Intuition of units: 6 chocolates $\div$ 2 friends = 6 $\times \frac 1 2$ units of what?

Question

I understand that $ \frac {6 \text{ chocolates}}{2 \text{ friends}} = 3 \text { chocolates per friend}$, but what are the units involved if I rearrange this as follows?

$$6 \times \frac 1 2 = 3$$

$6$ still seems to be chocolates, $3$ seems to be $\frac{\text{chocolates}}{\text{friends}}$, but $\frac 1 2$ doesn't seem to be friends.

Answer 1

In the second expression the units of $1/2$ are $\text{friends}^{-1}$.

In general $$ \frac{1}{x \text{ units}} = \frac{1}{x} \text{ units}^{-1}. $$ The $1$ in that expression is dimensionless.

Answer 2

The half would be in units of per friend or $(\mathrm{friend})^{-1}$. When you separate it out like this, the meaning isn't as clear, but this term would be cancelled out if you multiplied your answer by something with units of friends to obtain how many chocolates you would need to have.

Answer 3

First we must consider what this intuitively means. Let's say I have 10 chocolates, and 5 friends, and I want to distribute my chocolates evenly to each friend. Well, division provides us a clean and efficient method as to the number of chocolates each friend receives. More so, units are applied as to simply give meaning to the question, chocolates/friend implies I am giving some $constant$ real amount of chocolate to $each$ friend.

More mathematically, we can compute this by dividing the number of chocolates by the amount of friends, that is $\frac {10 \ chocolates}{2 \ friends}=\frac{5 \ chocolates}{1 friend}$. As you can see, the given units, chocolates and friends respectively, are in their own sense algebraic, as they obey the fundamental axioms/properties of math. For example, the division of units, follows analogously to the division of numbers, but without a numerical result.

In regards to your specific question, $\frac {6 \ chocolates}{2friends} = \frac 62(\frac {chocolates}{friends})=\frac {3 \ chocolates}{1 \ friends}$.

As explained above, units and numbers act analogously with eachother, yet don't effect one another when preforming computation. Therefore, the final result should be thought of as: you are giving 3 of your 6 chocolates, to $one$ friend $each$.