# An operation on a whole is equal to an operation on each part?

I've been pondering this question as to how and when you can perform an operation on a complete "unit" and the answer is the same when performing the operation on the individual parts of the "unit"

For example:

$$\sqrt{25\div4} = \sqrt{25}\div\sqrt{4}.$$ We took the square root operation and applied it to $25$ and $4$ separately to get the same result.

However this is not the case for:

$$(4+1)^2 \neq 4^2 + 1^2$$

However this is true again for:

$$\frac{2(7-6)}{2^2} = \frac{2(7)}{2^2} - \frac{2(6)}{2^2}.$$

We took multiplication by $2$ and division by $2^2$ and applied them to $7$ and $6$ individually.

My question is: Can you always distribute operations to the parts of a unit and get the same result as when you would perform the operation on the whole unit.

• Operations which allow this are called homomorphisms; you study them in group theory and abstract algebra. – Michael Burr Apr 19 '16 at 17:29
• Possibly helpful, possible duplicate: math.stackexchange.com/questions/1718013/… – Ethan Bolker Apr 19 '16 at 17:29
• I see, I haven't taken group theory or abstract algebra yet, but thank you for pointing me in the right direction – HDE K Apr 19 '16 at 17:31
• Doesn't your own counterexample $(4+1)^2\ne 4^2+1^2$ already give us a negative answer? – Hagen von Eitzen Apr 19 '16 at 17:32
• I meant to say not equal to each other, just learned how to use latex. So 4+1 squared = 25 while 4^2 + 1^2 = 17 – HDE K Apr 19 '16 at 17:33

As has been hinted in the comments, there is a deeper underlying structure here. But since you haven't been introduced to abstract algebra yet, I won't go down that road.

What I can say is that this does not hold for all operations. I would give a counter example, but you already gave one: $$(4+1)^2\neq4^2+1^2.\tag{1}$$

There is also no general rule for when it does hold. One of the reasons is that there is no definition of what a part of an expression is. For instance, in $(1)$ you consider "$4$" and "$1$" to be different part, but in your first example you also consider "$25$" and "$4$" to be differnt parts of ${25}\div4$. But in this case we do have that $$(25\div 4)^2=25^2\div4^2.\tag{2}$$

But when the operation is "taking powers" (I don't know if you learned this already), then we have: $$2^{25\div 4}\neq2^{25}\div2^{4}.\tag{3}$$

You wanted to know if we can

distribute operations to the parts of a problem and get the same result as performing an operation on a whole unit.

Examples $(1)$, $(2)$ and $(3)$ show us that this depends on

• what the operation is
• what we consider to be the "parts" of a unit.

The answer to your question will be different for each opereation and for each definition of "a part".

So you have to figure this out each time you learn about a new operation.

• Excellent response. I like to say that IN MATHEMATICS, NOTHING IS TRUE. Except when we have a proof that it’s true. OP’s teacher should have explained why $\sqrt a/\sqrt b=\sqrt{a/b}$. If(s)he didn’t, (s)he abdicated all responsibility for teaching. – Lubin Apr 19 '16 at 18:50