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I'm dealing with Paul Halmos' Linear Algebra Problem Book and I've found a problem already

The fourth exercise asks me to determine whether the following operation is compliant with the associative principle:

$$(α, β) · (γ, δ) = (αγ − βδ, αδ + βγ)$$

The answer says that it is, because:

$$(αγ − βδ)ε − (αδ + βγ)ϕ,(αγ − βδ)ϕ + (αδ + βγ)ε = α(γε − δϕ) − β(γϕ + δε), α(γϕ + δε) + β(γε − δϕ)$$

And the author adds: "By virtue of the associativity of the ordinary multiplication of real numbers the same eight triple products, with the same signs, occur in both these equations."

The thing is that I'm not being able to understand why this claim is true. I don't see "the same eight triple products with the same sign" occurring on both sides.

What am I taking wrong?

I tried to work through it with Latin letters: $$(a,b . x,y) . f,g\\ = (ax-by,ay+bx) . (f,g)\\ = f(ax-by)-g(ay+bx),g(ax-by)+f(ay+bx)\\ = afx-bfy-agy+bgx,agx-bgy+afy+bfx\\ \\~\\ a,b . (x,y . f,g)\\ = a,b . (xf-yg,xg+yf)\\ = a(xf-yg)-b(xg+yf),a(xg+yf)+b(xf-yg)\\ = afx-agy-bgx+bfy,agx+afy+bfx-bgy$$

But on the left element the summations seem to be different... I don't know whether I'm messing it up with the computations or whether there is a conceptual misunderstanding.

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    Apply the distributive property and then check term by term -- eg start with $\alpha \gamma \epsilon$ and go from there. – Alex K Mar 31 '24 at 21:16
  • For instance the 2nd of these 8 products in the left handside, $-(\beta\delta)\epsilon$, equals the 4th of the right handside, $ -\beta(\delta\epsilon)$. – Anne Bauval Mar 31 '24 at 21:28
  • To answer "why" it's associative, you might want to look again and see if you recognise the operation in the question. – Chris Lewis Mar 31 '24 at 21:45
  • Your $xg+yg$ is wrong. It should be $xg+yf$. – Ted Mar 31 '24 at 22:22
  • Check your algebra, you have flipped see some incorrect distributions of the minus signs in there. – user317176 Mar 31 '24 at 23:24

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It's best to look at each component, one at a time. I will also use English letters. The operation is defined as $$(a,b) \cdot (c,d) = (ac - bd, ad + bc).$$ Then $$\bigl((a,b) \cdot (c,d)\bigr) \cdot (e,f) = (ac - bd, ad + bc) \cdot (e,f).$$ The first component is $$(ac-bd)e - (ad+bc)f = ace-bde-adf-bcf. \tag{1}$$ The second component is $$(ac-bd)f + (ad+bc)e = acf - bdf + ade + bce. \tag{2}$$ Now we look at $$(a,b) \cdot \bigl((c,d) \cdot (e,f)\bigr) = (a,b) \cdot (ce - df, cf + de),$$ again by component. The first is $$a(ce - df) - b(cf+de) = ace - adf - bcf - bde. \tag{3}$$ The second is $$a(cf+de) + b(ce-df) = acf + ade + bce - bdf. \tag{4}$$ It is now easy to see that $(1) = (3)$ and $(2) = (4)$.

It so happens that this binary operation corresponds to multiplication in the field of complex numbers.

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