Is This Imaginary?

Discussion in 'Community Discussion' started by Squilly, Apr 5, 2013.

1. Apr 5, 2013
Last edited: Apr 5, 2013

Squilly macrumors 68020

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#1
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?

2. wordoflife macrumors 604

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Jul 6, 2009
#2
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

3. Squilly thread starter macrumors 68020

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#3
The ^2 is outside the square root. i^2 doesn't exist in this scenario...

4. wordoflife macrumors 604

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#4
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.

5. jav6454 macrumors P6

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#5
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.

6. iStudentUK macrumors 65816

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#7
Real.

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

7. Apr 12, 2013
Last edited: Apr 16, 2013

Shaun.P macrumors 68000

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#8
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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You are wrong, it does.

8. DisplacedMic macrumors 65816

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May 1, 2009
#9
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.

9. Shrink macrumors G3

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#10
Hey...when you guys finish with this thread, could one of you help me balance my checkbook??

You know that math stuff pretty good!!

10. Brian Y macrumors 68040

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Oct 21, 2012
#11
That's true for single imaginary numbers, but not for complex numbers, which can be proved simply with:

√(√(-25))^2 = 5i

11. DisplacedMic macrumors 65816

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#12
i'm definitely out of my depth here!

12. Apr 13, 2013
Last edited: Apr 13, 2013

chown33 macrumors 604

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#13
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

13. Brian Y macrumors 68040

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Oct 21, 2012
#14
√(√(-25)) = 1.58113883 + 1.58113883 i

14. Happybunny macrumors 68000

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#15
I'm with Shrink and Barbie on this one.

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That's why I employe accountants.

15. Shrink macrumors G3

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#16
A remarkable resemblance between the pictured "person" and me when I get dressed up!

And god bless my accountant, too.

16. chown33 macrumors 604

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#17
Agreed, but you posted this:
√(√(-25))^2 = 5i

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

17. Brian Y macrumors 68040

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Oct 21, 2012
#18
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.

18. chown33 macrumors 604

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#19
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.

19. mobilehaathi macrumors G3

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#20
Why are you all assuming that number is >0? Or strictly real?

20. Brian Y macrumors 68040

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#21
Read my original post. It's called moving the discussion on, not arguing for the sake of arguing

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#22
Real.

22. mobilehaathi macrumors G3

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#23
Why are you making that assumption? You are assuming that number is real and > 0.

23. Shaun.P macrumors 68000

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#24
If number is less <0 or =0 the result is still real.

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

24. mobilehaathi macrumors G3

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#25
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.