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What is the property of addition that allows $(a_1, a_2) + (b_1, b_2) = (a_1 + b_1, a_2 + b_2)$?

My research indicates that this may be due to the associative property of addition, but I haven't encountered any examples showing associativity of addition with ordered pairs.

I would greatly appreciate it if someone could please clarify this for me.

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    I think it's from vector space definitions. – 2017-01-27
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    @MyGlasses But it would still need to follow some property of addition, right? – 2017-01-27
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    When working with vectors, you _define_ this to be addition. Then, using the fact that _regular_ addition is commutative, associative, and so on, you show that this is the case also for the new vector addition. – 2017-01-27
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    @Arthur oh, I see. Thanks. – 2017-01-27
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    Have a look at: http://mathworld.wolfram.com/VectorAddition.html – 2017-01-27
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    "regular" addition only works and numbers. An ordered pair is not a number. But we can define an operation called PAIRPLUS as $(a,b) PAIRPLUS (c,d) := (a +c, b+d)$. We want PAIRPLUS to act as a "kinda" addition and as ordered pairs are not number there no ambiguity, we notate PAIRPLUS as $(a,b) + (c,d)$ means $(a,b) PAIRPLUS(c,d)$. Then we formally describe what we want a "kinda" addition to mean exactly, and we show that PAIRPLUS does everything we want. Generally because $a+c$ and $b+d$ does "everything we want" $(a+c,b+d)$ will too. – 2017-01-27
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    @fleablood this makes sense. Thank you :) – 2017-01-27
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    Please don't close this question. It has insightful information, and I would like to keep it open for future reference. – 2017-01-27

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The property is "Definition", we define addition to work that way and to be done component wise. Why? Because it is convinient and useful to us. After that we use the properties of addition (in each component) to show that it has all the properties it does have.

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    Are the properties of addition (in this case, within each component) defined by the field (say, real numbers or complex numbers), or are the properties of addition/multiplication defined outside (independent) of fields? – 2017-01-27
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    Outside and independent of fields. You could define $(a,b)\oplus (c,d)$ to have instead been something like $(a+c,b\cdot d)$ and it worked just as well. – 2017-01-27
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    What he said, it is all about the definitions and can be through anything. – 2017-01-27
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    @JMoravitz Thank you! I understand now. :) – 2017-01-27