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In algebra what does: "Is defined for" mean?

I have a question posted:

$\sqrt{a+b}$ is defined for $-b \leq a$.

The question posed is: Is this true...

My question: WHAT DOES "Is Defined For" mean??

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I'd say it means that you have a function defined by a formula $\sqrt{a+b}$ whose domain is those pairs $(a,b)$ with $-b\leq a$. – tomasz Apr 20 '14 at 9:53
Just a remainder: you can vote up every answer, which you find helpful. Then you can mark (by clicking on the $\checkmark$ to the left) the best answer. – quapka Apr 20 '14 at 10:15

If $\sqrt{a+b}$ is said to be defined for ${-b}\leq{a}$, then $\sqrt{a+b}$ is not necessarily defined for ${a}<{-b}$.

The reason for this, is that if ${a}<{b}$, then $\sqrt{a+b}$ could be the square root of a negative number, which is undefined (at least, when restricted to Real numbers).

So simply put, if a function is said to be defined for a certain range, then that means the function will provide a value for that range. And if the range is not satisfied, the function will quite possibly be undefined (that is, the function does not give a legitimate value), though without any further information we cannot be sure.

So if you're answering a question and you're given an expression and the range for which it is defined, you can only use said equation if the defined ranged is satisfied, as the expression may not be true for other values outside the range.

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Your first statement is wrong. Just because it is defined for $-b \le a$ does not imply that it is undefined for $a \lt -b$. Example: $\sqrt{a+b}$ is defined for $a \gt 0, b \gt 0$ but not undefined for eg. $a=-1,b=2$. – Keba Apr 20 '14 at 10:12
@Keba You're right, that was a complete oversight on my part. I have corrected my explanation. Thanks :) – BSnapZ Apr 20 '14 at 10:17

In the real number system, square roots are not defined for negative numbers. Hence when we consider $\sqrt{a+b}$, we require that $a+b$ be greater than or equal to zero.

$$a+b \geq 0 \Rightarrow -b \leq a.$$

In this particular case, the reason follows from the definition of the square root of a real number. Let $x$ be a real number. Then $\sqrt{x}$ is the (positive) number that when multiplied by itself equals $x$.

Recall that a negative number times a negative number is always a positive number. Hence there is no negative number, that when multiplied by itself produces a negative number. Therefore the square root of a negative number is not a real number, and this means that the square root of a negative number is not defined on the reals.

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During philosophy class, and independent of that specific class, a math professor simply claimed that "is defined for" is best understood as an axiom and a bridge for us to understand conditions. We can't really use logic to define it, since we defined logic through the use of "is defined for".

Another way of putting it is to go the other way around; we can express the proposition "is defined for" through the use of logical symbols, and we define "is defined for" through those.

In this context, you could probably translate it to "is only true for the condition of".

Edit: it appears I missunderstood the question - yes, I'm trying to present the idea of how to define defined

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I don't see a reason to down vote, on the other hand, this answer tries to explain more about the topic. – quapka Apr 20 '14 at 9:54
Yes, but the key word is "tries". (Although I definately agree it is not worth a downvote.) Perhaps O113 could explain this some more? – user1729 Apr 20 '14 at 10:25
+1 for using the word $philosophy$. I think what $O113$ is asking is "how do we define $define$?" – John Joy Apr 20 '14 at 10:52

The definition of $\sqrt{x}$ is "a real number whose square is x". Since there is a real number like that when $x \geq 0$, we can say $\sqrt{x}$ is defined for $x \geq 0$.

However, there is not a real number like that when $x \lt 0$, so we say $\sqrt{x}$ is undefined for $x \lt 0$. It's undefined because our definition doesn't cover that case (if there were a real number whose square was negative, it couldn't be 0 because 0*0 isn't negative, and it couldn't be a nonzero real number, because then its square would be positive). By the definition I gave for square root, $\sqrt{-1}$ doesn't have a meaning/doesn't exist/is undefined; the definition just doesn't work in this case.

(It turns out we can use a slightly different definition for square root that doesn't restrict it to real numbers and make the problem go away, using complex numbers. Then $\sqrt{-1}$ is defined and has a meaning.)

The definition gives meaning to the symbols, so if the definition doesn't apply in a particular case, we just have a meaningless jumble of symbols. We generally worry about whether something is defined only when it has the possibility of being undefined.

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a function $f(x)$ is defined for such values of $x$ means that $f(x)$ exists for such a values of $x$.

Ex: $f(x) = \sqrt{x}$. In this case, $f(x)$ is defined for all $x \geq 0 $. Notice if $x < 0$, then $f(x) = \sqrt{x}$ is not a real number. SO, the function exists(is defined) for positive values of $x$.

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It means, that the square root function $\sqrt{a+b}$ is properly defined, when you satisfy the given condition, in your example: $$-b\leq a$$ E.g. for $b = 3$ and $a = -5$, then the inequality is not satisfied, because $$ -b = -3 \nleq -5 = a. $$ When we work in $\mathbb{R}$ then square root $f(x)=\sqrt{x}$is not defined for $x < 0$. So with $b = 3$ and $a = -5$ the expression would be $\sqrt{a + b} =\sqrt{-5 + 3} = \sqrt{-2} $.

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From 1 vote up to -1? I was not being very formal, but tried to show a simple explanation. – quapka Apr 20 '14 at 18:44
I don't know why you got a downvote your answer is perfectly acceptable. – Zachary Selk Oct 27 '15 at 4:38

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