*a*/

*b*squared equals

*a*squared over

*b*squared, we obtain i/1 = 1/i (where i is the square-root of -1). But then multiplying both sides by i would yield i squared = 1, or -1 = 1, whence 1 + 1 = 0.

......That "proof" is fallacious (and hence it's no reason to outlaw square-roots, for example) because non-zero numbers have 2 square-roots (e.g. +1 and -1 both square to 1, while +i and -i both square to -1) so that, in particular, i/1 = 1/-i. But nonetheless it's fairly compelling because, when square-rooting -1/1 = 1/-1, we could easily assume that

*both*instances of the square-root of 1, and also

*both*instances of the square-root of -1, would have

*the same*sign.

......Furthermore, although when we solve quadratics, for example, we give 2 solutions (arising from the square-root sign in the familiar formula) as a matter of course (it being noteworthy when they're equal), nonetheless we may lose the habit of

*thinking*of, for example, -2 when square-rooting 4. Maybe we lose that habit because we usually use (the very useful)

*functions*, which are

*one-to-one*(e.g. taking the non-negative square-root) rather than

*multifunctions*, which are

*one-to-many*(e.g. taking roots).

......So note that the use of functions is only a matter of convenience (it is not that 4 really does have only the one square-root). I think that it is worth noting that fact because, although some will rightly say that 1/0 is undefined (usually) and that 0/0 is an indeterminate form (many numbers yielding 0 when multiplied by 0), others will say that division by 0 is impossible (less accurately) and even that 0/0 is nonsense.

......The usual "proof" that division by 0 is

*impossible*goes something like this: 0 equals 0, so 0 times 1 (which is just 0) equals 0 times -1 (which is also 0), but if we could divide by 0 we could cancel out those zeroes and so obtain 1 = -1 (whence 1 + 1 = 0). But note that we would only obtain that contradiction if dividing those zeroes by zero gave us, not an indeterminate form (such as

*all*the finite numbers, since zero times any of those is zero) but 1, and

*why should 0/0 equal 1?*

......I can only think of 2 remotely plausible answers, neither of which is very compelling. Firstly we might extrapolate, to the case of

*a*= 0, from

*a*/

*a*= 1 for all non-zero numbers. That is not very compelling because such extrapolations are notoriously unreliable, e.g. think of

*a*to the power of 0, which equals 1 for all positive

*a*, and think of 0 to the power of

*a*, which equals 0 for all positive

*a*.

......Secondly, since 'division by

*' means 'multiplication by the multiplicative inverse of*

**x***' within number fields, and since the multiplicative inverse of*

**x***is whatever yields 1 when multiplied by*

**x***, hence 0/0 should, if allowed, equal 1. But that would only be the case were division by 0 being allowed*

**x***within number fields*; whereas it is certainly

*not*allowed within fields!

......Nonetheless, division by 0 is allowed

*within number pitches*, which contain number fields in an algebraically strong, and maybe even a physically applicable way (and which were defined in my 2005

**:**)

## 27 comments:

Yes, but:

'0!/0^0 = 1'

From a comment by someone named 'Christian Burnham' on PZ Myers' Pharyngula blog.

0^0 (presumably 0 to the power of 0) may indeed be defined to be 1, e.g. in the interests of tidiness; and 0! is certainly 1 by definition (in the interests of tidiness, I imagine). So yes, 0!/0^0 = 1. But the extrapolation from 0 to the power of a, for positive a, would be just as wrong.

1 + 1 = 0 with {0,1} out of |F_2

Greetz

It depends of the type of field you're computing in. I assume you mean 1+1=0 in |R.

(I'm german. I hope field is the correct word)

or you can write it like this:

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

√(-1)(-1)=√1=1

√(-1)*√(-1)=i*i=-1

=>

√(-1)(-1) != √(-1)*√(-1)

("!=" means "isn't equals")

=>

1+1=1+√1=1+√(-1)(-1) != 1+√(-1)*√(-1)=1+i*i=1+(-1)=0

1+1=0 if u r using the binary system

The problem is like what Anonymous said.

√-1 != 1

The reason is simple: 1^2 equals 1 and not -1. It is i^2 that equals -1.

√-1 = i and i = √-1.

√-1 is an imaginary number and imaginary number only.

Am I right on this or did I miss anything?

If 0=a*0 for all a

then doesn't 0/0=a where a is any number...

I know this maths isn't really allowed, but is this what is meant by 0/0 is indeterminate, in that it can be 'equal' to any number?

Does this even count as proof that it's indeterminate?

I think there is an error.

you mean that i=square-root -1 .

but this is not true because I know that square-root function has the domain [0,inf) and not R so your demonstration is not so true I'm not so sure that I'm right but this is what I know about square-root function.

Manu

r^344/23 -1t4 5^65 lmn [3,29/7*!o^]

that is the proper equation if u dont understand send me an email at optimusdenis@aol.com so i may explain. i know the secrets of the universe.

1+1=o can only be true in the C filed the filed of complex numbers

and 1+1=0 in Z2 field, fields are algebric course witch is thought at the universites so i guess that explains all.

Prof.Hmody

1+1 = windows you fags

but one apple plus one apple is two apples right. I dunno I'm juss a kid.

1+1=0

nice

but what is the actual value of "i"

1+1=0

2+2=0

3+3=0

4+4=2

5+5=0

6+6=2

7+7=0

8+8=4

9+9=2

10+10=?

User 'Cem I' It is not like 2+2=0...10+10=0. This problem uses complex number theory that Square root of -1 is i because i^2=-1

If i/1 = 1/i and we multiply by i we get i^2 = 1 so strangely 1 = 1

Sorry to bump this, but I couldn't help noticing nobody had mentioned the distributive laws for the square root function.

For all NON-NEGATIVE real numbers x and y,

\sqrt{xy} = \sqrt x \sqrt y

It's okay if one of x and y is negative as long as the other one isn't. If they're both negative, the answer needs to be multiplied by (-1).

A similar restriction holds for distribution across division. Distribution of the square root function during division is invalid when there is a negative number on the bottom and a positive number on the top.

E.g. RT((-36)/9) = RT(-36)/RT(9) = 6i/3 = 2i

BUT RT(36/(-9)) does NOT equal RT(36)/RT(-9) = 6/3i = 2/i

That's where the 'proof' goes wrong.

P.S. I don't like the way you dismiss materialsm/physicalism as implausible. I understand you may disagree with it and there are worthy aguments against it, but the majority of philosophers are physicalists. This doesn't mean they're right, of course, but it is evidence that the theory is a plausible one and has a good deal of merit.

yes but i dohnt care

I don't really have a picked side in the argument. I am just overwhelmed that there are still some intellectual debate existing in blogs. Regards. :)

Psychic Phone Readings

Here is my belief about the "proof" that 1+1=0...The only thing it truly "proves" is that it is not possible to have more than 1 of anything. Sure you can have multiple variations of any item, however, there are subtle differences between them (except on an atomic level but it is nearly impossible to have anything pure for an extended period of time). So, sure 1+1 can= 0 in certain situations.

by taking log we have,

L.H.S= log(1)+log(1)

= log(1*1)

= log(1)

= 0 = R.H.S

In marriage.....1+1=1......if you are as smart as u think u are......u know this is indeed true.......if u don't believe this is true.....you're probably still single......

There's only one "place" when 1 + + = 0, and that's inside a singularity...

Sometimes, in marriage, 1+1=3. 3=2+a or 3=1+b. a=consensual third party whereas b=baby. Hope this helps!

its simple, 1 + 1 = 2

One unmatched sock plus another one exactly like it equals zero unmatched socks.

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