2
$\begingroup$

Let $\mathbf{S} = \{ (a_1, a_2): a_1, a_2 \in \mathbf{R} \}$. Where for $a = \left(a_1, a_2\right), b = \left(b_1, b_2\right) \in \mathbf{S}$ and $c \in \mathbf{R}$ $$a+b = (a_1, a_2) + (b_1, b_2) = (a_1 + b_1, 0)$$ and $$ca=(ca_1, 0)$$

In the book I am reading now it is written that $\mathbf{S}$ is not a vector space, because other axioms including the 4th axiom, i.e. "inverse element of addition", doesn't hold. But I can show that there exists such a vector. Let $a=(a_1, a_2) \in \mathbf{S}$ then $v = (v_1, v_2) \in \mathbf{S}$ where $v_1 = -a_1$ is an additive inverse of $a$: $$(a_1, a_2) + (v_1, v_2) = (a_1 + v_1, 0 ) = (a_1 - a_1, 0 ) = (0, 0)$$

Is it a typo in the book? Or I am missing something? Thanks in advance.

$\endgroup$
  • 3
    $\begingroup$ I suspect that the main problem is that there's no element which can act as zero: $(0,0)$ does not, in fact, act as an additive identity. Without an additive identity, we can't even really state the fourth axiom, never-mind decide whether it is true or false. How you choose to interpret that in context is up to you, though. $\endgroup$ – Chessanator Feb 10 '18 at 14:51
  • $\begingroup$ @Chessanator, yes, the same thing is written there. But I can't figure out why if 0 doesn't act as an identity, then inverse also doesn't exist $\endgroup$ – Turkhan Badalov Feb 10 '18 at 14:54
  • 1
    $\begingroup$ As far as the example given: $(-a_1,5)$ and $(-a_1,66)$ are both additive inverses of $(a_1,a2)$. The additive inverse ought to be unique. $\endgroup$ – zoli Feb 10 '18 at 14:54
  • $\begingroup$ @zoli, aha, thanks! But in wikipedia and even in the book it is not written that it should be unique. It is said that "there exists". Doesn't it imply that there should be at least one element acting as inverse? $\endgroup$ – Turkhan Badalov Feb 10 '18 at 14:56
  • 1
    $\begingroup$ @TurkhanBadalov What do you mean by "additive inverse"? The only thing I've ever used it for is "thing that you add to something to get additive identity." If there's no additive identity, then that definition doesn't even make sense. $\endgroup$ – Chessanator Feb 10 '18 at 15:01
1
$\begingroup$

I'll turn my comment into an answer:

I suspect that the main problem is that there's no element which can act as zero: $(0,0)$ does not, in fact, act as an additive identity. And once you've accepted that, what do you mean by "additive inverse"? The only thing I've ever used it for is "thing that you add to something to get additive identity." If there's no additive identity (nothing that we can actually call $0$, no matter how much like a $0$ it looks like) then that definition doesn't even make sense.

At this point there are three ways to interpret the fourth axiom and it's truth status:

  • $0$, as part of the signature of the logic, does refer to some particular element. It's just that, due to the failure of the earlier axiom, that element $0$ doesn't have the property we expect of it. Under this interpretation, axiom 4 is true but irrelevant.
  • $0$ refers to the element designated by the additive identity axiom. Since that axiom failed, $0$ was never even specified and the fourth axiom isn't properly stated, never-mind true or false.
  • The fourth axiom should be understood as stating that there is some element $-a$ such that $a + (-a)$ is an additive identity. Under this interpretation, axiom 4 is just false: there are no additive identities.

I lean towards the second interpretation. I guess your book prefers the third.

$\endgroup$
  • $\begingroup$ Could you please specify what is a "signature of the logic" and "never-mind true or false"? $\endgroup$ – Turkhan Badalov Feb 10 '18 at 15:21
  • $\begingroup$ "Signature" is something that you will come across later if you study formal logic, but it basically refers to the set of symbols that the logic uses. Under the first interpretation, $0$ is one of the symbols we're using so it has to mean something. I don't like this approach, because it tricks people into thinking that because you've called something $0$ it actually is zero, but I'll admit it's probably the most rigorous way to do it. $\endgroup$ – Chessanator Feb 10 '18 at 15:38
  • $\begingroup$ Thank you for the reply. I study symbolic logic but haven't met the term yet. But what about "never-mind true or false"? Is it something related to "vacuous truth"? $\endgroup$ – Turkhan Badalov Feb 10 '18 at 15:41
  • 1
    $\begingroup$ "Not true or false" might become clearer if I give you a more extreme example. Question: is "$++ \forall 0010 - (\exists =)$" true or false? Answer: Neither, it's just garbage. In the second interpretation, $0$ was never properly defined, so it's arguable that an axiom trying to use it is as meaningless as the garbage I just wrote. $\endgroup$ – Chessanator Feb 10 '18 at 15:41
  • 1
    $\begingroup$ I'm pretty inclined to swap that around: something is a vector if it lives in a vector space, whether or not it is a tuple. "$sin(x)$" is a vector, if you're doing Fourier analysis and focused on the vector space of wave-forms. $\endgroup$ – Chessanator Feb 10 '18 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.