Basic modular arithmetic question
$begingroup$
If for two integers $a,b$, $a=b mod x$, what is $x$ in terms of $a$ and $b$?.
I think the answer is $a-b$, but I'm not sure how to prove it without modular arithmetic, which I don't really understand.
Is there a simple way to prove and this without mods?
elementary-number-theory
asked Dec 7 '18 at 4:22
James PalmerJames Palmer
34
$endgroup$
add a comment |
$begingroup$
If for two integers $a,b$, $a=b mod x$, what is $x$ in terms of $a$ and $b$?.
I think the answer is $a-b$, but I'm not sure how to prove it without modular arithmetic, which I don't really understand.
Is there a simple way to prove and this without mods?
elementary-number-theory
asked Dec 7 '18 at 4:22
James PalmerJames Palmer
34
$endgroup$
add a comment |
$begingroup$
If for two integers $a,b$, $a=b mod x$, what is $x$ in terms of $a$ and $b$?.
I think the answer is $a-b$, but I'm not sure how to prove it without modular arithmetic, which I don't really understand.
Is there a simple way to prove and this without mods?
elementary-number-theory
asked Dec 7 '18 at 4:22
James PalmerJames Palmer
34
$endgroup$
If for two integers $a,b$, $a=b mod x$, what is $x$ in terms of $a$ and $b$?.
I think the answer is $a-b$, but I'm not sure how to prove it without modular arithmetic, which I don't really understand.
Is there a simple way to prove and this without mods?
elementary-number-theory
elementary-number-theory
asked Dec 7 '18 at 4:22
James PalmerJames Palmer
34
asked Dec 7 '18 at 4:22
James PalmerJames Palmer
34
edited Dec 7 '18 at 4:58
James Palmer
asked Dec 7 '18 at 4:22
James PalmerJames Palmer
34
asked Dec 7 '18 at 4:22
James PalmerJames Palmer
34
asked Dec 7 '18 at 4:22
James PalmerJames Palmer
34
34
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This follows directly from the definition of congruence with respect to modular arithmetic (there's not much you need to know):
$$a equiv b pmod{x} ;;; Leftrightarrow ;;; x | (a-b)$$
(This latter expression, if you're not familiar, just means "$x$ evenly divides $(a-b)$," or, equivalently, "$(a-b)$ has no remainder when divided by $x$.")
The problem is that, in reality, this means that, for some integer $k$,
$$frac{a-b}{x} = k ;;; Leftrightarrow ;;; x = frac{a-b}{k}$$
To say that $x = a-b$ (i.e. that $k=1$) means that there's some sort of context we're missing that would apply in this context. As presented, however, I think this is as far as one can get.
answered Dec 7 '18 at 4:28
Eevee TrainerEevee Trainer
7,67221338
$endgroup$
$begingroup$
Usepmod{x}to get $aequiv b pmod{x}$.bmod, will produce the binary operator:bbmod xproduces $bbmod x$. What you want to avoid ismod, because the spacing is wrong.
$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:30
$begingroup$
I just was straight up using text{mod}, but noted.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:33
$begingroup$
The three different modular commands are the second most common LaTeX mistake people make, I try to spread the knowledge where I can. (The most common is to use<and>instead oflangleandrangle)
$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:34
$begingroup$
Thanks! I think a better phrasing is "If two integers a,b have the same remainder when divided by x, what is x in terms of a and b?" I think I used the mod wrong, sorry!
$endgroup$
– James Palmer
Dec 7 '18 at 4:35
$begingroup$
I honestly just don't know much LaTeX, really. Most of what I know is self-taught or seen from experience, like using WYSIWYG editors and such. This site has at least been helpful in improving my LaTeX lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:35
add a comment |
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1 Answer
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1 Answer
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$begingroup$
This follows directly from the definition of congruence with respect to modular arithmetic (there's not much you need to know):
$$a equiv b pmod{x} ;;; Leftrightarrow ;;; x | (a-b)$$
(This latter expression, if you're not familiar, just means "$x$ evenly divides $(a-b)$," or, equivalently, "$(a-b)$ has no remainder when divided by $x$.")
The problem is that, in reality, this means that, for some integer $k$,
$$frac{a-b}{x} = k ;;; Leftrightarrow ;;; x = frac{a-b}{k}$$
To say that $x = a-b$ (i.e. that $k=1$) means that there's some sort of context we're missing that would apply in this context. As presented, however, I think this is as far as one can get.
answered Dec 7 '18 at 4:28
Eevee TrainerEevee Trainer
7,67221338
$endgroup$
$begingroup$
Usepmod{x}to get $aequiv b pmod{x}$.bmod, will produce the binary operator:bbmod xproduces $bbmod x$. What you want to avoid ismod, because the spacing is wrong.
$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:30
$begingroup$
I just was straight up using text{mod}, but noted.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:33
$begingroup$
The three different modular commands are the second most common LaTeX mistake people make, I try to spread the knowledge where I can. (The most common is to use<and>instead oflangleandrangle)
$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:34
$begingroup$
Thanks! I think a better phrasing is "If two integers a,b have the same remainder when divided by x, what is x in terms of a and b?" I think I used the mod wrong, sorry!
$endgroup$
– James Palmer
Dec 7 '18 at 4:35
$begingroup$
I honestly just don't know much LaTeX, really. Most of what I know is self-taught or seen from experience, like using WYSIWYG editors and such. This site has at least been helpful in improving my LaTeX lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:35
add a comment |
$begingroup$
This follows directly from the definition of congruence with respect to modular arithmetic (there's not much you need to know):
$$a equiv b pmod{x} ;;; Leftrightarrow ;;; x | (a-b)$$
(This latter expression, if you're not familiar, just means "$x$ evenly divides $(a-b)$," or, equivalently, "$(a-b)$ has no remainder when divided by $x$.")
The problem is that, in reality, this means that, for some integer $k$,
$$frac{a-b}{x} = k ;;; Leftrightarrow ;;; x = frac{a-b}{k}$$
To say that $x = a-b$ (i.e. that $k=1$) means that there's some sort of context we're missing that would apply in this context. As presented, however, I think this is as far as one can get.
answered Dec 7 '18 at 4:28
Eevee TrainerEevee Trainer
7,67221338
$endgroup$
$begingroup$
Usepmod{x}to get $aequiv b pmod{x}$.bmod, will produce the binary operator:bbmod xproduces $bbmod x$. What you want to avoid ismod, because the spacing is wrong.
$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:30
$begingroup$
I just was straight up using text{mod}, but noted.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:33
$begingroup$
The three different modular commands are the second most common LaTeX mistake people make, I try to spread the knowledge where I can. (The most common is to use<and>instead oflangleandrangle)
$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:34
$begingroup$
Thanks! I think a better phrasing is "If two integers a,b have the same remainder when divided by x, what is x in terms of a and b?" I think I used the mod wrong, sorry!
$endgroup$
– James Palmer
Dec 7 '18 at 4:35
$begingroup$
I honestly just don't know much LaTeX, really. Most of what I know is self-taught or seen from experience, like using WYSIWYG editors and such. This site has at least been helpful in improving my LaTeX lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:35
add a comment |
$begingroup$
This follows directly from the definition of congruence with respect to modular arithmetic (there's not much you need to know):
$$a equiv b pmod{x} ;;; Leftrightarrow ;;; x | (a-b)$$
(This latter expression, if you're not familiar, just means "$x$ evenly divides $(a-b)$," or, equivalently, "$(a-b)$ has no remainder when divided by $x$.")
The problem is that, in reality, this means that, for some integer $k$,
$$frac{a-b}{x} = k ;;; Leftrightarrow ;;; x = frac{a-b}{k}$$
To say that $x = a-b$ (i.e. that $k=1$) means that there's some sort of context we're missing that would apply in this context. As presented, however, I think this is as far as one can get.
answered Dec 7 '18 at 4:28
Eevee TrainerEevee Trainer
7,67221338
$endgroup$
This follows directly from the definition of congruence with respect to modular arithmetic (there's not much you need to know):
$$a equiv b pmod{x} ;;; Leftrightarrow ;;; x | (a-b)$$
(This latter expression, if you're not familiar, just means "$x$ evenly divides $(a-b)$," or, equivalently, "$(a-b)$ has no remainder when divided by $x$.")
The problem is that, in reality, this means that, for some integer $k$,
$$frac{a-b}{x} = k ;;; Leftrightarrow ;;; x = frac{a-b}{k}$$
To say that $x = a-b$ (i.e. that $k=1$) means that there's some sort of context we're missing that would apply in this context. As presented, however, I think this is as far as one can get.
answered Dec 7 '18 at 4:28
Eevee TrainerEevee Trainer
7,67221338
edited Dec 7 '18 at 4:33
answered Dec 7 '18 at 4:28
Eevee TrainerEevee Trainer
7,67221338
answered Dec 7 '18 at 4:28
Eevee TrainerEevee Trainer
7,67221338
answered Dec 7 '18 at 4:28
Eevee TrainerEevee Trainer
7,67221338
7,67221338
$begingroup$
Usepmod{x}to get $aequiv b pmod{x}$.bmod, will produce the binary operator:bbmod xproduces $bbmod x$. What you want to avoid ismod, because the spacing is wrong.
$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:30
$begingroup$
I just was straight up using text{mod}, but noted.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:33
$begingroup$
The three different modular commands are the second most common LaTeX mistake people make, I try to spread the knowledge where I can. (The most common is to use<and>instead oflangleandrangle)
$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:34
$begingroup$
Thanks! I think a better phrasing is "If two integers a,b have the same remainder when divided by x, what is x in terms of a and b?" I think I used the mod wrong, sorry!
$endgroup$
– James Palmer
Dec 7 '18 at 4:35
$begingroup$
I honestly just don't know much LaTeX, really. Most of what I know is self-taught or seen from experience, like using WYSIWYG editors and such. This site has at least been helpful in improving my LaTeX lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:35
add a comment |
$begingroup$
Usepmod{x}to get $aequiv b pmod{x}$.bmod, will produce the binary operator:bbmod xproduces $bbmod x$. What you want to avoid ismod, because the spacing is wrong.
$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:30
$begingroup$
I just was straight up using text{mod}, but noted.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:33
$begingroup$
The three different modular commands are the second most common LaTeX mistake people make, I try to spread the knowledge where I can. (The most common is to use<and>instead oflangleandrangle)
$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:34
$begingroup$
Thanks! I think a better phrasing is "If two integers a,b have the same remainder when divided by x, what is x in terms of a and b?" I think I used the mod wrong, sorry!
$endgroup$
– James Palmer
Dec 7 '18 at 4:35
$begingroup$
I honestly just don't know much LaTeX, really. Most of what I know is self-taught or seen from experience, like using WYSIWYG editors and such. This site has at least been helpful in improving my LaTeX lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:35
$begingroup$
Use
pmod{x} to get $aequiv b pmod{x}$. bmod, will produce the binary operator: bbmod x produces $bbmod x$. What you want to avoid is mod, because the spacing is wrong.$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:30
$begingroup$
Use
pmod{x} to get $aequiv b pmod{x}$. bmod, will produce the binary operator: bbmod x produces $bbmod x$. What you want to avoid is mod, because the spacing is wrong.$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:30
$begingroup$
I just was straight up using text{mod}, but noted.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:33
$begingroup$
I just was straight up using text{mod}, but noted.
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:33
$begingroup$
The three different modular commands are the second most common LaTeX mistake people make, I try to spread the knowledge where I can. (The most common is to use
< and > instead of langle and rangle)$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:34
$begingroup$
The three different modular commands are the second most common LaTeX mistake people make, I try to spread the knowledge where I can. (The most common is to use
< and > instead of langle and rangle)$endgroup$
– Arturo Magidin
Dec 7 '18 at 4:34
$begingroup$
Thanks! I think a better phrasing is "If two integers a,b have the same remainder when divided by x, what is x in terms of a and b?" I think I used the mod wrong, sorry!
$endgroup$
– James Palmer
Dec 7 '18 at 4:35
$begingroup$
Thanks! I think a better phrasing is "If two integers a,b have the same remainder when divided by x, what is x in terms of a and b?" I think I used the mod wrong, sorry!
$endgroup$
– James Palmer
Dec 7 '18 at 4:35
$begingroup$
I honestly just don't know much LaTeX, really. Most of what I know is self-taught or seen from experience, like using WYSIWYG editors and such. This site has at least been helpful in improving my LaTeX lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:35
$begingroup$
I honestly just don't know much LaTeX, really. Most of what I know is self-taught or seen from experience, like using WYSIWYG editors and such. This site has at least been helpful in improving my LaTeX lol
$endgroup$
– Eevee Trainer
Dec 7 '18 at 4:35
add a comment |
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