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The book is now explaining the multiplicative inverses in modular arithmetic, how if a number s has an inverse,t,modulo n, then it follows that $s*t≡ 1$mod(n).

From that definition the book shows that there are some numbers in certain modulos that have inverses for example 8's inverse is 2 in mod(15).

However, it also shows that there are numbers which don't have an inverse, for example 3 in mod(15). The book proves it doing this:

Suppose j is the inverse of 3 mod(15).

$$3j ≡ 1 mod(15)\implies 5*3j≡5mod(15)$$ Since the remainders of each side of the modulo equation is the same: $$rem(15j,15)=rem(5,15)$$ And from this we have $0=5$ a contradiction.

From this I have this question, if I multiply both sides by 15, would it be the same as in a normal equation multiplying by 0 and then not proving anything?

Jery Lazman
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  • Yes, $,\color{#c00}{n\equiv 0},\Rightarrow, \color{#c00}n\cdot a\equiv \color{#c00}0\cdot a\equiv 0,$ by the Congruence Product Rule in the linked dupe. $\ \ $ – Bill Dubuque Jul 20 '24 at 07:17
  • Or $,\ \color{#c00}3j = 1+\color{#c00}3\cdot 5\ \overset{!\bmod \color{#c00}3}\Longrightarrow\ 0\equiv 1\pmod{!\color{#c00}3},,$ i.e. $,3\mid 1,,$ i.e. if $,3,$ is invertible mod $,\color{#0af}3\cdot 5,$ then $,3,[\equiv\color{darkorange} 0:!],$ would be invertible $!\bmod\color{#0af} 3,,$ i.e. congruences (so inverses) persist mod factors of the modulus. $\ \ $ – Bill Dubuque Jul 20 '24 at 07:36
  • More generally this is a special case of polynomial root persistence mod factors of the modulus - see here. Here the polynomial is $,f(x) = 3x-1.\ \ $ – Bill Dubuque Jul 20 '24 at 07:53

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