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Let $G$ be a set of ordered pairs $(a, b), \, a \neq 0$, where $a, b \in \mathbb{R}$, and the binary operation $*$ on $G$ is given by $$(a, b) * (c, d) = (ac, bc + d)$$ Then, the identity element $e = (g, h)$, is such that $$(a, b) * (g, h) = (ag, bg + h) = (a, b)$$ Let $*'$ denoted the binary operation $*$ on the individual coordinates in the ordered pair. Then, $ag = a \Rightarrow g = 1$ and $bg + h = b \Rightarrow b = -\dfrac{h}{g - 1}$, which implies that $e = \left(1, -\dfrac{h}{g - 1}\right)$.

My question would be, is the value of $e$ above the identity element of $G$ under $*$? Another thing I noticed is that if $g = 1$, then the second coordinate will be undefined. If the value of $e$ is correct, then do we have to apply this restriction?

  • Insert $g = 1$ into $bg + h = b$ before you start solving it. – Arthur Nov 13 '23 at 14:03
  • This is very hard to follow. The right identity element for your operation is $(1,0)$. Is that what you are asking. – lulu Nov 13 '23 at 14:04
  • @Arthur I was thinking about that, but I also thought the value of the $g$ in the two coordinates mayn't be the same. –  Nov 13 '23 at 14:08
  • @lulu I just wanted to know if the conclusion is wrong, for the value of $e$, and what I could've done to get the right answer. –  Nov 13 '23 at 14:10
  • $bg + h = b$ doesn't imply $b = -\frac{h}{g - 1}$, it implies that $b = -\frac{h}{g - 1}$ or $g = 1$, $b + h = b$. – mihaild Nov 13 '23 at 14:15

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You began with variables $g,h$ representing the coordinates of the identity. To find the identity, you must explicitly find $g$ and $h$; they must be specific real numbers.

You found two linear equations which $g$ and $h$ must satisfy: $$ag=a, bg+h=b$$ You correctly solved the first one, getting $g=1$. Now you have more information with which to solve the second one! Since $g=1$ you may (and should) plug in $1$ everywhere you see $g$. This transforms the second equation into $b+h=b$ which clearly implies $h=0$. Thus $e=(1,0)$.

Malady
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  • Thanks. I was thinking that the value of $g$ in the first coordinate mayn't be the same value of $g$ in the second coordinate –  Nov 13 '23 at 14:15
  • This is a failure of mathematical reasoning. $g$ is simply a placeholder here for a number. It takes one value. Practice solving systems of linear equations. – Malady Nov 13 '23 at 14:17