Determine if S = $\{[a,b] | a, b \in \mathbb Q\}$ is a field under operations $+$ and $\cdot$ where $+$ is given by $[a_1, b_1] + [a_2 + b_2] = [a_1 + a_2, b_1+b_2]$ and $\cdot$ is given by $[a_1, b_1] \cdot [a_2, b_2] = [a_1a_2 - b_1b_2, a_1b_2 + b_1a_2], a_1, a_2, b_1, b_2 \in \mathbb Q$

Clearly closed under addition and multiplication

Since the components are $\in \mathbb Q$ then associative, commutative, and destributive laws follow.

additive identity: $[0,0]$

$[a_1, b_1] + [0,0] = [a_1, b_1]$, holds

multiplicative identity: $[1,0]$

$[a_1, b_1]\cdot[1,0] = [a_1, b_1]$, holds

additive inverse: $[-a_1, -b_1]$

$[a_1, b_1]+[-a_1, -b_1] = [0, 0]$, holds

multiplicative inverse: ??

Unsure how to do this part. I think I have to do this:

$$[a_1, b_1] \cdot [x, y] = [1, 0]$$

Not sure how to though.

  • 3
    $\begingroup$ Does $[a,b]$ mean closed interval? Also, your definition of $+$ is garbled. $\endgroup$ – Eric M. Schmidt Sep 17 '17 at 1:54
  • $\begingroup$ What do you mean? I put a line between the question and my attempt at it. I'm just trying to satisfy the properties for a field. $\endgroup$ – Tinler Sep 17 '17 at 1:58
  • $\begingroup$ By $[a,b]$ do you mean a closed interval from analysis, i.e., something like $[a,b]=\{x\in\mathbb{R}:a\leq x\leq b\},$ or do you mean an ordered pair of numbers, usually denoted $(a,b)$ (though, admittedly, this notation is also used to mean "open interval" in analysis)? $\endgroup$ – Will R Sep 17 '17 at 1:58
  • $\begingroup$ I'm not really sure, the question never labeled it. $\endgroup$ – Tinler Sep 17 '17 at 2:05
  • 1
    $\begingroup$ I think, here $[a,b]$ means ordered pair. In that case it is a field, (isomorphic to) $\mathbb{Q}(i)$. If this is true, then think $[a,b]$ as a complex number $a+ib$, and then try. You can easily findout the inverse in this way. $\endgroup$ – Krish Sep 17 '17 at 2:19

You need to solve $[a,b]*[x,y]=[ax-by,bx+ay]=[1,0] $

So solve $ax-by=1$


$bx=-ay $ so (assuming $b\ne 0$) $x=-\frac aby $

So $-\frac {a^2}by -by=1$ and $y=\frac -1 {\frac {a^2}b-b} =-\frac b {a^2-b^2} $.

And $x=-\frac ab*-\frac b {a^2-b^2}=\frac a {a^2-b^2} $

So $[a,b]^{-1}=[\frac a {a^2-b^2},-\frac b {a^2-b^2}] $

That should look very familiar.


If $b=0$ we get $ax=1$ and $ay=0$

And assuming $a\ne 0$ we get $x =\frac 1a =\frac a {a^2-b^2} $, and $y=0=-\frac b {a^2-b^2} $.

So the inverse formula still holds.


If $a=0,b=0$ then $[a,b] $ is the additive identity and can not have an inverse.

And indeed $[0,0]*[x,y]=[0,0]\ne [1,0] $


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