Surds help please? Watch

XShmalX
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#1
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How would I justify that surds in the form a+b√2+c√3+d√6 are closed under + - * /. All letter being rational of course!
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generalebriety
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Well, + and - are fairly easy. (Given two numbers in that forum, show that their sum / difference is of that form.) * also isn't too hard by the same method. / is a lot harder, but all you need to show is that 1/(a+b√2+c√3+d√6) is of that form (because dividing by a+b√2+c√3+d√6 is the same as multiplying by 1/(a+b√2+c√3+d√6)) - can you see how you might do that? How do you normally rationalise a denominator? Fiddle around with it, it's not easy.
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nuodai
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You have to show that the sum, difference, product and quotient of two numbers in the form a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} can also be written in the form a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}.

I'll do the easiest (addition) for you:
\newline (a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) + (p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6})\newline

= (a+p) + (b+q)\sqrt{2} + (c+r)\sqrt{3} + (d+s)\sqrt{6}

Now you have to do the same for the other three operators.

Simples *squeak*
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XShmalX
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(Original post by generalebriety)
Well, + and - are fairly easy. (Given two numbers in that forum, show that their sum / difference is of that form.) * also isn't too hard by the same method. / is a lot harder, but all you need to show is that 1/(a+b√2+c√3+d√6) is of that form (because dividing by a+b√2+c√3+d√6 is the same as multiplying by 1/(a+b√2+c√3+d√6)) - can you see how you might do that? How do you normally rationalise a denominator? Fiddle around with it, it's not easy.
Normally rationalise the denominator by timesing by the congregate.
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generalebriety
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(Original post by XShmalX)
Normally rationalise the denominator by timesing by the congregate.
Conjugate. Yes. The conjugate isn't so obvious here, but there are a few obvious guesses you can make - try a few things and see what comes out. You are basically looking for an expression (of the form a+b√2+c√3+d√6) whose product with (p+q√2+r√3+s√6) is rational - instead of trying to do all that in one go, just try to get rid of one or two of the surds.
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XShmalX
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(Original post by nuodai)
You have to show that the sum, difference, product and quotient of two numbers in the form a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} can also be written in the form a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}.

I'll do the easiest (addition) for you:
\newline (a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}) + (p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6})\newline

= (a+p) + (b+q)\sqrt{2} + (c+r)\sqrt{3} + (d+s)\sqrt{6}

Now you have to do the same for the other three operators.

Simples *squeak*
So - is basically the same right??
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nuodai
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(Original post by XShmalX)
So - and x are basically the same right??
More or less yes. Minus is just as easy as plus really. Times is slightly more complicated because you have to simplify \sqrt{2}\sqrt{3} and \sqrt{6}\sqrt{6} (etc.) when you expand the brackets, but it's still relatively straightforward. Division's more complicated though, because you have to rationalise the denominator.
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XShmalX
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I think I'm just stuck on division now and getting that conjugate :confused:
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XShmalX
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Any clues??? :confused:
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ghostwalker
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(Original post by XShmalX)
Any clues??? :confused:
As previously suggested, don't try and eliminate them all at once.

And, this might be too vague, but I can't think of another. Remember
\sqrt{6}=\sqrt{2}\times\sqrt{3} when considering conjugacy.
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nuodai
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(Original post by XShmalX)
I think I'm just stuck on division now and getting that conjugate :confused:
Say you have \dfrac{a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}}{p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6}}. I'd start by taking (p + q\sqrt{2}) + (r\sqrt{3} + s\sqrt{6}) as if it were A + B\sqrt{C} (although strictly not true). Multiplying top and bottom by (p + q\sqrt{2}) - (r\sqrt{3} + s\sqrt{6}) gives a denominator of (p² + 2pq \sqrt{2} + 2q²) - (3r² + 6rs \sqrt{2} + 6s²), which is (p² + 2q² - 3r² - 6s²) + (6rs - 2pq)\sqrt{2}. As you can see, this is now in the form A + B\sqrt{C}, so finding the conjugate of this is more simple (but the algebra that ensues is tedious -- you have to multiply top and bottom by its conjugate).
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XShmalX
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(Original post by ghostwalker)
As previously suggested, don't try and eliminate them all at once.

And, this might be too vague, but I can't think of another. Remember
\sqrt{6}=\sqrt{2}\times\sqrt{3} when considering conjugacy.
Oh so like do two separate conjugations??
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XShmalX
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I'm sorry, I'm just completely lost on this division one...
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ghostwalker
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Nuodai has actually posted you a full working of the first part.
Reducing it to the form a+b\sqrt{2} in the denominator. From there it's the standard conjugacy method, multiply top and bottom by a-b\sqrt{2}.

Slightly different to what I'd intended, but in form only. And as he's taken the trouble to do all the Latex, I'd go with that.

Edit:

I was going for (p + q\sqrt{2}) + (r\sqrt{3} + s\sqrt{6})

=(p + q\sqrt{2}) + \sqrt{3}(r + s\sqrt{2})

and then using:

=(p - q\sqrt{2}) + \sqrt{3}(r - s\sqrt{2}) as the multiplier.
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generalebriety
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The theory behind why this works is actually quite advanced...

Spoiler:
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But I'll give it a go anyway, just in case anyone's interested.

We have taken a set Q (the set of all rational numbers), and appended some extra numbers to it - namely, √2, √3 and √6 - in a special way, so that we're allowed to add them and multiply them and so on. The resultant set (i.e. the set of all numbers that can be made from products and sums of rational numbers, √2, √3 and √6, which is the set you're dealing with) can be written Q[√2, √3, √6].

However, as ghostwalker points out, this can all actually be generated using √2 and √3 alone, because √6 = √3 * √2, and so a + b√2 + c√3 + d√6 can equally legitimately be written (a + b√2) + (c + d√2)√3. This means that we can write this number as p + q√3, for some p and q in the set Q[√2] (where the notation means the obvious thing from what I explained before - the set of rationals with √2 appended in the same way). Following me so far?

So, in essence, we've taken Q, appended a √2 to it to get the set Q[√2], and then appended a √3 to that to get the set (Q[√2])[√3], which is equal to the set Q[√2, √3, √6] - but this observation will come in handy to us in a minute. What we want to show is that Q[√2, √3, √6] is closed under division, and so it's enough to show that any number of the form 1/(a + b√2 + c√3 + d√6) can be written in the form (p + q√2 + r√3 + s√6) (do you see why?). That is, we want to show that any number in (Q[√2])[√3] also has its reciprocal in (Q[√2])[√3].

Now, it's easy to show that Q[√2] is closed under the four arithmetic operations - in particular, any number in Q[√2] also has its reciprocal in Q[√2]. So now let's look at (Q[√2])[√3], i.e. the set of numbers p + q√3 where p and q are in Q[√2]. Showing this set is closed under +, -, * is easy as always; showing it's closed under / is just a case of showing that (p - q√3) is a 'conjugate' of (p + q√3), i.e. (p + q√3)(p - q√3) is some number not involving √3, i.e. it's some number in Q[√2]. So we can divide by (p + q√3) if we can divide by numbers in Q[√2], because we can always multiply top and bottom by (p - q√3) to reduce it to something of that form. But we already know Q[√2] is closed - so we're done.


Now, what consequences does this have? Well, we want to know that we can divide by (a + b√2 + c√3 + d√6). When we multiply top and bottom by (a + b√2 - c√3 - d√6), we get back something of the form (e + f√2) - and then we can just multiply top and bottom by (e - f√2) to get something rational. And we're done. The algebra is horrendous, but it's a simple two-step process.
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