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Dividing by variable when solving algebraic equation
#41
RE: Dividing by variable when solving algebraic equation
No, no! You can't...

My equations! My precious equations, ruined!
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#42
RE: Dividing by variable when solving algebraic equation
3 = 2 => The Pope is a space squirrel
The fool hath said in his heart, There is a God. They are corrupt, they have done abominable works, there is none that doeth good.
Psalm 14, KJV revised edition

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#43
RE: Dividing by variable when solving algebraic equation
(October 27, 2016 at 2:41 am)Alex K Wrote: 3 = 2  => The Pope is a space squirrel

Bzzzt. Wrong answer. It's raccoon.

And you call yourself a math teacher ...
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#44
RE: Dividing by variable when solving algebraic equation
For anyone who hasn't yet seen the huge problem with the divide by zero error, and the fact that you can't say 0/0 = 1:

3 * 0 = 2 * 0 [this is true, both sides are just 0]

(3 * 0) / 0 = (2 * 0) / 0

3 * (0 / 0) = 2 * (0 / 0)

[this is a legitimate rearrangement; the order you multiply and divide doesn't matter]

3 * 1 = 2 * 1

3 = 2 = Pope = Squirrel

Obviously if it can produce incorrect results, it's not valid. So you simply can't divide by zero; the rules of mathematics forbid it.
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#45
RE: Dividing by variable when solving algebraic equation
1=sqrt(1)=sqrt(-1 * -1)=sqrt(-1)*sqrt(-1)= i*i=-1
The fool hath said in his heart, There is a God. They are corrupt, they have done abominable works, there is none that doeth good.
Psalm 14, KJV revised edition

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#46
RE: Dividing by variable when solving algebraic equation
That's some tricky shit.
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#47
RE: Dividing by variable when solving algebraic equation
(October 27, 2016 at 3:05 am)Alex K Wrote: 1=sqrt(1)=sqrt(-1 * -1)=sqrt(-1)*sqrt(-1)= i*i=-1

Ah, yeah, I remember something about the rule: no splitting the radicals when both factors are negative. Is that correct?
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#48
RE: Dividing by variable when solving algebraic equation
(October 27, 2016 at 3:39 am)Irrational Wrote:
(October 27, 2016 at 3:05 am)Alex K Wrote: 1=sqrt(1)=sqrt(-1 * -1)=sqrt(-1)*sqrt(-1)= i*i=-1

Ah, yeah, I remember something about the rule: no splitting the radicals when both factors are negative. Is that correct?

I want to know why actually.

Is it because of the product of two negative numbers being positive?
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#49
RE: Dividing by variable when solving algebraic equation
The essential problem is that mapping 1 to sqrt(1) isn't one to one, unless you force the result to be positive. -1 also maps to sqrt(1). To keep the mapping consistent, when evaluating the split radicals it should be -i * i = 1

I'm not actually sure if there's a formal rule for dealing with this, it's not come up for me.

It's one of the easiest ways to slip up and one of the more confusing aspects of maths.
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#50
RE: Dividing by variable when solving algebraic equation
(October 27, 2016 at 4:02 am)robvalue Wrote: The essential problem is that mapping 1 to sqrt(1) isn't one to one, unless you force the result to be positive. -1 also maps to sqrt(1). To keep the mapping consistent, when evaluating the split radicals it should be -i * i = 1

I'm not actually sure if there's a formal rule for dealing with this, it's not come up for me.

It's one of the easiest ways to slip up and one of the more confusing aspects of maths.

Yes, exactly, the value of the square root in the complex plane is not unique, because if you go once round the origin in the complex plane for the values you stick in, you only circle half round the origin in the image set, so it depends on which way you go from 1 to -1 what the result for sqrt(-1) is. The branch sqrt(-1)=i is simply the standard convention, but everyone can see that -i*-i=-1 as well.
Splitting the square root into a factor of two square roots lets us suddenly jump into the wrong half of the image set if we stick with the standard convention for both of them.

This is kind of for the specialists:

Basically, the prescription has something to do with analytic continuation: We have to go from 1=1*1 to 1=-1*-1 in a continous fashion.

It would go something like this: you first split the 1 into two factors of 1=1*1, put them under sqrts, so you get 1=sqrt(1)*sqrt(1). Then, to get a continuous well-defined result for the expression sqrt(-1)*sqrt(-1), what you'd have to do is under both sqrts, you have to go from 1 to -1 on a circle around the origin. In order for the product of the two numbers under the sqrts to remain =1 during this continuous process, you have to go clockwise around the origin in the complex plane for one of them, and counter-clockwise for the other. Then, the continuous process of wandering from 1=1*1 ----> -1*-1 , the product always remains 1. So what if we add the square roots?

What will then happen is that with one sqrt(-1) where you went from 1 to -1 counter-clockwise you end up on the standard branch with sqrt(-1)=+i, and with the other sqrt(-1) where you went from 1 to -1 clockwise, you end up with the sqrt(-1)=-i, and you get "sqrt(-1)*sqrt(-1)"=i*-i = 1, the desired result.
The fool hath said in his heart, There is a God. They are corrupt, they have done abominable works, there is none that doeth good.
Psalm 14, KJV revised edition

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