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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.

gebruiker
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HDE K
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    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: http://math.stackexchange.com/questions/1718013/why-is-the-square-root-of-a-sum-not-equal-to-the-square-root-of-each-its-addends/1718049#1718049 – 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
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    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
  • @HDEK Exactly. And therefore "Can you always distribute operations to the parts of a problem and get the same result as performing an operation on a whole unit?" - "No" – Hagen von Eitzen Apr 19 '16 at 17:46
  • You mentioned you just leaned LateX. If you are ever unsure on how to use LateX on this site you can look here. – gebruiker Apr 19 '16 at 18:18
  • If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this. – gebruiker Apr 26 '16 at 08:42

1 Answers1

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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.

gebruiker
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    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