Question

Докажите что если каждое из натуральных чисел a и b делится на натуральное число c то верно равенство (a+b):c=a:c+b:c

Answer 1

(9+6)÷3=9÷3+6÷3=5 ясно

Answer 2

Proof of the Equality

To prove that if each of the natural numbers a and b is divisible by the natural number c, then the equality (a+b):c = a:c + b:c holds, we can use the properties of divisibility.

Let's start by expressing a and b in terms of c: - Since a is divisible by c, we can express a as a = k*c for some integer k. - Similarly, since b is divisible by c, we can express b as b = m*c for some integer m.

Now, let's substitute these expressions into the equality (a+b):c = a:c + b:c: - (a+b):c becomes (k*c + m*c):c which simplifies to ((k+m)*c):c. - a:c + b:c becomes (k*c):c + (m*c):c which simplifies to k + m.

So, we have ((k+m)*c):c = k + m.

This equality holds true because k + m is the same as (k+m)*c divided by c. Therefore, the equality (a+b):c = a:c + b:c is proven.

[[SOURCE #3]]