Quote from O)FaRTy1billion[MM]

Quote from Vrael

but the actual field of real numbers is composed completely of just multiplication and addition.

I don't understand this and never have. Why just multiplication and addition? Where do exponents fit in to this? It is my understanding that multiplication is to addition as exponents are to multiplication ... i.e.: 4*3 = 4+4+4; 4^3 = 4*4*4. I would say addition is the simplest operation.

Then, on top of that, you get derivatives and integrals, which are to exponents what exponents are to multiplication.

Quote from O)FaRTy1billion[MM]

I don't know. D: My main concern is why don't exponents fit in to this. Do decimal exponents with just multiplication.
I'm just kinda oversimplifying going both ways, anyway...


My only way to do decimal multiplication is to simplify the decimals out ... but, of course, you can't represent that mathematically without multiplication itself.

I'm confused at what you're saying...you mean that you can't multiply x^.5 and x³ without simplifying? You're right...and it doesn't get any simpler. You keep the base, add the exponents (x^{.5 + 3}, and get x^3.5.

No, that would be incorrect. Integrals and derivatives are not numerical operations. You'd want this: http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation

I didn't quote the post I was responding to (It was Roy's).. but read it and then re-read mine. Context is helpful.



Post has been edited 6 time(s), last time on Apr 9 2011, 1:59 am by FaRTy1billion.

That's it baby. Almost a whole page of pointless discussion while the truth being glaringly obvious, once proof is found. Way to be a boss guys.

You cannot get an exact representation of sqrt(2) with a finite number of ANY operations.

The exponent should not be included as an elementary operator because it can be computed by using only multiplications and additions.

If I give you just a pen and paper and tell you to calculate 3^pi, how would you do it? There's only one way to do this and it involves an infinite number of multiplications and additions (the same goes for any irrational exponent).

Computers do it the same way. Whether it's log, exponent, the gamma function or whatever you want, they're all computed as multiplications and additions, albeit in finite numbers to obtain a good approximation but if an infinite number was used, you'd get the exact representation.

For example, exp(x)=sum(x^k/k!) for k=0 to infinity.

And multiplication can be computed only using addition. So how is it different?

Quote

Show an equivalent of 4.5*3.5 using only addition without any form of multiplication... I am only allowing you to expand things, not simplify them (so you can't just assume or do 3.5*.5 and write the results down). You'd get 4.5 + 4.5 + 4.5 + something. But what?

I mean that, but with exponents and multiplcation (4.5^3.5).

That's inherently impossible. It would be 4.5*4.5*4.5*2.asdkadsmalskn, but you would never, ever be able to show it by multiplication alone.

Quote from O)FaRTy1billion[MM]

Quote from Chia-Tyrant

The exponent should not be included as an elementary operator because it can be computed by using only multiplications and additions.

And multiplication can be computed only using addition. So how is it different?

Because an exponent operates on a whole 'nother level. Using multiplication or additioni, you can find the rate at which a function changes, while it has an exponent of one or zero. As soon as you change the exponent, you must use derivatives to find the rate of change. It makes the function much more dynamic: If I told you to tell me what the rate of change at any point in a f(x) = ax+c formula, you could easily do that. It will always be a.

However, when you have a different exponent, you'll have an x in the rate itself. For example, x^.5 will always change at the rate of .5x^-.5. I It will require both addition and multiplication to get the rate, and that's what makes it more advanced. can't explain it any better than changing an exponent from 1 or 0 will make it require both precursors, rather than just one.

Post has been edited 1 time(s), last time on Apr 9 2011, 3:25 am by poison_us.

Well, for one, it would complexify computations drastically.

I don't know whether or not it's impossible to completely exclude multiplication as an elementary operator for real numbers but here's an argument that should be more or less satisfactory:

How would you compute the product of, say, 0.5*4.5 in terms of additions? Sure, you could argue that it's 1+1.25, 3.25-2, ... but there exists a x+y representation for ALL real numbers; you'd have no "good" manner(i.e. algorithm) of picking out which number you'd need to use in your addition.

Wait... maybe it's just late and I'm forgetting something basic here.

Multiplication and exponentiation can be reduced to a lower-order operation if and only if the operands are integers. 4 * 4 = 4 + 4 + 4 + 4, but 4 * 3.5 cannot be expressed in terms of its operands using only addition; you must perform the multiplication 4 * 0.5 at some point. Similarly, you cannot reduce fractional exponents like 4^3.5 to just multiplications; you must perform an exponentiation at some point with a fractional part.

Obv. we can use Taylor series to express everything, but that is irrelevant as it is an infinite series.

That was the same argument I was using. xD

Except I'm trying to extend it to exponents instead of multiplication... I'm claiming exponents are to multiplication as multiplication is to addition. How would you do 0.5^4.5 with only multiplication?

... So why are exponents excluded from the "addition and multiplication are the basic operators!"?

Read my first post in this thread. The ONLY way to compute x^i where i is an irrational number is by using an infinite number of multiplications and additions.

Because all exponents can be computed as additions and multiplications (i.e. it's the ONLY way) whereas not all multiplications can be expressed as additions.

Think of it fundamentally: what's an operation in the first place? Let us consider a binary representation for the sake of simplicity. For addition, you'd need to "memorise" the following "associations":
1+0 makes 1
1+1 makes 10
0+0 makes 0
You can then repeat that process for numbers that are as large as you like.

For multiplication it would be:
1*0 makes 0
0*0 makes 0
1*1 makes 1

However, you do not need to "memorise" any associations for exponents or anything else as they are simply extensions of those two operations.

Note, however, that if you were to "memorise" more such associations, you could also considerably reduce computation time.

You clearly did not read my post since you stipulated that I read your post when in fact I have actually responded to your post within the body of mine.

The short answer: No, sir, Taylor series are horribly inefficient for fast computation requirements and are definitely not the only way to compute those things. This is why we use alternatives such as variations of Newton's method to do things like Fast invSqrt. Please do some research before claiming these things.

Where did I ever say that a taylor series was the only way to compute it? I said that you needed to use additions and multiplications. You could also memorise the numbers but that's not an operation.

You don't need to get pissy, it certainly won't get us anywhere.

48÷2(9+3)
48/2*9+2*3
48/(18+6)
48/(24)
= 2

Distributive property, guys.
trollface.jpg

I feel stupid for voting 2...
I just realized.

dats wut I said

There is only one correct answer here:



Seriously, you guys make me sad they don't teach more math.

I don't even understand what you guys are trying to argue about operators. The real numbers are just a set with two fundamentally defined operations (addition and multiplication) which satisfy a set of axioms. Just because you can think of an operation (or equivalently, a function of) two real numbers which cannot be repeated in a finite number of additions or multiplications means nothing. The real numbers are complete by these axioms, in convergent infinite (or finite) Cauchy series of various types [1]. It turns out that the fundamental definition of exponentiation to non-integer powers is based on this. For anyone who seriously cares, I highly recommend a basic course in real analysis.

Actually, you would be distributing 48/2, because it is a fraction and cannot be separated in the manner you've shown.

48/2(9+3)
(48/2)*9 + (48/2)*3
24*9 + 24*3
216 + 72
= 288

Distributive property, guys.

Not reading the whole mess, I assume 288 was agreed upon because anything else is just making an associative assumption.

Saw this thread on TeamLiquid too, was astounded how many people don't understand the reasoning behind rules and just blindly memorize them.

The education system isn't doing a good job.