Mixing
Help with Division Rules in Modular Arithmetic
$begingroup$
I was following a proof, the crux of which came down to understanding that
$$ frac{10^k}{10^h}equiv frac{-1}{1} mod(p)$$ for some $ h < k $ and prime $p$, provided that
$ {10^k}equiv {-1} mod(p)$ and $ {10^h}equiv {1} mod(p)$
are already known to be true.
I know that there are very specific rules for modular division (although I am ignorant to them), so I tried out another example
$$ frac{4^3}{4^2}equiv frac{-1}{1} mod(5)$$
in which, sure enough, the "division" turned out to be true. Please help me out, what the heck is this sorcery?
number-theory
edited Jan 6 at 5:42
El borito
575216
asked Jan 6 at 5:09
Math EnthusiastMath Enthusiast
1168
$endgroup$
add a comment |
$begingroup$
I was following a proof, the crux of which came down to understanding that
$$ frac{10^k}{10^h}equiv frac{-1}{1} mod(p)$$ for some $ h < k $ and prime $p$, provided that
$ {10^k}equiv {-1} mod(p)$ and $ {10^h}equiv {1} mod(p)$
are already known to be true.
I know that there are very specific rules for modular division (although I am ignorant to them), so I tried out another example
$$ frac{4^3}{4^2}equiv frac{-1}{1} mod(5)$$
in which, sure enough, the "division" turned out to be true. Please help me out, what the heck is this sorcery?
number-theory
edited Jan 6 at 5:42
El borito
575216
asked Jan 6 at 5:09
Math EnthusiastMath Enthusiast
1168
$endgroup$
add a comment |
$begingroup$
I was following a proof, the crux of which came down to understanding that
$$ frac{10^k}{10^h}equiv frac{-1}{1} mod(p)$$ for some $ h < k $ and prime $p$, provided that
$ {10^k}equiv {-1} mod(p)$ and $ {10^h}equiv {1} mod(p)$
are already known to be true.
I know that there are very specific rules for modular division (although I am ignorant to them), so I tried out another example
$$ frac{4^3}{4^2}equiv frac{-1}{1} mod(5)$$
in which, sure enough, the "division" turned out to be true. Please help me out, what the heck is this sorcery?
number-theory
edited Jan 6 at 5:42
El borito
575216
asked Jan 6 at 5:09
Math EnthusiastMath Enthusiast
1168
$endgroup$
I was following a proof, the crux of which came down to understanding that
$$ frac{10^k}{10^h}equiv frac{-1}{1} mod(p)$$ for some $ h < k $ and prime $p$, provided that
$ {10^k}equiv {-1} mod(p)$ and $ {10^h}equiv {1} mod(p)$
are already known to be true.
I know that there are very specific rules for modular division (although I am ignorant to them), so I tried out another example
$$ frac{4^3}{4^2}equiv frac{-1}{1} mod(5)$$
in which, sure enough, the "division" turned out to be true. Please help me out, what the heck is this sorcery?
number-theory
number-theory
edited Jan 6 at 5:42
El borito
575216
asked Jan 6 at 5:09
Math EnthusiastMath Enthusiast
1168
edited Jan 6 at 5:42
El borito
575216
asked Jan 6 at 5:09
Math EnthusiastMath Enthusiast
1168
edited Jan 6 at 5:42
El borito
575216
edited Jan 6 at 5:42
El borito
575216
edited Jan 6 at 5:42
El borito
575216
575216
asked Jan 6 at 5:09
Math EnthusiastMath Enthusiast
1168
asked Jan 6 at 5:09
Math EnthusiastMath Enthusiast
1168
asked Jan 6 at 5:09
Math EnthusiastMath Enthusiast
1168
1168
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $10^hequiv1pmod p$ and $10^kequiv-1pmod p$ where $h<k$ then
$$-1equiv 10^k=(10^h)(10^{k-h})equiv(1)(10^{k-h})=10^{k-h}
pmod p.$$
Thus $10^{k-h}equiv-1pmod p$.
Note that I have used standard facts about modular multiplication
here, but nothing about "division".
answered Jan 6 at 5:16
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
Thank you very much! That looks much nicer haha
$endgroup$
– Math Enthusiast
Jan 6 at 5:19
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
If $10^hequiv1pmod p$ and $10^kequiv-1pmod p$ where $h<k$ then
$$-1equiv 10^k=(10^h)(10^{k-h})equiv(1)(10^{k-h})=10^{k-h}
pmod p.$$
Thus $10^{k-h}equiv-1pmod p$.
Note that I have used standard facts about modular multiplication
here, but nothing about "division".
answered Jan 6 at 5:16
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
Thank you very much! That looks much nicer haha
$endgroup$
– Math Enthusiast
Jan 6 at 5:19
add a comment |
$begingroup$
If $10^hequiv1pmod p$ and $10^kequiv-1pmod p$ where $h<k$ then
$$-1equiv 10^k=(10^h)(10^{k-h})equiv(1)(10^{k-h})=10^{k-h}
pmod p.$$
Thus $10^{k-h}equiv-1pmod p$.
Note that I have used standard facts about modular multiplication
here, but nothing about "division".
answered Jan 6 at 5:16
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
$begingroup$
Thank you very much! That looks much nicer haha
$endgroup$
– Math Enthusiast
Jan 6 at 5:19
add a comment |
$begingroup$
If $10^hequiv1pmod p$ and $10^kequiv-1pmod p$ where $h<k$ then
$$-1equiv 10^k=(10^h)(10^{k-h})equiv(1)(10^{k-h})=10^{k-h}
pmod p.$$
Thus $10^{k-h}equiv-1pmod p$.
Note that I have used standard facts about modular multiplication
here, but nothing about "division".
answered Jan 6 at 5:16
Lord Shark the UnknownLord Shark the Unknown
103k1160132
$endgroup$
If $10^hequiv1pmod p$ and $10^kequiv-1pmod p$ where $h<k$ then
$$-1equiv 10^k=(10^h)(10^{k-h})equiv(1)(10^{k-h})=10^{k-h}
pmod p.$$
Thus $10^{k-h}equiv-1pmod p$.
Note that I have used standard facts about modular multiplication
here, but nothing about "division".
answered Jan 6 at 5:16
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Jan 6 at 5:16
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Jan 6 at 5:16
Lord Shark the UnknownLord Shark the Unknown
103k1160132
answered Jan 6 at 5:16
Lord Shark the UnknownLord Shark the Unknown
103k1160132
103k1160132
$begingroup$
Thank you very much! That looks much nicer haha
$endgroup$
– Math Enthusiast
Jan 6 at 5:19
add a comment |
$begingroup$
Thank you very much! That looks much nicer haha
$endgroup$
– Math Enthusiast
Jan 6 at 5:19
$begingroup$
Thank you very much! That looks much nicer haha
$endgroup$
– Math Enthusiast
Jan 6 at 5:19
$begingroup$
Thank you very much! That looks much nicer haha
$endgroup$
– Math Enthusiast
Jan 6 at 5:19
add a comment |
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