Dialogue de sourds about mathematics: Are exponents substractions in disguise?
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Patryn of Elvenshae
First Post
Rystil Arden said:
Agreed.
All *I'm* saying is that exponents are a form of short-hand for a series of multiplications (or divisions).
Therefore, there is no *real* difference when calculating
(2-x)^2 vs. (2-x)*(2-x)
As you yourself said, they are identical statements. Therefore, there is no difference between them.
Any exponent can be expressed as a group of multiplications. Therefore, there's no real difference between them.
ThirdWizard
First Post
Thanee said:
Actually, I'm saying that they are different.
EDIT: while admitting that one is derived from the other
Patryn of Elvenshae said:
The exact same thing can be said of addition and multiplication, though.
Rystil Arden
First Post
What ThirdWizard, Hypersmurf, and I are saying is that you are in fact not doing exponents at the same step as multiplication in your simplification example. While the simplification is completely correct, you aren't following your own rule because, you'll notice, you are expanding the exponent into multiplication *before* you multiply in every case.
In 2y^2, you expand to 2*y*y before multiplying
Int 2^2*y, you expand to 2*2*y before multiplying
Since exponents are *always* expanded before multiplying it out, they are not the same step.
If they were the same step, you could proceed in strict left to right order associatively, and always get the same answer. ThirdWizard and Hypersmurf's use of parentheses was *not* an attempt to bring parentheses into the mix, but simply a disambiguating citing of the failure of the Associative Property between exponents and multiplication.
In 2y^2, you expand to 2*y*y before multiplying
Int 2^2*y, you expand to 2*2*y before multiplying
Since exponents are *always* expanded before multiplying it out, they are not the same step.
If they were the same step, you could proceed in strict left to right order associatively, and always get the same answer. ThirdWizard and Hypersmurf's use of parentheses was *not* an attempt to bring parentheses into the mix, but simply a disambiguating citing of the failure of the Associative Property between exponents and multiplication.
Patryn of Elvenshae
First Post
Rystil Arden said:
I disagree that this is an important distinction.
2 * y * y == 2y * y = 2y^2
In other words, given an equation in the form AB^X, you may calculate AB first, and multiply that result by B^X-1. There is no requirement that you first calculate B^X before considering A (except that, usually, it is easier to do so).
Some examples:
2(4)^2 = (2*4) * 4^2-1 = 8 * 4 = 32 == 2(4^2) = 2(16) = 32
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
First Post
But here you once again worked to allow the exponent step before you multiplied. In this case, the way you did this is by precalculating the B^X-1 before you multiplied A*B.
If you have a memoryless linear time invariant system for doing your arithmetic (and you do in any formulaic order of operations system), then you can't perform A*B until you have stored a copy of B somewhere else to be used in the calculation of B^X-1 (you can decrement X at any time, of course). Otherwise, you'll find that there is no instance of B in existence that can be raised to the X-1 power. So you must have gone ahead and worked on the exponent before doing the multiplication.
Whether the exponent gets to go first in some preliminary parsing step, or in the actual calculation itself, the fact is that exponents go first. You can't just read the equation from left to right and blindly multiply out without if there are exponents later on without foreknowledge of the exponents that will later appear.
If you have a memoryless linear time invariant system for doing your arithmetic (and you do in any formulaic order of operations system), then you can't perform A*B until you have stored a copy of B somewhere else to be used in the calculation of B^X-1 (you can decrement X at any time, of course). Otherwise, you'll find that there is no instance of B in existence that can be raised to the X-1 power. So you must have gone ahead and worked on the exponent before doing the multiplication.
Whether the exponent gets to go first in some preliminary parsing step, or in the actual calculation itself, the fact is that exponents go first. You can't just read the equation from left to right and blindly multiply out without if there are exponents later on without foreknowledge of the exponents that will later appear.
ThirdWizard
First Post
Having... architecture... flashbacks...
AB^X
N=1
AB * B^X-N
N-1
ABB * B^X-N
etc
until you reach your base case (N=0)
So you're basically expanding the exponent into multiplication. Again, same thing as above.
Now I'm having flashbacks to inductive proofs and recursion. I shake my fist at you Rystil.
AB^X
N=1
AB * B^X-N
N-1
ABB * B^X-N
etc
until you reach your base case (N=0)
So you're basically expanding the exponent into multiplication. Again, same thing as above.
Now I'm having flashbacks to inductive proofs and recursion. I shake my fist at you Rystil.
Patryn of Elvenshae
First Post
Rystil Arden said:
No, I didn't.
Don't tell me what I did or did not do.
I calculated A * B first. Then, and only then, did I consider the effects of that calculation upon X.
Note that this still works even if X is 1.
An exponent is a shorthand. It is designed to make things easier to visualize. It *is* multiplication.
EDIT:
And, I'm sorry, but wasn't the point of all of this that one "Had to calculate an exponent before you can do any multiplication, because they are separate steps?"
I calculated the exponent after I started multiplication. Therefore, your statement is false.
Last edited:
Rystil Arden
First Post
Shaking your fist? What no gratitude for backing you upThirdWizard said:
ThirdWizard
First Post
Rystil Arden said:Shaking your fist? What no gratitude for backing you up
I hate to admit that I do love recursion. My hat for inductive proofs know no limit however.
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