides $k$" it is supposed to be "$d$ divides $k$". That fits with $(m-l)d = knimplies m-l = frac kd n$ which assumes $frac kd$ is an integer.... on the other hand, that is one heck of a type to make!
$endgroup$
– fleablood
yesterday