Is This Imaginary?

[Squilly]

A friend and I are having a debate about this over Skype. I saw it on Facebook and can't figure out if it's an imaginary number or real. If all else fails, ask MR community.

My logic behind it is that no matter what, there's a negative under a square root, forcing it to be imaginary. Everything else is a distraction due to PEMDAS. Therefore, must be imaginary, yes? My friend is arguing that it's real because the ^2 cancels out the square root and a negative * negative is a positive, making it real. Which is it?

 

[wordoflife]

Haha are you allowed to post this?
Anyways

i^2 = -1
i = √-1

Looks like it turns imaginary and then turns back to real, because its squared.

(√(-1))^2
(i)^2
real

 

[Squilly]

Haha are you allowed to post this?
Anyways

i^2 = -1
i = √-1

Looks like it turns imaginary and then turns back to real, because its squared.

(√(-1))^2
(i)^2
real

The ^2 is outside the square root. i^2 doesn't exist in this scenario...

 

[wordoflife]

The ^2 is outside the square root. i^2 doesn't exist in this scenario...

Yes, I know.
Simplify everything inside the parenthesis first. sqrt of a negative give you an imaginary. Square it at the end, goes back to real.

 

[jav6454]

A friend and I are having a debate about this over Skype. I saw it on Facebook and can't figure out if it's an imaginary number or real. If all else fails, ask MR community.
Image

My logic behind it is that no matter what, there's a negative under a square root, forcing it to be imaginary. Everything else is a distraction due to PEMDAS. Therefore, must be imaginary, yes? My friend is arguing that it's real because the ^2 cancels out the square root and a negative * negative is a positive, making it real. Which is it?

It's real... Assume the number in the center is X. Then, we know a square is the inverse or opposite of a root. Hence, both of those cancel out, leaving just the number X alone. Whatever number X is, it is definitely real.

 

[Peace]

http://www.regentsprep.org/Regents/math/algtrig/ATO6/SquareRootLes.htm

There.

 

[iStudentUK]

Real.

It us quite clear, this should not be one of those arguable ones that goes on for pages and pages on here!

 

[Shaun.P]

A friend and I are having a debate about this over Skype. I saw it on Facebook and can't figure out if it's an imaginary number or real. If all else fails, ask MR community.
Image

My logic behind it is that no matter what, there's a negative under a square root, forcing it to be imaginary. Everything else is a distraction due to PEMDAS. Therefore, must be imaginary, yes? My friend is arguing that it's real because the ^2 cancels out the square root and a negative * negative is a positive, making it real. Which is it?

Taking the square root of a negative number is imaginary. Now that you've performed the calculation inside the brackets, you must now square. This reveals a real answer.

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The ^2 is outside the square root. i^2 doesn't exist in this scenario...

You are wrong, it does.

 

[DisplacedMic]

The ^2 is outside the square root. i^2 doesn't exist in this scenario...

sorry mate - i think you're wrong here.

√(-1)= i
i^2 = -1

typically this will reduce down to: √(-x) = i√x
so for example √(-25) = √(25) x √(-1) = 5√(-1) = 5i

5i^2 = -5

an imaginary number squared will always give a real negative number. that's the point of imaginary numbers.

 

[Shrink]

Hey...when you guys finish with this thread, could one of you help me balance my checkbook??

You know that math stuff pretty good!!

 

[Brian Y]

sorry mate - i think you're wrong here.

√(-1)= i
i^2 = -1

typically this will reduce down to: √(-x) = i√x
so for example √(-25) = √(25) x √(-1) = 5√(-1) = 5i

5i^2 = -5

an imaginary number squared will always give a real negative number. that's the point of imaginary numbers.

That's true for single imaginary numbers, but not for complex numbers, which can be proved simply with:

√(√(-25))^2 = 5i

 

[DisplacedMic]

That's true for single imaginary numbers, but not for complex numbers, which can be proved simply with:

√(√(-25))^2 = 5i

i'm definitely out of my depth here!

 

[chown33]

That's true for single imaginary numbers, but not for complex numbers, which can be proved simply with:

√(√(-25))^2 = 5i

I see no complex numbers here. I see an additional square root when compared to the original problem statement, which yields an unsurprising imaginary result.

Simplifying:
√(√(-25))^2 = (√(-25)) = √(-25) = i√(25) = 5i

 

[Brian Y]

I see no complex numbers here. I see an additional square root when compared to the original problem statement, which yields an unsurprising imaginary result.

Simplifying:
√(√(-25))^2 = (√(-25)) = √(-25) = i√(25) = 5i

√(√(-25)) = 1.58113883 + 1.58113883 i

 

[Happybunny]

Hey...when you guys finish with this thread, could one of you help me balance my checkbook??

You know that math stuff pretty good!!

I'm with Shrink and Barbie on this one.

[/URL]

That's why I employe accountants.

 

[Shrink]

I'm with Shrink and Barbie on this one.


Image[/URL]

That's why I employe accountants.

A remarkable resemblance between the pictured "person" and me when I get dressed up!

And god bless my accountant, too.

 

[chown33]

√(√(-25)) = 1.58113883 + 1.58113883 i

Agreed, but you posted this:
√(√(-25))^2 = 5i

That is, you have an additional squaring operation, when compared to:
√(√(-25))

 

[Brian Y]

Yes - that's what I was saying. The square root of the square root of -25 is the complex number ~1.58(1+i) (sorry I'm on a mobile). I was trying to show how the given statement about squaring imaginary numbers resulting in a negative real number didn't apply to complex numbers - hence squaring that complex number and resulting in the imaginary 5i.

I didn't post the intermediate complex number since it was a one line well established proof.

 

[chown33]

Yes - that's what I was saying. The square root of the square root of -25 is the complex number ~1.58(1+i) (sorry I'm on a mobile). I was trying to show how the given statement about squaring imaginary numbers resulting in a negative real number didn't apply to complex numbers - hence squaring that complex number and resulting in the imaginary 5i.

I didn't post the intermediate complex number since it was a one line well established proof.

The given statement about squaring imaginary numbers was this:

an imaginary number squared will always give a real negative number
​

I see nothing in the post suggesting it might apply to complex numbers in general. AFAICT it's entirely accurate when applied to pure imaginary numbers, i.e. complex numbers of the form 0 + bi.

In my experience, I've never heard a complex number called "imaginary" unless it had zero for its real component. "Imaginary" is reserved solely for the complex numbers lying on the imaginary axis in the complex plane, just as "real" is reserved for the complex numbers lying on the real axis of the complex plane. "Complex" covers the entire plane, of which "real" and "imaginary" are subsets. This is coming mainly from an electrical engineering background, though perhaps the preferred terminology is different now.

 

[mobilehaathi]

Why are you all assuming that number is >0? Or strictly real?

 

[Brian Y]

The given statement about squaring imaginary numbers was this:

an imaginary number squared will always give a real negative number
​

I see nothing in the post suggesting it might apply to complex numbers in general. AFAICT it's entirely accurate when applied to pure imaginary numbers, i.e. complex numbers of the form 0 + bi.

In my experience, I've never heard a complex number called "imaginary" unless it had zero for its real component. "Imaginary" is reserved solely for the complex numbers lying on the imaginary axis in the complex plane, just as "real" is reserved for the complex numbers lying on the real axis of the complex plane. "Complex" covers the entire plane, of which "real" and "imaginary" are subsets. This is coming mainly from an electrical engineering background, though perhaps the preferred terminology is different now.

Read my original post. It's called moving the discussion on, not arguing for the sake of arguing

 

[Squilly]

Why are you all assuming that number is >0? Or strictly real?

Real.

 

[mobilehaathi]

Real.

Why are you making that assumption? You are assuming that number is real and > 0.

 

[Shaun.P]

Why are you all assuming that number is >0? Or strictly real?

If number is less <0 or =0 the result is still real.

----------

sorry mate - i think you're wrong here.

√(-1)= i
i^2 = -1

typically this will reduce down to: √(-x) = i√x
so for example √(-25) = √(25) x √(-1) = 5√(-1) = 5i

5i^2 = -5

an imaginary number squared will always give a real negative number. that's the point of imaginary numbers.

For OP's question, not 5i^2, but (5i)^2!

 

[mobilehaathi]

If number is less <0 or =0 the result is still real

If number is real, the answer is real. On the other hand if it is complex the result is complex. Aside from the fact that Squilly's reasoning was embarrassingly wrong, no one seemed to realized that number could be negative or non-real. The answer to his question depends on the properties of number.