Dialogue de sourds about mathematics: Are exponents substractions in disguise?

[ThirdWizard]

Where's the recursion, I ask? I certainly never advocated recursion. I merely said that it is not required that the exponent be calculated first, and that, rather, it may be done during the same step as multiplication.

Here's the crux of Rystil's argument.

When you use order of operations to do, say X*Y*Z, you don't have to know either X nor Y to calculate the multiplication of Z. When you calculate XY^Z (with your method) and do XY first, then you must still know what Y is to finish the calculation. Thus, you are doing two separate calculations still. You're doing XY and Y^(Z-1). You can in no way calculate them in order and expect to get the correct calculations. The exponent must still be calculated independantly.

 

[Rystil Arden]

Really? X^-1 = 1/2 X? I don't think it does, but then maybe my math is off ...

Do you?

AB^-1 = A / B, eh?

Therefore,

AB^-1 = AB * B^-2 = AB / B^2 = A / B

Where's the recursion, I ask? I certainly never advocated recursion. I merely said that it is not required that the exponent be calculated first, and that, rather, it may be done during the same step as multiplication.

X^-1 doesn't equal 1/2 X, but you can use it to calculate 1/2 X. It isn't easy though. You need to add together X X times first (call it Y). Then add together X + X add take that to ^-1 power. Take that and add it to itself Y times. Bingo, X/2.

Its a really dumb way to do it though

 

[ThirdWizard]

To all arguing: I'm sorry I brought it up. Please forgive me. It looks like once again I've unintentionally hijacked a thread.

See my sig.

The origional questions's resolution was found too darn quickly.

 

[Patryn of Elvenshae]

When you use order of operations to do, say X*Y*Z, you don't have to know either X nor Y to calculate the multiplication of Z.

And here's mine.

Exponents are shorthand notation.

Writing 2x^3 is exactly the same as writing 2*x*x*x.

The only reason the Order of Operations makes Exponents a separate step is to make using the shorthand easier.

However, if you understand what the shorthand actually means, then you'll see that they aren't actually separate steps except for matters of convenience.

 

[ThirdWizard]

Again, you expanded the exponential notation before multiplying. Showing that the exponent must in fact be handled before multiplication.

Because (2x)^3 is different than 2(x^3) just like (2X)+Y is different than 2(X+Y).

That's what order of operations is about. You can say that multiplication is just a matter of convenience for addition, but it doesn't change the fact that you multiply before you add. Just like you do exponents before you multiply.

 

[Rystil Arden]

Again, you expanded the exponential notation before multiplying. Showing that the exponent must in fact be handled before multiplication.

Because (2x)^3 is different than 2(x^3) just like (2X)+Y is different than 2(X+Y).

That's what order of operations is about. You can say that multiplication is just a matter of convenience for addition, but it doesn't change the fact that you multiply before you add. Just like you do exponents before you multiply.

Agreed. And fractional exponents prove that not all exponents can be rewritten as mulitiplication.

 

[ThirdWizard]

Order of operations is really only implied perenthasis after all.

 

[Patryn of Elvenshae]

Again, you expanded the exponential notation before multiplying. Showing that the exponent must in fact be handled before multiplication.

I think I came up with a better explanation.

The "real" equation is:

2 * y * y

Then, later on, someone came along and came up with a fancy form of shorthand, collapsing like terms and adding an "exponent."

2 * y^2

Thus, new ways of handling this shorthand had to be devised.

So, if you insist on using shorthand notation, you're going to have to fiddle around with things a bit.

However, if you recognize what the shorthand actually says, there's no order problems.

Does that make sense?

 

[Patryn of Elvenshae]

Agreed. And fractional exponents prove that not all exponents can be rewritten as mulitiplication.

Eh?

Prove it.

Given:

2(4)^1/2 = (2*4) * 4^1/2-1 = 8 * (1/(4^1/2)) = 8 * (1/2) = 4 == 2 * (4^1/2) = 2 * 2 = 4

 

[Rystil Arden]

I think I came up with a better explanation.

The "real" equation is:

2 * y * y

Then, later on, someone came along and came up with a fancy form of shorthand, collapsing like terms and adding an "exponent."

2 * y^2

Thus, new ways of handling this shorthand had to be devised.

So, if you insist on using shorthand notation, you're going to have to fiddle around with things a bit.

However, if you recognize what the shorthand actually says, there's no order problems.

Does that make sense?

This is true, but recognising and expanding the shorthand requires the use of the order of operations first, and you have to do it if you are given an equation with exponents rather than iterated multiplications when either would apply (which is true over 99% of the time). Also, some exponents can't be expanded this way.