surely
should be 1 not 1? because the negatives cancel?

timebent
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 19072016 16:02

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 19072016 16:06

timebent
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 19072016 16:08
(Original post by oShahpo)
sqrt(1) is not a defined quantity in terms other than i. Thus, the algebraic rules of surd multiplication do not apply here. 
an_atheist
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 19072016 16:08
x=1 in this case. 
HFancy1997
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 19072016 16:09
Sqrtx*Sqrtx=(Sqrtx)^2=x=1
Hope that helps
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 19072016 16:09
(Original post by timebent)
so different ball game so for example where the 1 is could be subbed in for something else right? so let 1 be x and the root of x squared is still x so where x is 1 the answer is 1 right? or am i wrong?
Sqrt(1) means i, that is the only definition we have of this quantity in mathematics.Last edited by oShahpo; 19072016 at 16:15. 
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 19072016 16:11
(Original post by timebent)
so different ball game so for example where the 1 is could be subbed in for something else right? so let 1 be x and the root of x squared is still x so where x is 1 the answer is 1 right? or am i wrong?
The square root of 1, squared, is equal to 1.
More simply put, i^2 = 1
In reference to your initial post, i is neither positive nor negative, as it does not lie on the real number line.
It's coefficients can be positive or negative, but never i itself. 
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 19072016 16:13
You cannot do it that way. We don't write like that for good reason. We use the imaginary (complex with real part 0) number . So for what you have put e would actually write it as by definition. So . Using it this way is consistent.

timebent
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 19072016 16:13
(Original post by oShahpo)
Wrong(Original post by JLegion)
Correct.
rip my brain 
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 19072016 16:18
(Original post by oShahpo)
Wrong, because if you use sqrt(x) then you must also specify the condition x>= 0.
Sqrt(1) means i, that is the only definition we have of this quantity in mathematics.
i^2 = 1
I think the wording of the post was open to misinterpretation. 
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 19072016 16:21
(Original post by JLegion)
The square root of 1, squared, is equal to 1.
More simply put, i^2 = 1(Original post by JLegion)
The OP was asking (or seemed to for me) if the square root of 1 multiplied by itself is 1, which is true.
i^2 = 1 
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 19072016 16:25
So, taking renders that formula obsolete and your approach fails. This is due to a more deeply rooted issue with having to give up certain 'privileges' of the sort when working in the algebraic closure of , as a extension to this, you lose commutativity when you move to the quaternion system, but I digress. 
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 19072016 16:25
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 19072016 16:27

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 19072016 16:29
(Original post by timebent)
rip my brain
Square root of x^2 is not x but + or  x. The way to deal with this is to understand that sqrt of 1 means absolutely nothing if you don't treat as i, because it is not defined in terms of anything but i.
Therefore, sqrt(x) * sqrt(x) = sqrt(x^2) indeed, but then if you plug x = 1 you get sqrt(1) which is 1 or 1. You see the confusion here?
The reason for this confusion is that sqrt(x) is not defined for negative numbers, if you are using negative numbers you must break it down in terms of i. So in this case it becomes sqrt(x^2) = sqrt(x*x)= sqrt(x) * sqrt(x) = sqrt(1) ^2 = i^2 = 1 only. 
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 19072016 16:30
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timebent
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 19072016 16:31
(Original post by Zacken)
To address, your issue, I know you're thinking by using the rule . However, something that your teachers neglected to mention is that that rule, that nice little operator nicety we take for granted only works over the real field. Indeed, the true formula should read
So, taking renders that formula obsolete and your approach fails. This is due to a more deeply rooted issue with having to give up certain 'privileges' of the sort when working in the algebraic closure of , as a extension to this, you lose commutativity when you move to the quaternion system, but I digress.
However it seems that if you use negative numbers you still use the negative regardless
so in this case then should then equal right?
so if i'm right
then what is the outcome of ??Last edited by timebent; 19072016 at 16:38. Reason: corrected some mistakes 
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 19072016 16:32
(Original post by Zacken)
No. Not "the square root of 1", there isn't a single square root, there are two square roots of 1, following that, it doesn't make sense to say "the square root of 1 squared is 1". 
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 19072016 16:36
(Original post by oShahpo)
The answer is correct, but the method is wrong.
Square root of x^2 is not x but + or  x. The way to deal with this is to understand that sqrt of 1 means absolutely nothing if you don't treat as i, because it is not defined in terms of anything but i.
Therefore, sqrt(x) * sqrt(x) = sqrt(x^2) indeed, but then if you plug x = 1 you get sqrt(1) which is 1 or 1. You see the confusion here?
The reason for this confusion is that sqrt(x) is not defined for negative numbers, if you are using negative numbers you must break it down in terms of i. So in this case it becomes sqrt(x^2) = sqrt(x*x)= sqrt(x) * sqrt(x) = sqrt(1) ^2 = i^2 = 1 only.
Sorry. 
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 19072016 16:36
(Original post by oShahpo)
The answer is correct, but the method is wrong.
Square root of x^2 is not x but + or  x. The way to deal with this is to understand that sqrt of 1 means absolutely nothing if you don't treat as i, because it is not defined in terms of anything but i.
Therefore, sqrt(x) * sqrt(x) = sqrt(x^2) indeed, but then if you plug x = 1 you get sqrt(1) which is 1 or 1. You see the confusion here?
The reason for this confusion is that sqrt(x) is not defined for negative numbers, if you are using negative numbers you must break it down in terms of i. So in this case it becomes sqrt(x^2) = sqrt(x*x)= sqrt(x) * sqrt(x) = sqrt(1) ^2 = i^2 = 1 only.
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