# Why is dividing by an irrational a no-no in algebra? **URL:** \ **Category:** Factual Questions\ **Created:** [February 22, 2022, 11:12pm UTC](https://boards.straightdope.com/t/why-is-dividing-by-an-irrational-a-no-no-in-algebra/959975 "2022-02-22T23:12:18Z")\ **Posts on this page:** 5\ **Page:** 2
**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\ **Post date:** [February 24, 2022, 4:40am UTC](https://boards.straightdope.com/t/why-is-dividing-by-an-irrational-a-no-no-in-algebra/959975/21 "2022-02-24T04:40:17Z")
> [@DPRK](#): > > or eg Q(√2,√3) and Q(√2−√3) are isomorphic Neat. This was a little surprising at first, but with some playing around I see that (\sqrt{2} - \sqrt(3))^3 = 11\sqrt{2} - 9\sqrt{3}. And then (11\sqrt{2} - 9\sqrt{3}) - 9(\sqrt{2} - \sqrt{3}) = 2\sqrt{2}, and from there it’s obvious how to get the rest. ---
**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\ **Post date:** [February 24, 2022, 12:02pm UTC](https://boards.straightdope.com/t/why-is-dividing-by-an-irrational-a-no-no-in-algebra/959975/22 "2022-02-24T12:02:34Z")
> [@Francis\_Vaughan](#): > > Q(√2)\mathbf{Q}(\sqrt{2}) and \mathbf{Q}( \frac{1}{\sqrt{2} }) Q(1√2)\mathbf{Q}( \frac{1}{\sqrt{2} }) are the same, in that you have added the same irrational to Q, but have done so in a manner in which it is recoverable with the operations defined on Q. Since division is available in Q, an extending number, or its reciprocal give you the same extension. > > However are \mathbf{Q}(\sqrt{2}, i) Q(√2,i)\mathbf{Q}(\sqrt{2}, i) and \mathbf{Q}(e^{i \frac{\pi}{4}}) Q(eiπ4)\mathbf{Q}(e^{i \frac{\pi}{4}}) also the same? Do e ee and \pi π\pi leak into Q once we let i ii and \sqrt{2} √2\sqrt{2} into the extension? I can’t see how they can, so in that sense there are aspects of shared value and different notated value. Basically because the notation includes operations that can’t be found in other contexts. This is just the same game. On one hand, e^{\frac{\pi i}{4}} = \frac{1+i}{\sqrt{2}} by Euler’s formula. In the other direction, let j = e^{\pi i/4}. Then \begin{align} i &= j^2\\ \sqrt{2} &= \frac{(j+1)^4}{4j^2}-\frac{3}{2} \\ \end{align} ---
**Author:** ![Francis\_Vaughan](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/francis_vaughan/32/3093_2.png) [@Francis\_Vaughan](https://boards.straightdope.com/u/Francis_Vaughan)\ **Post date:** [February 24, 2022, 12:13pm UTC](https://boards.straightdope.com/t/why-is-dividing-by-an-irrational-a-no-no-in-algebra/959975/23 "2022-02-24T12:13:29Z")
> [@DPRK](#): > > This is just the same game. I agree. I thought that is what I meant. What I thought was interesting is that e and \pi don’t make it into the extended field. At least I couldn’t see how. So we can express \mathbf{Q}({\sqrt 2}, i) using i, e and \pi, but the last two are not part of the resultant extension of Q. So in some manner the notation we use is prevented from leaking into the field extension by the set of defined operations of the field. Which, in some manner makes the values and notated values distinct. We are required to evaluate the value in some sense before it passes into the extension. Yet we can pass the notated value unevaluated so long as the notation is restricted to operations defined on Q. Perhaps this is just plain obvious, but it was in answer to the question about equivalence of notated values. ---
**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\ **Post date:** [February 24, 2022, 12:24pm UTC](https://boards.straightdope.com/t/why-is-dividing-by-an-irrational-a-no-no-in-algebra/959975/24 "2022-02-24T12:24:21Z")
You are right; to hammer in this point we could say it does not matter what the element generating the field extension “is” in this case, only that it satisfies the minimal polynomial x^4+1. Exponentiation does not come into it; we could simply call it “j” and do all the algebra, the same way complex numbers use “i” such that “i² = -1” ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [February 24, 2022, 6:10pm UTC](https://boards.straightdope.com/t/why-is-dividing-by-an-irrational-a-no-no-in-algebra/959975/25 "2022-02-24T18:10:09Z")
If you are thinking of Q(\sqrt2) and Q(1/\sqrt2) as subfields of the reals, then of course they are the same. If you think of them abstractly as Q[x]/(x^2-2) and Q[x]/(2x^2-1), then they are different, isomorphic but different because the ideal (x^2-2) is not the same as the ideal (2x^2-1). It reminds me of the distinction between rational numbers and fractions. As fractions 1/2 and 2/4 are not the same, but they represent the same rational number. Is it a useful distinction to make? No. [Previous page](https://boards.straightdope.com/t/why-is-dividing-by-an-irrational-a-no-no-in-algebra/959975.md?page=1)