M2C2A: Episode 2, Irrational problems, rational answer

When you talk about irrational numbers, you normally think about π... or less normally, 1/3.

But 1/3 shouldn't exist for one thing: THERE IS NEVER A REMAINDER EQUAL TO 0 (zero)!

You can get 0.3333333333333..., but there´s always an extra three.
Another thing that is invalid: (1/3) x 3 ≠ 1, because (1/3) x 3 it´s only 0.999999999999...

But why do people use 1/3?

Because the difference between (1/3)x3 and 1 is of an extremely puny 1/∞!
So really, it doesn't actually matter a lot.
But it really isn't valid, and contradicts that (y/n) x n=y
And so do 6, 7, 9, 11, 12, 13, 14... In fact, any number bigger than 1 that isn't a power of 2, a power of 5, or divisible by 10 will contradict it!
…

But why powers of 2?

That depends on the sole fact that n/2 when n is an integer will have no more than 1 decimal place, and n/2 ^ x will have no more than x decimal places.
...

Why?

Because of the way divisions are made, and because:
Every even integer is divisible by 2, but with odd integers the only remainder possible is 1, you add the invisible 0 to get 10, divide by 2 to get 5 remainder 0.

 

4 is divisible by every 2 even numbers, and the possible remainders are 1, 2, 3 (and 0).
If the remainder is 1, add the invisible 0, 10/4=2 remainder 2, add the invisible 0, 20/4=5 remainder 0
If the remainder is 2, add the invisible 0, 20/4=5 remainder 0.
If the remainder is 3, add the invisible 0, 30/4=7 remainder 2, add the invisible 0, 20/4=5 remainder 0.
etc....
Try it yourself, see as far as you can get.
- The Roaring Thunder