My online lessons insist that one reformulates answers so there are no irrationals in the divisor. So for example 3/ √2 gets rewritten as 3√2/ 2. Is this just a convention or is there a logical/mathematical reason why one should not have irrationals in the divisor?
ETA: “rationalizing the denominator”
It’s not even convention, really. There’s nothing wrong with an irrational in the denominator.
Knowing how to rationalize the denominator is a useful skill, to make some types of simplification easier. And in the case of math classes, I suspect that making it easier for the teacher to grade the tests/homework is no small advantage, and this is one way of forcing the students to present the results in a standard way.
But in general? 1 \over \sqrt{2} is totally fine.
According to blackpenredpen on YouTube, the practice started because it makes for easier calculation by hand. You can’t really do long division with an irrational number as the divisor (the number on the outside), as you’d have round it off. But you can have it as the dividend (the number on the inside) as you can just keep adding more digits.
Here’s a video he has on it:
Yes, @BigT is correct. With a slide rule or a calculator, it makes no difference. But you mostly do it with square roots. You can rationalize 1 divided by 1 + the cube root of 2 (mathjax failed me here), but it is probably not worth the effort.
Also, computing cube roots by “long division” seems pretty masochistic. I don’t remember the algorithm being taught in school beyond square roots, and even that was obviously more for fun.
mathjax test
Just iterate:
x' = \frac{1}{3}(2x + \frac{C}{x^2})
Ok, it’s not a “pure” long division algorithm, but the C \over x^2 requires a divide.
Yes, it’s not a no-no, but sometimes it’s desirable to “rationalize the denominator.”
As @BigT and his video point out, it made calculation easier in the pre-calculator days. You could look up, say, the square root of two on a table, to as many decimal places as you need, but it’s easier to divide a long decimal by an integer than to divide an integer by a long decimal. And even with a calculator available, there are still times when you want your denominators, specifically, to be as simple as possible.
Another thing is that sometimes you need a way to see if two differently-written numbers are really equal, and you want a “standard” form to put numbers into so you can say they’re the same number iff they look the same in standard form.
And sometimes, though not always, rationalizing the denominator genuinely does make the result look simpler.
The OP’s online lessons may insist on rationalizing the denominator because they’re too dumb to recognize all the multiple possible forms an answer might be in. Or they may want him to be able to rationalize the denominator because that’s arguably still a useful skill.
I didn’t learn the reason until I did my master’s in math.
In algebra, there are extension fields and they are important in how algebra works. For example Q(√2) is the extension field of all the numbers a + b √2 where a and b are rational numbers. If I have a fraction 5 ÷ (3 √2), that is not in the right form as the extension field requires but if I rewrite it as (5 √2) ÷ 6 or rather (5/6) √2 then it is in the form making it a member of Q(√2)
That’s what I was wondering, if somewhere down the road to higher mathematics it would eventually make a difference. Similar to how some pedants drill you that the commutative property notwithstanding, one distinguishes between a∙b versus b∙a because eventually in matrixes it does make a difference,
You could say something similar about the complex numbers (of the form a + b\sqrt{-1}), which are probably better known than Q (√2).
Any extension field really. The complex numbers are simply R(√-1). I used the square root of two to go along with the OP.
I think that this is the biggest reason. It’s nice to have a standard form to convert everything into. Exactly which form you choose to be the standard doesn’t matter much, but you have to pick something. It’s like asking if there’s a fundamental reason why we drive on the right: Ultimately, there probably is some reason, but any standard would work basically just as well, as evidenced by England doing just fine driving on the left.
All that said, once you get to calculus, you’ll find that sometimes, rationalizing the numerator will make it much easier to calculate some limits.
There’s a philosophical question in here. If two expressions share a value, are they the same number? Or, do they represent an isomorphism? That is, is Q(√2) the “same” as Q(1/√2)?
I think that would be verboten since the extension involves division.
That’s just the notation. The value of 1/√2 is just another irrational number.
But the notation involves division so I don’t think it would be allowed. Kind of like writing the word Þe. Yes it is technically the word “the” but it is not in the right form to be accepted as an English word.
You can just design your number system that way. For instance, take the set of all ordered pairs of integers (a, b). Then the rational numbers are the equivalence class of pairs such that (a, b) and (c, d) are equivalent if ad=bc. Probably not too much more work to do the same for a quadratic field.
Then your philosophical alignment appears to be “two expressions may share a value, but are not the same number”.
Yep, things like that. It’s interesting to me how much our concept of numbers is tied to the notations we use.
Er. \mathbf{Q}(\sqrt{2}) and \mathbf{Q}(1/\sqrt{2}), or eg \mathbf{Q}(\sqrt{2},\sqrt{3}) and \mathbf{Q}(\sqrt{2}-\sqrt{3}) are isomorphic, and in this case you can obviously identify them. But, generally speaking, an isomorphism is not enough for two abstract objects to be the “same”. For instance, if V is a finite-dimensional vector space then it is isomorphic to its dual V^*, e.g they have the same dimension, but there is no canonical way to identify then; whereas V is naturally isomorphic to V^{**} so they are the “same” as far as doing linear algebra.
\mathbf{Q}(\sqrt{2}) and \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) and \mathbf{Q}(e^{i \frac{\pi}{4}}) also the same? Do e and \pi leak into Q once we let i and \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.
I think.