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

Did I make a mistake by ccing email to boss to others?

How do you justify more code being written by following clean code practices?

Echo with obfuscation

Do you waste sorcery points if you try to apply metamagic to a spell from a scroll but fail to cast it?

Proving a complicated language is not a CFL

How much do grades matter for a future academia position?

Unable to disable Microsoft Store in domain environment

SOQL query causes internal Salesforce error

Is there a distance limit for minecart tracks?

Consistent Linux device enumeration

Should I assume I have passed probation?

How do I prevent inappropriate ads from appearing in my game?

Identifying "long and narrow" polygons in with Postgis

Can I cause damage to electrical appliances by unplugging them when they are turned on?

Difference between shutdown options

Why is participating in the European Parliamentary elections used as a threat?

Can I run 125khz RF circuit on a breadboard?

What the heck is gets(stdin) on site coderbyte?

Mimic lecturing on blackboard, facing audience

What is this high flying aircraft over Pennsylvania?

How would a solely written language work mechanically

Is stochastic gradient descent pseudo-stochastic?

Review your own paper in Mathematics

Personal or impersonal in a technical resume



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













2












$begingroup$


There are two propositions in the chapter of Number Theory in my book, the proofs of which I am having trouble to understand.



enter image description here



enter image description here



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?










share|cite|improve this question









$endgroup$











  • $begingroup$
    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
    5 hours ago















2












$begingroup$


There are two propositions in the chapter of Number Theory in my book, the proofs of which I am having trouble to understand.



enter image description here



enter image description here



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?










share|cite|improve this question









$endgroup$











  • $begingroup$
    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
    5 hours ago













2












2








2





$begingroup$


There are two propositions in the chapter of Number Theory in my book, the proofs of which I am having trouble to understand.



enter image description here



enter image description here



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?










share|cite|improve this question









$endgroup$




There are two propositions in the chapter of Number Theory in my book, the proofs of which I am having trouble to understand.



enter image description here



enter image description here



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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









MrAPMrAP

1,18721432




1,18721432











  • $begingroup$
    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
    5 hours ago
















  • $begingroup$
    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
    5 hours ago















$begingroup$
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
5 hours ago




$begingroup$
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
5 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

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 $a_jb^j$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    @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
    5 hours ago











  • $begingroup$
    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
    27 mins ago



















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    $(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
    $endgroup$
    – MrAP
    4 hours ago











  • $begingroup$
    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
    1 hour ago










Your Answer





StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3155648%2fnumber-theory-problem-with-proofs%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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 $a_jb^j$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    @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
    5 hours ago











  • $begingroup$
    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
    27 mins ago
















4












$begingroup$

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 $a_jb^j$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    @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
    5 hours ago











  • $begingroup$
    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
    27 mins ago














4












4








4





$begingroup$

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 $a_jb^j$.






share|cite|improve this answer











$endgroup$



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 $a_jb^j$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 5 hours ago

























answered 5 hours ago









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
    5 hours ago










  • $begingroup$
    @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
    5 hours ago











  • $begingroup$
    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
    27 mins ago

















  • $begingroup$
    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
    5 hours ago










  • $begingroup$
    @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
    5 hours ago











  • $begingroup$
    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
    27 mins ago
















$begingroup$
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
5 hours ago




$begingroup$
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
5 hours ago












$begingroup$
@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
5 hours ago





$begingroup$
@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
5 hours ago













$begingroup$
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
27 mins ago





$begingroup$
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
27 mins ago












3












$begingroup$

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.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    $(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
    $endgroup$
    – MrAP
    4 hours ago











  • $begingroup$
    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
    1 hour ago















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    $(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
    $endgroup$
    – MrAP
    4 hours ago











  • $begingroup$
    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
    1 hour ago













3












3








3





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 5 hours ago

























answered 5 hours ago









fleabloodfleablood

72.8k22788




72.8k22788











  • $begingroup$
    $(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
    $endgroup$
    – MrAP
    4 hours ago











  • $begingroup$
    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
    1 hour ago
















  • $begingroup$
    $(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
    $endgroup$
    – MrAP
    4 hours ago











  • $begingroup$
    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
    1 hour ago















$begingroup$
$(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
$endgroup$
– MrAP
4 hours ago





$begingroup$
$(m-l)d=kn$ implies that $d$ divides $k$ and also $n$ divides $m-l$. Correct?
$endgroup$
– MrAP
4 hours ago













$begingroup$
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
1 hour ago




$begingroup$
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
1 hour ago

















draft saved

draft discarded
















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3155648%2fnumber-theory-problem-with-proofs%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Францішак Багушэвіч Змест Сям'я | Біяграфія | Творчасць | Мова Багушэвіча | Ацэнкі дзейнасці | Цікавыя факты | Спадчына | Выбраная бібліяграфія | Ушанаванне памяці | У філатэліі | Зноскі | Літаратура | Спасылкі | НавігацыяЛяхоўскі У. Рупіўся дзеля Бога і людзей: Жыццёвы шлях Лявона Вітан-Дубейкаўскага // Вольскі і Памідораў з песняй пра немца Адвакат, паэт, народны заступнік Ашмянскі веснікВ Минске появится площадь Богушевича и улица Сырокомли, Белорусская деловая газета, 19 июля 2001 г.Айцец беларускай нацыянальнай ідэі паўстаў у бронзе Сяргей Аляксандравіч Адашкевіч (1918, Мінск). 80-я гады. Бюст «Францішак Багушэвіч».Яўген Мікалаевіч Ціхановіч. «Партрэт Францішка Багушэвіча»Мікола Мікалаевіч Купава. «Партрэт зачынальніка новай беларускай літаратуры Францішка Багушэвіча»Уладзімір Іванавіч Мелехаў. На помніку «Змагарам за родную мову» Барэльеф «Францішак Багушэвіч»Памяць пра Багушэвіча на Віленшчыне Страчаная сталіца. Беларускія шыльды на вуліцах Вільні«Krynica». Ideologia i przywódcy białoruskiego katolicyzmuФранцішак БагушэвічТворы на knihi.comТворы Францішка Багушэвіча на bellib.byСодаль Уладзімір. Францішак Багушэвіч на Лідчыне;Луцкевіч Антон. Жыцьцё і творчасьць Фр. Багушэвіча ў успамінах ягоных сучасьнікаў // Запісы Беларускага Навуковага таварыства. Вільня, 1938. Сшытак 1. С. 16-34.Большая российская1188761710000 0000 5537 633Xn9209310021619551927869394п

Беларусь Змест Назва Гісторыя Геаграфія Сімволіка Дзяржаўны лад Палітычныя партыі Міжнароднае становішча і знешняя палітыка Адміністрацыйны падзел Насельніцтва Эканоміка Культура і грамадства Сацыяльная сфера Узброеныя сілы Заўвагі Літаратура Спасылкі НавігацыяHGЯOiТоп-2011 г. (па версіі ej.by)Топ-2013 г. (па версіі ej.by)Топ-2016 г. (па версіі ej.by)Топ-2017 г. (па версіі ej.by)Нацыянальны статыстычны камітэт Рэспублікі БеларусьШчыльнасць насельніцтва па краінахhttp://naviny.by/rubrics/society/2011/09/16/ic_articles_116_175144/А. Калечыц, У. Ксяндзоў. Спробы засялення краю неандэртальскім чалавекам.І ў Менску былі мамантыА. Калечыц, У. Ксяндзоў. Старажытны каменны век (палеаліт). Першапачатковае засяленне тэрыторыіГ. Штыхаў. Балты і славяне ў VI—VIII стст.М. Клімаў. Полацкае княства ў IX—XI стст.Г. Штыхаў, В. Ляўко. Палітычная гісторыя Полацкай зямліГ. Штыхаў. Дзяржаўны лад у землях-княствахГ. Штыхаў. Дзяржаўны лад у землях-княствахБеларускія землі ў складзе Вялікага Княства ЛітоўскагаЛюблінская унія 1569 г."The Early Stages of Independence"Zapomniane prawdy25 гадоў таму было аб'яўлена, што Язэп Пілсудскі — беларус (фота)Наша вадаДакументы ЧАЭС: Забруджванне тэрыторыі Беларусі « ЧАЭС Зона адчужэнняСведения о политических партиях, зарегистрированных в Республике Беларусь // Министерство юстиции Республики БеларусьСтатыстычны бюлетэнь „Полаўзроставая структура насельніцтва Рэспублікі Беларусь на 1 студзеня 2012 года і сярэднегадовая колькасць насельніцтва за 2011 год“Индекс человеческого развития Беларуси — не было бы нижеБеларусь занимает первое место в СНГ по индексу развития с учетом гендерного факцёраНацыянальны статыстычны камітэт Рэспублікі БеларусьКанстытуцыя РБ. Артыкул 17Трансфармацыйныя задачы БеларусіВыйсце з крызісу — далейшае рэфармаванне Беларускі рубель — сусветны лідар па дэвальвацыяхПра змену коштаў у кастрычніку 2011 г.Бядней за беларусаў у СНД толькі таджыкіСярэдні заробак у верасні дасягнуў 2,26 мільёна рублёўЭканомікаГаласуем за ТОП-100 беларускай прозыСучасныя беларускія мастакіАрхитектура Беларуси BELARUS.BYА. Каханоўскі. Культура Беларусі ўсярэдзіне XVII—XVIII ст.Анталогія беларускай народнай песні, гуказапісы спеваўБеларускія Музычныя IнструментыБеларускі рок, які мы страцілі. Топ-10 гуртоў«Мясцовы час» — нязгаслая легенда беларускай рок-музыкіСЯРГЕЙ БУДКІН. МЫ НЯ ЗНАЕМ СВАЁЙ МУЗЫКІМ. А. Каладзінскі. НАРОДНЫ ТЭАТРМагнацкія культурныя цэнтрыПублічная дыскусія «Беларуская новая пьеса: без беларускай мовы ці беларуская?»Беларускія драматургі па-ранейшаму лепш ставяцца за мяжой, чым на радзіме«Працэс незалежнага кіно пайшоў, і дзяржаву турбуе яго непадкантрольнасць»Беларускія філосафы ў пошуках прасторыВсе идём в библиотекуАрхіваванаАб Нацыянальнай праграме даследавання і выкарыстання касмічнай прасторы ў мірных мэтах на 2008—2012 гадыУ космас — разам.У суседнім з Барысаўскім раёне пабудуюць Камандна-вымяральны пунктСвяты і абрады беларусаў«Мірныя бульбашы з малой краіны» — 5 непраўдзівых стэрэатыпаў пра БеларусьМ. Раманюк. Беларускае народнае адзеннеУ Беларусі скарачаецца колькасць злачынстваўЛукашэнка незадаволены мінскімі ўладамі Крадзяжы складаюць у Мінску каля 70% злачынстваў Узровень злачыннасці ў Мінскай вобласці — адзін з самых высокіх у краіне Генпракуратура аналізуе стан са злачыннасцю ў Беларусі па каэфіцыенце злачыннасці У Беларусі стабілізавалася крымінагеннае становішча, лічыць генпракурорЗамежнікі сталі здзяйсняць у Беларусі больш злачынстваўМУС Беларусі турбуе рост рэцыдыўнай злачыннасціЯ з ЖЭСа. Дазволіце вас абкрасці! Рэйтынг усіх службаў і падраздзяленняў ГУУС Мінгарвыканкама вырасАб КДБ РБГісторыя Аператыўна-аналітычнага цэнтра РБГісторыя ДКФРТаможняagentura.ruБеларусьBelarus.by — Афіцыйны сайт Рэспублікі БеларусьСайт урада БеларусіRadzima.org — Збор архітэктурных помнікаў, гісторыя Беларусі«Глобус Беларуси»Гербы и флаги БеларусиАсаблівасці каменнага веку на БеларусіА. Калечыц, У. Ксяндзоў. Старажытны каменны век (палеаліт). Першапачатковае засяленне тэрыторыіУ. Ксяндзоў. Сярэдні каменны век (мезаліт). Засяленне краю плямёнамі паляўнічых, рыбакоў і збіральнікаўА. Калечыц, М. Чарняўскі. Плямёны на тэрыторыі Беларусі ў новым каменным веку (неаліце)А. Калечыц, У. Ксяндзоў, М. Чарняўскі. Гаспадарчыя заняткі ў каменным векуЭ. Зайкоўскі. Духоўная культура ў каменным векуАсаблівасці бронзавага веку на БеларусіФарміраванне супольнасцей ранняга перыяду бронзавага векуФотографии БеларусиРоля беларускіх зямель ва ўтварэнні і ўмацаванні ВКЛВ. Фадзеева. З гісторыі развіцця беларускай народнай вышыўкіDMOZGran catalanaБольшая российскаяBritannica (анлайн)Швейцарскі гістарычны15325917611952699xDA123282154079143-90000 0001 2171 2080n9112870100577502ge128882171858027501086026362074122714179пппппп

ValueError: Expected n_neighbors <= n_samples, but n_samples = 1, n_neighbors = 6 (SMOTE) The 2019 Stack Overflow Developer Survey Results Are InCan SMOTE be applied over sequence of words (sentences)?ValueError when doing validation with random forestsSMOTE and multi class oversamplingLogic behind SMOTE-NC?ValueError: Error when checking target: expected dense_1 to have shape (7,) but got array with shape (1,)SmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?solving multi-class imbalance classification using smote and OSSUsing SMOTE for Synthetic Data generation to improve performance on unbalanced dataproblem of entry format for a simple model in KerasSVM SMOTE fit_resample() function runs forever with no result