The real math folks generally ignore them I guess because the easy ones are not worth their time, leaving said easy ones to dilettantes like me, who then look smarter-than-I-really-am to the noobies.
Love the easy ones.
So, to the question: How many two-fifths are in one?
Well now. How shall we handle this? Hmm…
I know! Let's count the fifths (just numbers, no booze), because it's probably as good a way of getting to an answer as any other way, the question is about fifths, and counting the fifths will at least give us a picture of what we're dealing with here.
Let's jump ahead, to see where we want to end up:
We want to end up with five fifths, also known as 5/5, because 5/5 is the same as 1. How do I know that 5/5 is the same as 1? Well, I know (you'll just have to trust me on this) because saying “5/5” is the same as saying “5 divided by 5”, and saying “1/2” is the same as saying “1 divided by 2", and so on. This is true for all fractions.
So: if 5/5 is the same as 5 divided by 5, then 5/5 equals 1. (You can see why, right? Any number divided into itself is equal to one; also, five goes into five exactly one time!).
And counting the fifths,
1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 5/5 = 1
Or, to count them another way,
1/5 + 1/5 = 2/5
2/5 + 1/5 = 3/5
3/5 + 1/5 = 4/5
4/5 + 1/5 = 5/5
Now that we can see what's inside of that 5/5, let's count how many two-fifths are in five-fifths by adding 2/5 together until we get as close to 5/5 as possible without going over:
2/5 + 2/5 = 4/5
leaving just 1/5 remainder, meaning that there are two 2/5 in 5/5.
We can see that that is true and correct by looking at our little counting exercise above.
Now to make sure our reasoning is clear let's count our fifths again, this time slightly rearranging the way we write the numbers, and then taking things one step further:
(1/5 + 1/5) + (1/5 + 1/5) + 1/5 = 5/5
(2/5) + (2/5) + (1/5) = 5/5
And there it is: there are two 2/5 in 1!
So let's complete the addition for the sake of those like myself who need to see these things through to their furthest conclusion:
