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In the complex number system, why can't 1+1 = 0 ?

by pondzo
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pondzo
#1
Feb14-14, 01:17 AM
P: 70
in the following i will demonstrate a 'proof' that 1+1=0

1+1=√1 +1
=√(-1)(-1) +1
= (√(-1))(√(-1)) +1
= (i)(i) +1
= i2 +1
= -1 + 1
= 0

I know I'm not the first to come up with this 'proof', and i have been told that the problem lies with splitting the radical between the 2nd and 3rd lines of working. But in the complex number system there is no problem with going from, say, √(-4) = √(-1)(4) = 2i so why is there a problem with the above working, isn't it the same line of thought?

thanks, Michael.
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DrClaude
#2
Feb14-14, 03:27 AM
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There's a FAQ on the subject: http://www.physicsforums.com/showthread.php?t=637214
phinds
#3
Feb14-14, 06:38 AM
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Quote Quote by pondzo View Post
in the following i will demonstrate a 'proof' that 1+1=0
Doesn't simple common sense tell you this is nonsense?

pondzo
#4
Feb14-14, 09:25 AM
P: 70
In the complex number system, why can't 1+1 = 0 ?

Quote Quote by phinds View Post
Doesn't simple common sense tell you this is nonsense?
yes, which is why i asked the question. it just seemed mathematically sound to me so i wanted to see why we disregard it. But it seems i have a lot to catch up on with complex exponentiation!
phinds
#5
Feb14-14, 09:48 AM
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Quote Quote by pondzo View Post
yes, which is why i asked the question. it just seemed mathematically sound to me so i wanted to see why we disregard it. But it seems i have a lot to catch up on with complex exponentiation!
Ah ... yeah, I see from rereading your question that you aren't quite saying that you believe it, just that you are puzzled by it.
crownedbishop
#6
Feb18-14, 07:45 PM
P: 19
Quote Quote by pondzo View Post
in the following i will demonstrate a 'proof' that 1+1=0

1+1=√1 +1
=√(-1)(-1) +1
= (√(-1))(√(-1)) +1
= (i)(i) +1
= i2 +1
= -1 + 1
= 0

I know I'm not the first to come up with this 'proof', and i have been told that the problem lies with splitting the radical between the 2nd and 3rd lines of working. But in the complex number system there is no problem with going from, say, √(-4) = √(-1)(4) = 2i so why is there a problem with the above working, isn't it the same line of thought?

thanks, Michael.
The problem is that for complex numbers, √(a+bi) has two complex roots. With real numbers, one is positive and the other is negative. We can assign the positive one to equal the positive square root. However, with complex numbers it isn't so simple. A complex root is neither negative nor positive, sense it doesn't make any sense to say a complex number is positive. So, it doesn't make sense to have a "positive" square root. This is where the issue arises.
Seydlitz
#7
Feb18-14, 09:33 PM
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Quote Quote by pondzo View Post
yes, which is why i asked the question. it just seemed mathematically sound to me so i wanted to see why we disregard it. But it seems i have a lot to catch up on with complex exponentiation!
Here's the explanation:

In our Example A, we did
[tex]\sqrt{(-1)(-1)}=\sqrt{-1}\cdot \sqrt{-1}[/tex]
This is just applying the rule
[tex](xy)^a=x^ay^a[/tex]
to the situation x=y=-1 and a=1/2. But this rule only holds for positive x and y!!
The fun thing is you can go 'crazy' with the rule if you know it.

say:

##i^{-i} = e^{\frac{\pi}{2}}##
Curious3141
#8
Feb18-14, 10:12 PM
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Quote Quote by Seydlitz View Post
Here's the explanation:



The fun thing is you can go 'crazy' with the rule if you know it.

say:

##i^{-i} = e^{\frac{\pi}{2}}##
Umm, that actually is valid. One of the values of ##i^{-i}## is ##e^{\frac{\pi}{2}}##.
pondzo
#9
Feb19-14, 02:19 AM
P: 70
okay, why are we allowed to go from ((4)*(-1))0.5 to (4)0.5*(-1)0.5 (which is equal to 2i) but we aren't allowed to go from ((-1)(-1))0.5 to (-1)0.5*(-1)0.5?
pondzo
#10
Feb19-14, 02:24 AM
P: 70
and on another note, in my most recent post, isn't the first example a negative number and the second a positive (ie (4)(-1)= -ve and (-1)(-1)= +ve)
Seydlitz
#11
Feb19-14, 05:15 AM
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Quote Quote by Curious3141 View Post
Umm, that actually is valid. One of the values of ##i^{-i}## is ##e^{\frac{\pi}{2}}##.
Yes it is valid, I don't think I imply otherwise.
Curious3141
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Feb19-14, 08:17 AM
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Quote Quote by Seydlitz View Post
Yes it is valid, I don't think I imply otherwise.
Oh, OK. Sorry that I misunderstood your post.
lendav_rott
#13
Feb19-14, 09:46 AM
P: 223
It's not possible. If 1+1 = 0, then +1 would have to be equal to -1, but it can't be equal to -1 and +1 at the same time. Same way you could try and prove for any "A" that A+A = 0
gopher_p
#14
Feb19-14, 10:55 AM
P: 471
Quote Quote by pondzo View Post
okay, why are we allowed to go from ((4)*(-1))0.5 to (4)0.5*(-1)0.5 (which is equal to 2i) but we aren't allowed to go from ((-1)(-1))0.5 to (-1)0.5*(-1)0.5?
Alright, here's the rub. This isn't actually a "valid" operation. It just looks like what's going on. The notation convention used by every mathematician and math educator that I know is that ##\sqrt{a}## denotes the non-negative solution of ##x^2=a## if ##a## is a non-negative real number and ##\sqrt{a}## denotes the root ##i\sqrt{-a}## of ##x^2=a## when ##a## is a negative real number (so we've already agreed on what ##\sqrt{-a}## means), where ##i## denotes either (a) some made-up thing with the property that ##i^2=-1## or (b) the principle complex root (we singled one of them out and said "you're it") of ##x^2+1=0## depending on where you are in your math studies. There are no rules of exponents or multiplicative identities of roots or anything of that sort going on. It's all just definitions of notation; these are the symbols that we use to talk about these abstract mathematical concepts.

In other words, ##\sqrt{-4}=i\sqrt{-(-4)}=i\sqrt{4}=2i## because somebody a long time ago decided that's what those symbols mean, and everybody else went along with it. It only looks like we're using ##\sqrt{-4}=\sqrt{-1\cdot 4}=\sqrt{-1}\cdot\sqrt{4}=i\cdot2## because either (a) you were clever enough to notice that might be the case based on your experience using rules that you learned in precalc or (b) you had a teacher (or tutor or helper on the interwebs) that told you a little white lie to convince you that the answer they gave was correct without realizing (or caring) that it might later cause serious strife for you and other teachers (or tutors or helpers on the interwebs) or (c) you had a teacher (or tutor or helper on the interwebs) who was incompetent tell you this is what was going on without understanding that it really wasn't.
pondzo
#15
Feb19-14, 11:03 AM
P: 70
Quote Quote by lendav_rott View Post
It's not possible. If 1+1 = 0, then +1 would have to be equal to -1, but it can't be equal to -1 and +1 at the same time. Same way you could try and prove for any "A" that A+A = 0
if the same exponential laws applied to negative numbers as they do positive.

a+a=√(a^2 )+a
=√((-a)(-a) )+a
=√(-a) √(-a)+a
=√((-1)(a) ) √((-1)(a) )+a
=(√(-1))(√a)(√(-1))(√a)+a
=(i)(√a)(i)(√a)+a
=i^2 a+a
=-a+a
=0
pondzo
#16
Feb19-14, 11:21 AM
P: 70
Quote Quote by gopher_p View Post

In other words, ##\sqrt{-4}=i\sqrt{-(-4)}=i\sqrt{4}=2i## ... It only looks like we're using ##\sqrt{-4}=\sqrt{-1\cdot 4}=\sqrt{-1}\cdot\sqrt{4}=i\cdot2## .
Isn't that pretty much equivalent to what your doing when you say √(-4) = i√(-(-4)) because your just taking √(i)2 out of the radical, which is the same as √(-1). so really your doing the exact same thing as me?
Nugatory
#17
Feb19-14, 01:07 PM
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Quote Quote by pondzo View Post
Isn't that pretty much equivalent to what your doing when you say √(-4) = i√(-(-4)) because your just taking √(i)2 out of the radical, which is the same as √(-1). so really your doing the exact same thing as me?
No, he is not doing the same thing. He is saying that we use the sequence of symbols "##\sqrt{a}##" as a convenient shorthand for "the non-negative solution of ##x^2=a## if ##a## is a non-negative real number; and the root ##i\sqrt{-a}## of ##x^2=a## when ##a## is a negative real number, where ##i## denotes some made-up thing with the property that ##i^2=-1##"

That's a definition, and we can use that definition to show that if ##a## and ##b## are both non-negative reals, then ##\sqrt{ab} = \sqrt{a}\sqrt{b}##. However, we can also use this definition to show that, in general, this equality does not hold for negative numbers.
gopher_p
#18
Feb19-14, 01:09 PM
P: 471
Quote Quote by pondzo View Post
Isn't that pretty much equivalent to what your doing when you say √(-4) = i√(-(-4)) because your just taking √(i)2 out of the radical, which is the same as √(-1). so really your doing the exact same thing as me?
No it's not. Read again what I wrote regarding the definition of the meaning of the collection of symbols ##\sqrt{a}## when ##a## is a negative real number:

If ##a## is a negative real number, the ##\sqrt{a}=i\sqrt{-a}##.

There is nothing in between the left and right hand side that explains why it is true. There is not proof of this identity. It is not a matter of mathematics, but a matter of mathematical notation. It is simply true by definition of the notation being used. I'm not "taking √(i)2 out of the radical". The ##i## essentially materializes out of thin air, just like the ##\times## does in the definition of the exponential notation ##x^2=x\times x##.


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