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

 Follow
 1
 19072016 16:02

 Follow
 2
 19072016 16:06

 Follow
 3
 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. 
 Follow
 4
 19072016 16:08
x=1 in this case. 
 Follow
 5
 19072016 16:09
Sqrtx*Sqrtx=(Sqrtx)^2=x=1
Hope that helps
Posted from TSR Mobile 
 Follow
 6
 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. 
 Follow
 7
 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. 
 Follow
 8
 19072016 16:13

 Follow
 9
 19072016 16:13
(Original post by oShahpo)
Wrong(Original post by JLegion)
Correct.
rip my brain 
 Follow
 10
 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. 
 Follow
 11
 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 
 Follow
 12
 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.Post rating:1 
 Follow
 13
 19072016 16:25
Posted from TSR Mobile 
 Follow
 14
 19072016 16:27

 Follow
 15
 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. 
 Follow
 16
 19072016 16:30
Posted from TSR Mobile 
 Follow
 17
 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 
 Follow
 18
 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". 
 Follow
 19
 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. 
 Follow
 20
 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.
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: July 19, 2016
Share this discussion:
Tweet
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.