#Is this correct?
45 messages · Page 1 of 1 (latest)
yes because modulo can be distributed and becomes congruent too eg
(nx10^a) mod 10 becomes congruent too n mod 10 . 10^a mod 10 whole mod 10 now as 10^a would be a multiple of 10 for any natural number a the whole mod becomes 0 mod 10
or 0
If n is a rational number but not an integer, is there at least one such n that holds for all natural numbers a?
Would hold for n with denominator of lower order of magnitude than 10^a in general, so if the denominator is > 1 then it won't hold for a = 1
In most reduced form
it does not matter if n is a rational number if a is a natural number it would always be valid
Otherwise any single digit denominator rational will work for a>2
modulo would get distributed and become congruent to
wait if denominator of n is 10^a then things might change
they would cancel
If it's of the same order of magnitude in general things change
10/3 = 3.3 for example
yes
yes
if denom(n) is of the form 10^k where k is a natural number >=a then things will change mod can become undefined or numer(n)mod 10
eg check this
,w (3/10^3)*(10^2) mod 10
Is 0.3 it's correct
let me explain if n is a rational number of the form a/b then a/b mod some number k can be written as ( (a mod k) (b mod inverse k) ) whole mod k if gcd(a,b) = 1 then this exists
hmm then 0.3is correct?
(3/10) mod 10 is not defined because in modular arithmetic k mod inverse k does not exist
do you need help in #help-6 ?
(3/10) mod 10 as gcd being 1 of (3,10) can be written as 3 mod 10 10 mod inverse 10 whole mod 10 but as 10 mod inverse 10 is not defined this modulo is not defined
Yes but (x) mod m is not modular arithmetic it's a binary operation yielding the remainder of division
You're thinking of x (mod m)
Which is congruence
man he has asked modular arithmetic
Where
For any integer n and any natural number a, (n x 10^a) mod 10 = 0.
see the bracket
Right, that's mod operation
(x) mod 10 or x % 10