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I can't find the multiplicative inverse of 3? This is what I did.

6/3 * (MI of 3) (mod 6)
 6/3 (mod 6)
 A= 3
 B= ?
 M = 6
 (A * B) % M = 1
 (3 * B) % 6 =1

Some body please guide me how can I solve this?

Zulfi.

zak100
  • 177

2 Answers2

1

A number $m$ will have a multiplicative inverse mod $n$ if and only if $m$ and $n$ are coprime (have no common divisors except 1). This is clearly not the case for 3 and 6.

See section Multiplicative group of integers modulo m for example.

Student
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Let $w$ be the multiplicative inverse modulo $6$ of $3$.

That means $3w \equiv 1 \pmod 6$.

That means i) $6|3w - 1$ (definition) or alternatively that ii) $3w = 6k + 1$ for some integer $k$ (not quite the definition but something equivalent to the definition that IMO is more intuitive)

i) As $3|6$ that means $3|3w -1$ and $3|-1$.

ii) $3w = 6k + 1$ means $w = 2k + \frac 13$.

i) is clearly false. ii) means $w$ is not an integer??????

....

So what do we conclude?

We conclude ..... $3$ does not have a multiplicative inverse modulo $6$.

Which raises the two questions.

a) Why did we think it did? and

b) When does a number $m$ have a multiplicative inverse modulo $n$?

....

I think the answer to a) is Because we can say the words "multiplicative inverse o $3$ modulo $6$" in the same way we can say "a triangle with four sides", "the real square root of $-1$", "$7$ divided by $0$" or "the first seven even primes greater than $49$" . Just because we can say something doesn't mean it exists.

b) If $m$ has a multiplicative inverse modulo $n$ then there is an integer $w$ so that $wm \equiv 1 \pmod n$. Or in other words there is an integer $k$ so that $wm = kn + 1$ or $wm -kn = 1$.

By Bezout's Lemma, this occurs if and only if $\gcd(m,n) = 1$.

So......

$m$ has a multiplicative inverse modulo $n$ if and only if $m$ and $n$ are relatively prime.

fleablood
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