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Commentator

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# 1=2

06/16/2007 10:40 AM

How can be 1=2?

The ans to the question was:

if, a=b

then, ab=b2

or, ab-a2=b2-a2

or, a(b-a)=(b-a)(b+a)

or, a=b+a

taking, a=1,

1=2.

Is there any error that is disguised, in here?

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The Engineer

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#1

### Re: 1=2

06/16/2007 1:48 PM

From this step:

a(b-a)=(b-a)(b+a)

you go to this step

a=b+a

by dividing both sides by (b-a)

But if a=b as is stated in the beginning, then (b-a)=0. Since dividing by 0 in math is undefined, your step of dividing by (b-a) is invalid.

Any conclusions reached after taking an invalid step are also invalid.

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#8

### Re: 1=2

06/17/2007 7:29 AM

thanks, that makes some sense to me!!

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Anonymous Poster
#18

### Re: 1=2

06/18/2007 1:07 PM

also, since a-b=0 in this statement, you have a*0=(b+a)*0, which is 0=0 not 1=2.

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#2

### Re: 1=2

06/16/2007 3:42 PM

It just goes to show...

ignore the maths...do it by eye!

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Associate

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#3

### Re: 1=2

06/16/2007 4:44 PM

Allright. I put it this way.

1=sqrt(1)

= sqrt(-1X-1)

=sqrt(-1)Xsqrt(-1)

= j X j j-complex number

= j^2

= -1

therefore 1+1=0 or 2=0. How is this possible?

The Engineer

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#4

### Re: 1=2

06/16/2007 6:15 PM

That's incorrect because your first step isn't valid, You wrote:

1=sqrt(1)=sqrt(-1X-1)

How? How is the square root of one equal to the sqrt of negative x minus 1? I'll give you a hint, it isn't.

Anonymous Poster
#10

### Re: 1=2

06/17/2007 5:34 PM

That's a nice formulation of the paradox, Omkar.

for Roger Pink:
Quadratic equations generally have two solutions. The solutions to x^2=1 are indeed +1 and -1. This is pretty elementary, and extensively used in the analogies between logic and arithmetic.

The first part of setting up the error is in the following step:
sqrt(-1x-1) = sqrt(-1) x sqrt(-1)
That type of statement is true if the solutions of each section are unique. In this case, you can choose values for sqrt(-1)* that give the correct answer - but the statement itself is incorrect.

* i.e. j, and -j

The Engineer

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#12

### Re: 1=2

06/18/2007 12:33 AM

Ok, I'm going to pretend you guys didn't just try to explain 5th grade math to me and assume you realize now that I thought x was a variable (My last sentence of my original comment should have made that clear). I didn't realize all he was saying was:

1=√(-1)(-1)=√-1√-1= i x i = -1

What you all seem to miss in your in your redundant explanations, and why it isn't a 'paradox', is that there is also two square roots of -1, they are i and -i. You must consider both for the complete solution.

Which means

(sqrt (-1)) x (sqrt (-1))=

can be any of the following combinations-

(i x i) = -1
(i x -i) = 1
(-i x i) = 1
(-i x -i) = -1

since

i x i = -i x -i = -1 and

-i x i = i x -i = -(-1)= 1

thus

(sqrt (-1)) x (sqrt (-1))= ±1

Cleary no paradox beyond the paradox of people being amazed by their own mathematical inability.

Anonymous Poster
#15

### Re: 1=2

06/18/2007 5:28 AM

I read that presentation to be humour - the author clearly knew the error was there.

The short version of this is:

-1 = sqrt(1) = 1

The humour is the method of "hiding" the ambiguous definition step.

Fyz

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#17

### Re: 1=2

06/18/2007 10:11 AM

I'm pretty sure he just didn't realize that the sqrt(-1) has two solutions, ±i. Whether he was joking or not is irrelevant.

Elika's example is more subtle. All the algebraic steps were done correctly, she just failed to realize that her constraint (a-b=0) made one of the algebraic steps invalid.

Anonymous Poster
#19

### Re: 1=2

06/18/2007 4:13 PM

Obviously, subtlety in this case is a matter of viewpoint.

The reason that I thought this was "nice" is that this illustration is based on the ambiguous way we use "equality". Generally, the dangers of this are under-appreciated, whereas the issue of multiply/divide by zero is pretty much worked to death in most high-school maths courses.

So my view is that it illustrates an important, and not widely enough appreciated, mathematical point with more than a little element of humour. Plus, I've seen more than one 'well-qualified' engineer producing a nonsense result for this very reason.

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#20

### Re: 1=2

06/18/2007 4:35 PM

You Wrote: "The reason that I thought this was "nice" is that this illustration is based on the ambiguous way we use "equality". Of course, the problem is it doesn't illustrate the point you're mentioning."

That's my point. This doesn't illustrate an ambiguous use of equality. Consider below:

1 = √(-1)(-1)=√-1√-1 = -i x i = 1

You see. I've made exactly the same incorrect assumption about equality in the beginning and still got a answer that looks correct. Do you see my point yet?

Anonymous Poster
#21

### Re: 1=2

06/18/2007 4:55 PM

I doh't know what you mean by ambiguity, but I mean something that is capable of taking multiple interpretations. In order to illustrate its importance, you can find an illustration that ends up with an incorrect result - preferably in an amusing way.
The fact the you can find a both an incorrect and a correct interpretation is precisely an illustration of ambiguity.
You have also missed the mathematical point - the final expression [effectively sqrt(1)=-1] is not incorrect, any more than that 1=sqrt(1) is incorrect. The error is in the interpretation of the statements - due to the ambiguity in the use of the equals sign.

Maybe the ambiguity in the use of equals wasn't carefully enough defined. The usual usage of = are:
1. that the two elements are intrinsically equal (e.g. 1+1=2)
2. to set two elements equal (e.g. [let] x=2)
3. as consequences of a setting as in 2

Here, = is used for "one of the possible values of () is". That is a different meaning, but is often used alternately (as here) with "are intrinsically equal" without any clarification. The results can be the sort of nonsense result you see here.

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#22

### Re: 1=2

06/18/2007 5:03 PM

No, I didn't miss the point which is why I wrote "still got a answer that looks correct". Looks correct is not the same as correct. I was careful in my language because I know who I'm arguing with here.

I used your definition of ambiguous.

I forgot rule number 1 which is of course "You're always right"

Roger

Anonymous Poster
#23

### Re: 1=2

06/18/2007 5:26 PM

I use Fyz when I think I'm certain, or it's clear I'm joking (the latter after persuasion by Kris) - hence the lack of signature here. If I wanted full disguise, I'd change my writing style and use a different system for the communication.

Avoiding the personal... I'm completely missing your point; so I need you either to re-explain it in terms that allow this, or try to explain where I've gone wrong.

My present position - still capable of being moved...

If you can write something that looks correct but isn't, isn't that ambiguity? So the question is - where did the ambiguity arise? My view is that there are the following possibilities:
The first expression 1=sqrt(1) is wrong
The first expression 1=sqrt(1) is ambiguous
There is an error or ambiguity somewhere else in the algebra

Each of the lines in the original formulation is indeed equal to a square root of one. So where is the error? Please explain.

Incidentally, you will find places in the various threads where I have admitted (even as Fyz) to being wrong, so that barb is inaccurate, which makes it all the more unworthy.

Not really Fyz

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#24

### Re: 1=2

06/18/2007 8:58 PM

Ok, I'll try to demonstrate why the humor wouldn't work by analogy.

Imagine someone writes:

Clearly "desturb" should be spelled "disturb".

So someone trying to be humorous writes in response:

"I'll bate the lure"

Pretty clever right? After all they wrote "bate" instead of "bait".

Well actually, not very clever at all, because it turns out there is nothing wrong with the sentence beyond the humorists lack of vocabulary. You see, bate is a word, it means among other things "to lessen" or "to remove". "I'll bate the lure" literally means "I'll remove the lure". Of course the humorist didn't know that when they wrote it.

So I come along and read the first sentence and point out that "desturb" is spelled incorrectly and everyone agrees. I get to the next sentence which reads "I'll bate the lure" and say there is nothing wrong with that sentence.

This is followed by people trying to explain to me that "bate" is misspelled and should be "bait". I try to explain that I understood what the second sentence was meant to do, it didn't accomplish its goal because bate is a word too and it works in that sentence. I'm then told in no uncertain terms that I'm missing the "subtlety" of the joke. Again it is explained to me that bate is a misspelling of bait. At this point I give up realizing no one is going to bother to look up "bate" in a dictionary.

That's what happened in this thread. The humorist wrote:

1 = √(-1) x (-1) =√-1 x √-1 = i x i = -1

I come along an point out that √-1 can equal -i so you could write it as:

1 = √(-1) x (-1) =√-1 x √-1 = -i x i = 1

Obviously 1=1 is not a problem. The math sentence is not a paradox, it's fine. It only appears to be a paradox is you lack the math vocabulary of -i.

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#25

### Re: 1=2

06/18/2007 9:35 PM

All right children, that's enough. Class will come to order.

To quote Sir Donald Francis Tovey (1935, p.195), "Theorists are apt to vex themselves with vain efforts to remove uncertainty [ambiguity] just where it has a high aesthetic value."

Aesthetic value in the present context is "humor"! It means ±humor, so laugh damnit!

QED

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#26

### Re: 1=2

06/18/2007 10:14 PM

Oh brother.

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#27

### Re: 1=2

06/19/2007 5:04 AM

Jonjohn - thanks for your intervention - but it would still be good if we could agree on the issue behind the error (I tend to agree that neither version of 1=2 is a fully-blown paradox)

Roger - By stating that puns are not funny if they have a relevant second meaning, you have at a stroke banished 90% of what is used for humour in my two prime languages. And of course, any humorists using the technique would need know the second meaning, as they could not frame a proper pun without that knowledge. (Nevertheless, it is generally considered unkind to build your pun on a mistype or accident of another communication). I find your tendency to assume other people do not know what they are doing to be quite worrying (except when applied to politicians, but that is a different matter).

(Ignoring your mistype) the fact remains that you have, as you previously stated, only appeared to correct the "spelling" *. It is correct usage (as Omkar did) to break the derivation down into self-contained steps. And each self-contained step should be correct.
Omkar's first line is correct by convention, although if you require equals to mean "is inevitably equal to" you have to write +/-1 = sqrt(1). The last line reapplies that convention in a way that is inappropriate. Obviously, the individual stage should be equivalent to
sqrt(-1) x sqrt(-1)=+/-i x +/-i
showing that the value needs to be defined elsewhere (in this case with reference to the initial conditions).

To my mind, using the closeness to a well-known "paradox" to highlight this dangerous (and not that uncommon) misusage is humorous. I can settle for differing on the humour (the world would be a dull place if we were all the same). But I hope we can agree that the problem comes down to the ignoring of an inherent range of options.

*Perhaps that is a better joke than the original?

Not quite Fyz

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#28

### Re: 1=2

06/19/2007 8:54 AM

an argument that apparently derives self-contradictory conclusions by valid deduction from acceptable premises

I've pointed out a number of times now that you can make the assumption made and still get the answer you expect, in other words, the answer isn't self-contradictory and thus this isn't a paradox. So again, starting with the original premise of 1=√1 I will show how you can avoid getting a self-contradictory answer:

1=√1=√(-1)(-1)=√(-1)√(-1)= i x -i = 1 (Not Self-Contradictory)

Would you please comment specifically on this.

You Wrote "I find your tendency to assume other people do not know what they are doing to be quite worrying"

I'm glad that we're avoiding making this personal. They should name the high road after you.

Roger

Anonymous Poster
#29

### Re: 1=2

06/19/2007 10:00 AM

In mathematical terms, it is normal to regard each sequential equation as a stage that can be self-contained. That is the reason that the divide-by-zero falls down - the process used assumes that the algebraic statements made are general statements, in which case you could validly divide by (a-b).
If we allow a similar independence for the stages of the square-rooting sequence, you have no reference that tells you that you should choose either specific solution. Just as, in the divide-by-zero there is no apparent local reason that you hould not divide by (a-b), in this case there is no local reason that you should select a specific sign for the answer.
Your formulation disguises the conventionial independence of successive stages by writing the sequential equations in a line - with the RHS of one equation serving as the LHS of the next. This does not, however, mean that the individual stages should not be self-contained. In that sense, the choice of -i x i is equally arbitrary locally to the selection of i x i (for example). The "correct" local solution remains +/- i x i (although this is expressible in different ways). Similarly, if we wish to be rigorous in the first equation, we ought to write +/-1=sqrt(1) - even though (for positive real numbers) there is a convention that "sqrt" is the positive root unless there is an apparent reason to do otherwise. I believe that it is this convention having become ingrained (and used even where +ve and -ve don't really have that meaning) that is at the root of these problems. As I tried to say before, I wouldn't care if I had not seen very closely analagous issues causing nonsense results in real design situations.

Regards

Definitely lacking Fyz (to say nothing of a spell-checker)

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#30

### Re: 1=2

06/19/2007 10:38 AM

Yes, that is a valid argument:

1=√1, -1=√1, i=√-1, -i=√-1 are all the same sort of mistake, they imply equality to a single value when in fact they should be written as:

±1= √1
±i = √1

Choosing a particular single value i or -i instead of ±i for √-1 is the same mistake as beginning the problem with 1 or -1 instead of ±1 equal to √1.

A very good point. But I never justified (specifically choosing i or -i to suit my purposes) what I did as mathematically correct. I just used it to illustrate that the method used to illustrate the problem of using single values for square roots doesn't work in this case. The problem is if you make the same mistake twice, which is reasonable since you should not be expected to learn the rule halfway through the equation, then you get an answer that is reasonable (1=1). This whole time all I have been saying is the logic of his proof, or as you call it, paradox, has been invalid.

In other words, I haven't been saying that what he wants to point out is incorrect, I mean anyone with some algebra knows there are 2 solutions to a square root (except 0). No, what I'm saying is that the method used to illustrate the problem, reductio ad absurdum, is used incorrectly and fails to make the point he wishes to make.

Can you comment on the Reductio ad absurdum method as it is used in this case and indicate why it is correct? What is the absurdity of 1=1? Surely you agree that the answer is only absurd if 1=-1? If that is the case, how can it be said, when 1=1 is a possible answer by making the same mistake again later in the derivation, that this equation creates a paradox?

Are you starting to see my point yet?

Anonymous Poster
#31

### Re: 1=2

06/19/2007 10:59 AM

Yes and no - yes with relation to the square-root route not being a paradox, and no w.r.t. it being different in that respect from the a^2-a.b method. I'm sure you could find a way to repeat the mistake in the divide-by-zero example in a way that recovered the 1=1 at the output (repeating a mistake in a symmetrical manner should generally do this, I think).

I already said, I don't think either version is a fully-fledged paradox - on the other hand, I'm not certain that such a thing (as I understand your strict definition) exists*.
In that context, I regard the square-root version as equivalent in logical merit (or otherwise) tp the divide-by-zero, but relevant because it illustrates confusions that can arise, and amusing because it is not as hackneyed as the original.

*On the basis that if every stage is correct of an argument is sound and applicable to the situation, then the result will also be correct and applicable.

I think you'll probably agree that we've done this bit of the topic well-and-truly to death. Hopefully also that our disagreement is of emphasis rather than substance. Finally, that you will appreciate that my reaction was mainly an attempt to counter the apparent negativity of the original response to what I thought was a potentially useful contribution.

Almost Fyz again?

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#33

### Re: 1=2

06/19/2007 11:20 AM

Fyz,

Not sure about the whole name thing. As far as I'm concerned, your Fyz (unless you want to be called something differently).

As for this argument. I don't agree (surprised?). In my analogy with the words I think you'd see that one sentence is definitely wrong and the other isn't. I think we're looking at the same thing here. I understand you don't agree and that we have beat this thing to death.

I hold no ill feelings, afterall, through the insight of our discussion, I found a way to put into words what was wrong with equation. Before that I just knew, like one knows that it is bright or dark, hot or cold, that the logic was flawed. Our discussion forced me to identify the technique used and the flaw. I know you weren't convinced but that's secondary since I learned something.

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#32

### Re: 1=2

06/19/2007 11:14 AM

Hello Roger

I'm not too sure he's using reductio ad absurdum. I understand this to consist of assuming the opposite of your proposition, and showing that this leads to an absurdity, which means your proposition is correct. Perhaps it's in there somewhere and I'm missing it.

Anyway, I once read a book about a mathematician (Dutch, I think he was) who didn't approve of reductio ad absurdum and argued that such proofs should be ruled out. To start the attack, an example quoted was "prove i < 1". For assuming the opposite, i > 1 and squaring, -1 > 1, which is clearly wrong, completing the proof.

The comment was made "we all know this is nonsense, but what is wrong with the proof?" Well, my reaction, as I expect most contributors to this thread would be, was - if he can't see what's wrong with that, he shouldn't be writing books about maths (or math if you prefer).

Not particularly relevant to the present discussion, which I've enjoyed following, but I thought I'd throw it in.

Cheers......Codey

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#34

### Re: 1=2

06/19/2007 11:46 AM

Codemaster,

I could of course be wrong, but I think it is reductio ad absurdum. Here is a definition, tell me what you think:

"is a type of logical argument where one assumes a claim for the sake of argument, derives an absurd or ridiculous outcome, and then concludes that the original assumption must have been wrong as it led to an absurd result."

As for this debate, it is entertaining in a "I can't believe I've spent this much time debating 1=-1" sort of way for me. It's sunny out, clearly I should get outdoors for a little bit.

Roger

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#35

### Re: 1=2

06/19/2007 12:52 PM

Roger - I think the wiki definition you quote is pretty much equivalent to mine, but perhaps with different emphasis. The wiki one seems to show something is false, but doesn't establish what the correct answer is (though it goes into greater depth further down the page).

My definition is meant to be how I understand reductio ad absurdum is used in maths, where you start out knowing what you want to prove, and assume the opposite of that. Showing this gives an absurd result in theory proves your proposition, though I'm sure philosophical objections could be raised, as the Dutch guy I mentioned was doing.

Codey

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#36

### Re: 1=2

06/19/2007 1:58 PM

I think what you're describing is called Proof by Contradiction which is similiar but slightly different from Reductio ad Absurdum. Please take a look at this link and tell me what you think:

http://www.math.csusb.edu/notes/proofs/pfnot/node6.html#SECTION00042000000000000000

Roger

Anonymous Poster
#37

### Re: 1=2

06/19/2007 4:17 PM

I think the issue here is that there is no explicit statement or theory that clearly leads to the absurd result via faultless logic. In fact, my interpretation is that in both cases the problem is of missing constraints and/or generality in the derivation - not a false explicit statement or theory as such.

Contradictory results -yes. Absurd? Absurdlutely. Reductio? Possibly not.

Incidentally, because mathematical systems can be self contained, statements that "common sense" would reject as absurd can be acceptable within specific mathematical systems (and indeed in physics at the subatomic level). For this reason, SFIK reductio ad absurdum is not commonly accepted as a method of mathematical proof. I think the closest thing to that is the use of contradiction of an opposite statement as proof - and even here, although this is accepted in most mathematical systems, the additional possibilities in systems where this method of proof is not accepted has led to some interesting insights.

Fyz

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#38

### Re: 1=2

06/19/2007 4:28 PM

You Wrote: "For this reason, SFIK reductio ad absurdum is not commonly accepted as a method of mathematical proof."

In this link from a website by California State University, San Bernardino Department of Mathematics it lists it as a method. Not to say they couldn't be wrong? What do you think? Is the Math Department from CSUSB wrong?

http://www.math.csusb.edu/notes/proofs/pfnot/node6.html#SECTION00042000000000000000

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#39

### Re: 1=2

06/19/2007 4:50 PM

What he is describing is one route to using a joint statement (p_AND_r=true) together with a contradiction if r is not true*. That is not merely using absurdity, it is using contradiction of a known-true statement.

Of course, I would not claim to be knowledgeable enough to state that CSUSB have got it wrong - just that this definition of reductio ad absurdum is indistinguishable from proof by contradiction. I admit I should have been more precise in my statements - I should have said that reductio ad absurdum is not regarded as a mathematical proof except for the special case where the "absurd" statement contradicts a statement that is known to be true within the specific mathematical system. This used to be taken as a very specific subset of absurd statements, and that in my day this meant that you would always have used the term "contradiction" rather than the longer and more ambiguous "reductio..." for such mathematical proofs.

*r AND p => q by assuming p ANDNOT q, and establishing a contradiction with r. That is just using NOTq => NOT(p_AND_r). As we start from p_AND_r is true, this is just a particular case of proof by contradiction.

Fyz

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#40

### Re: 1=2

06/19/2007 6:09 PM

Whatever.

Here is the definition from Wikipedia for Reductio ad Absurdum:

"Say we wish to disprove proposition p. The procedure is to show that assuming p leads to a logical contradiction. Thus, according to the law of non-contradiction, p must be false."

Proposition:

1=√1

Procedure
1=√1
1= √(-1)(-1)
1=√-1√(-1)
1= i x i
1=-1

Since 1=-1 is a logical contradiction, thus 1=√1 must be incorrect.

Seems like Reductio ad absurdum to me. I've looked around and despite what you may personally believe, Reductio ad Absurdum seems like a generally accepted mathematical method of proof. Please see links below.

http://www.mathpath.org/proof/proof.methods.htm

The rest here is just overkill:

http://www.mathpath.org/proof/sqrt2.irrat.htm (here's an example)

Roger

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#41

### Re: 1=2

06/19/2007 7:01 PM

Hi Roger,

I have not followed this thread completely through but just looking at the procedure that you show:

1=√1
1= √(-1)(-1)
1=√-1√(-1)
1= i x i
1=-1

Step 2: 1= √(-1)(-1) The property √a√b = √ab where a > 0 and b > 0, does not carry over to principal square roots of negative numbers:

Just as √(-4) √(-9) = (2i)(3i) = 6i² = -6

and √(-4)(-9) = √36 = 6 ≠ √(-4) √(-9)

Perhaps this is getting away from the debate over types of logical fallacies but this is just my take on this particular post.

Regards,

-John

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#42

### Re: 1=2

06/19/2007 7:40 PM

John,

Pretty much makes the debate over types of logical fallacies irrelevant, though now I know that Proof by Contradiction is a subset of Proof by Reductio ad Absurdum, not that I have any idea when that might come in handy.

Roger

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#45

### Re: 1=2

06/20/2007 2:22 PM

John,

I regret this, but I'm not as certain you were correct as I was a day ago. I've had some time to read up on the subject and unfortunately the introduction of complex numbers muddies the water a bit. I think deep down I was dying to agree with somebody about something in this thread. Anyway, if you're curious, have time and an open mind, please take a look at some of these links and you'll see what I mean.

http://mathforum.org/library/drmath/view/53873.html
http://mathforum.org/library/drmath/view/64430.html

If you're not familiar with mathforum.org, its very good, so you can be assured the quality of the answers are top notch.

Roger

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#46

### Re: 1=2

06/20/2007 3:22 PM

As I understand it, complex numbers don't have principal values in the accepted sense. But I'm not certain that this contradicts what John actually wrote - that the law does not apply - it merely gave a reason.

On the other hand, there are situations in the application of complex numbers where the difference between +i and -i is quite important. Granted that they are equivalent in principle; but once you set up a correspondence to one of them, the sign does acquire significance. Radio and transmission engineers will be very familiar with this situation, I think.

Fyz

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#47

### Re: 1=2

06/20/2007 10:13 PM

Hi Roger,

What I last posted was a direct quote from my old college algebra textbook:

College Algebra & Trigonometry, Gilbert & Gilbert, 1986, ISBN 0-534-06120-6

Allow me to present a little more of this to see if it un-muddies things a little bit (or perhaps muddies the water even more):

"Multiplication is defined by (a+bi)(c+di) = (ac-bd) + (ad+bc)i

Since i² = -1, we find that -1 has another square root.

Since (-bi)(-di) = -bd

we have

(-i)(-i) = -1 thus -i² = -1 just as i² = -1

Then i and -i are both square roots of -1. Similarly, 2i and -2i are square roots of -4. Every negative number is of the form -a, where a is a positive real number, and -a has two square roots, i√a and -i√a. The square root of -a which has the positive imaginary part is designated as the principal square root, and is denoted by √-a."

I'm not trying to give you [all] an algebra lesson but I just wanted to emphasize a couple of things. Since i is by definition √-1, we need to make it fit into the additive, multiplicative, and associative properties as has been done. However, I repeat what I said earlier: √a√b = √ab where a > 0 and b >0, does not carry over to principal square roots of negative numbers. To do so results in the absurdities that have thus far been pointed out.

This thread has been very helpful in that it has caused me to re-examine some long forgotten principles. To witt, question everything! Just because it's in a printed book doesn't necessarily make it gospel. I've gotten a little lazy over the years but I'm getting over it. I happen to agree with Gilbert, et. al. on this one but the debate going on in this thread is teriffic.

Regards,

-John

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#48

### Re: 1=2

06/20/2007 10:42 PM

John,

I hear you, but here's my problem.

As you stated:

√a√b = √ab where a > 0 and b >0

But I'm sure you'll agree that:

√-25 = 5i

In order for that to happen, I believe you have to follow the following steps:

√-25=√25√-1= 5i

where a=25 and b=-1, but b<0

So the example above has violated the rule:

√a√b = √ab where a > 0 and b >0

Please, take a look at the example I provide above and tell me what you think. It is the reason I'm doubting the distribution rule you provided.

Roger

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#49

### Re: 1=2

06/21/2007 1:58 AM

Roger,

You make a convincing argument. The way you write it makes it look legit. There is a falacy here though and I think it is best shown like this:

Using the format √a√b = √ab, and using the values a=(-25), b = -1 (not a= 25).

√(-25) √(-1) = (5i) i = 5i² = -5, but meanwhile, back at the ranch:

√(-25)(-1) = √(25) = 5. I don't think you're expressing this step in your example.

It appears to me that in the complex number form a + bi, you're intermixing b with i and that leads to problems.

Also if I may go back to your post #40:

1=√1
1= √(-1)(-1), this statement, as written, is indeed correct: 1 = √(-1)(-1) = √1 = 1, but even so, one cannot build on it to derive the following steps that you give:

1=√-1√(-1) [false]

1= i x i [false]

1 = -1 [false]

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#50

### Re: 1=2

06/21/2007 11:47 AM

John,

You Wrote: It appears to me that in the complex number form a + bi, you're intermixing b with i and that leads to problems.

But that's my point. You don't have an i until after distribution. You must first separate the -1. That seems to me like it violates the rule you've brought up.

One other point I'd like to make is that square roots always have two solutions (with the exception of √0). For instance √25=±5. I think in your example above you are merely finding the second root, not disproving the mathematical step used.

Consider:

√25 = ±5 since (-5)2=25 and (5)2=25

√-1 = ±i since (-i)2=-1 and (i)2=-1

So is it really so surprising that:

√(-25) √(-1) = ±5

To be clear, I'm not suggesting I'm correct here. I'm just saying that all the math seems to work out fine. I need an example that is absurd that makes me say "wait a second, that's definitely not right". In my mind:

√(-25) √(-1) = ±5

Isn't an absurd result in my mind.

My point isn't astonishing or ground breaking, which means some book or webpage should address it for the sake of thoroughness (either to confirm it or deny it), but I can't find any such webpage. I've seen webpages mention that -i is also a root of -1, and then completely ignore that when they calculate negative square roots, which seems inconsistent to me (and ignoring is not the same as refuting). I'm really not happy about this. I don't mind being wrong, and I love to be right, but if there is one thing I can't stand, it's not knowing.

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#51

### Re: 1=2

06/22/2007 12:41 AM

Hi Roger,

I'm still mulling all this over but in the meantime consider this:

You may think you can do this:

i² = (√-1)² = √(-1)² = √1 = 1

But this doesn't make any sense! You already have two numbers that square to 1; namely –1 and +1. And i already squares to –1. So it's not reasonable that i would also square to 1. This points out an important detail: When dealing with imaginaries, you gain something (the ability to deal with negatives inside square roots), but you also lose something (some of the flexibility and convenient rules you used to have when dealing with square roots). In particular, YOU MUST ALWAYS DO THE i-PART FIRST!

-John

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#52

### Re: 1=2

06/22/2007 4:35 AM

Johnjohn - perhaps I'm missing something, but I don't see a problem here.

√1 = 1 or -1, and taking the -1, the conclusion is correct. I think the apparent error arises from the various meanings of = , as pointed out by Guest in #21. Here, √1 = 1 is a true statement, but = is not exclusive, as √1 = -1 is also true.

Need to take the answer that makes sense, and ignore the other(s) (that's probably a bit too sweeping, but in general). If you're \$100 in the red, bank balance -\$100, you can't square that to get \$210000 (whatever a \$2 is!) and then √ it to get \$100 in the black, mores the pity.

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#53

### Re: 1=2

06/22/2007 10:47 AM

Hi Codey,

What you say: "√1 = 1 or -1" is, of course, correct. The error is in mapping i² across to +1. It's the same absurdity we've been talking about.

Seems to me what makes this thing so apparently ornery is that we're dealing with unity. Regardless whether you divide, multiply, or square root unity by unity, the only thing that changes is the sign. As my previous post pointed out, with the concept of "i" you gain something but you also lose something.

I'll take all the \$² you want to contribute.

-John

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#54

### Re: 1=2

06/22/2007 11:27 AM

John,

I think what your demonstrating illustrates that there are two answers to a square root, not that you can't distribute negative radicals. For instance:

You Wrote:

i² = (√-1)² = √(-1)² = √1 = 1

but isn't the √1 = ±1 since (-1)2= 1, so:

i² = (√-1)² = √(-1)² = √1 = -1

In essence, in your example, you chose the wrong root of 1. That doesn't seem to indicate to me that you can't distribute negative radicals.

I'm still not sure. I will look around today and see if I can't find some explanation for this. I'm still not convinced I'm right, but I'm not sure I'm wrong either, I just don't know for sure.

Roger

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#55

### Re: 1=2

06/22/2007 3:47 PM

Roger.

As you say i² = (√-1)² = √(-1)² = √1 = 1 does result in an incorrect conclusion (+1). What makes it wrong is, of course, the i² at the front. Otherwise, ±1 is the correct root and +1 at the end if perfectly valid. If one had never heard of complex numbers, the entire equation would make perfect sense.

Now as to the problem of distributing negative radicals, I can't give you a sound argument either for, or against, distribution because negative distribution (at times) seems to make sense. Using your earlier example √-25 = √25 √(-1) = 5i, where a = 25 and b = -1, but b < 0, then using the format: √-a √-b = √(-a)(-b), the thing that we're debating, we would have √-25 √-1 = √(-25)(-1) = 5i(-1) = -5i which can't be because the correct result is 5i. So in this case negative distribution does not work! Of course, by writing the equation a little differently it can be made to appear to correct.

Going back to your earlier post, #28 you said: "I've pointed out a number of times now that you can make the assumption made and still get the answer you expect, in other words, the answer isn't self-contradictory and thus this isn't a paradox. So again, starting with the original premise of 1=√1 I will show how you can avoid getting a self-contradictory answer:

1=√1=√(-1)(-1)=√(-1)√(-1)= i x -i = 1 (Not Self-Contradictory)"

In light of what's been discussed since then I'd like to hear you expand on this a little more since the conclusion is obviously incorrect (self-contradictory ?).

Anyway I have no conclusive answer to the problem of negative distribution, not yet anyway. The wheels are still churning though.

-John

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#56

### Re: 1=2

06/22/2007 6:43 PM

Roger,

As you've no doubt seen, I've made a couple of gross errors in the previous post. I said: "√-25 √-1 = √(-25)(-1) = 5i(-1) = -5i which can't be because the correct result is 5i." The -5i solution that I posted is, of course, wrong. Allow me to redo it using (illegal?) negative distribution: √(-25) √(-1) = √[(-25)(-1)] = √(25) = 5. Now, since I presumed, in this example, that the distributive property holds for negative roots, the solution; 5, is obviously incorrect because in the second step where we distribute the negative roots, "i" gets thrown out the window! We know the correct solution to be 5i.

I don't know of a formal proof that the distributive property does not hold for negative roots. It seems though that the examples alone deny its validity.

In the example you gave from your post #28, I said: (self-contradictory?). In step 2 you use the negative distribution we've been debating. So going by what I just said above, step 2 should be √[(-1)(-1)] = √1 which agrees with the step before it.

In your next to last step; i x -i = 1, I realize that -i² gives the same thing as i² but I don't think you can just arbitrarily throw either sign you want to on it because by doing so it allows you to choose the solution to be whatever you want it to be: either 1 or -1.

-John

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#57

### Re: 1=2

06/22/2007 7:47 PM

John,

I hear ya. At this point, I'm taking a step back. I'm going to try a new approach. All ambiguity is removed by this approach, however, the approach requires an understanding of complex numbers.

Complex Numbers- An Introduction (Please Skip if You Are Familiar)

Complex numbers are part of a general number system. All reals are complex numbers, but not all complex numbers are reals. All imaginary numbers are complex numbers, but not all complex numbers are imaginary. Basically complex numbers include real numbers, imaginary numbers, and combinations of the two. For instance

A complex number is defined to have the form (a +bi) where i=√-1.

The real number "4" written as a complex number would be (4 +0i)=4
The imaginary number "2i" written as a complex number would be (0 +2i)=2i
3+4i is also a complex number, but it has both real and imaginary parts. When you express a formula in terms of complex numbers, you are being very general and that formula is automatically true for real numbers (Since the reals are a subset of complex numbers)

(a+bi) is not the only way to write complex numbers. They can also be written as re. In order to understand why, you must recall that e= Cosφ + iSinφ. Thus if we call d=Sinφ and c=Cosφ, we can rewrite reas r(Cosφ + iSinφ)=r(c + di) and if we say a=rc and b=rd, we get (a + bi), the other form for complex numbers. In other words, the two forms are equivalent.

Some examples of complex numbers:

1ei0=1(1)=1

1e=1(-1)=-1
Since: e=Cos(π) + iSin(π)= -1 + i(0) = -1

1eiπ/2= i
Since: e=Cos(π/2) + iSin(π/2)= 0 + i(1) = i

3eiπ/2=3i (hopefully you can see why at this point)

Things I know for certain

1. All square roots have two unique solutions.

2. All cube roots have three unique solutions.

3. I know this because of the following algorithm that works for complex numbers (which is a number system that includes both real and imaginary numbers):

for k = 0, 1, 2, …, n − 1, where represents the principal nth root of r.

This equation holds the key, but if you are unfamiliar with complex math it is very confusing. Lets look at the equation in detail to figure out what it is telling us.

Essentially k tells you how many unique solutions there are to the root. Notice that k will be 0,1,....n-1, where n is the "root". So for n=2, a square root, k= 0, 1. In other words, there are two solutions (k=0, and k=1) for the square root. For a cubed root, n=3, so k=0,1,2 so there are 3 unique solutions to a cubed root. So on and so forth.

So if we want a thorough explanation of roots, that equation is the most thorough.

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#58

### Re: 1=2

06/22/2007 9:11 PM

So based on the equation in my last response,

for k = 0, 1, 2, …, n − 1, where represents the principal nth root of r.

Lets see if the rule

√a√b = √ab where a > 0 and b >0

is true

Proof (sort of):

If the number a is negative, ie. a<0, it means φ=π for the complex number re

So negative numbers are written as complex numbers as:

re=-1 where r=1,2,3,.....

To be more explicit:

-1=1e
-2=2e
-3=3e
etc.

Thus,

√a√b=√ce√de (where c=-a and d=-b are just the r part of re)

*note that if c=-a and a<0, c>0 must be true. Same for d. This is the key to this proof.

and

√ab=√cede

Thus the rule is rewritten in complex numbers as:

√ce√de ≠ √cede

So

√a√b=√ce√de=(ce)1/2(de)1/2=c 1/2 eiπ/2 d 1/2 eiπ/2 = (cede)1/2= √ab

please note that above c and d are positive, so there is no issue. All operations on the exponent were legal.

So I'm pretty sure that the rule:

√a√b = √ab where a > 0 and b >0

is only true for reals, which makes sense since √-1 is imaginary an outside the reals jurisdiction.

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#59

### Re: 1=2

06/22/2007 9:53 PM

Roger,

It's been a long time since I delved into higher math so on this I'll have to defer to you. At least your conclusion:

"So I'm pretty sure that the rule:

√a√b = √ab where a > 0 and b >0

is only true for reals, which makes sense since √-1 is imaginary an outside the reals jurisdiction."

agrees with what I've been saying. I'm glad you pursued it to a logical end though. If it comes up again, I'll just say, "Go ask Roger".

Anyway, even without a proof, every time distributing negative roots is applied, the end result is always wrong. So there's no contradictions in such equations, it's just that the = sign has to be replaced with the ≠ sign.

In my last post I was attempting to address the issue of the distributive property. I had the feeling you thought I was avoiding it. Even though I could not offer a proof of its invalidity, my contention was that every example that was posted using negative distribution always produced an incorrect solution thus it just about had to be invalid (still not nearly as good as a definitive proof).

Its been an interesting discussion and I've learned a lot from it.

Thanks,

-John

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#60

### Re: 1=2

06/22/2007 10:16 PM

John,

I can honestly say that at no point did I get the impression you were avoiding anything (believe me, I would have said something). I was unhappy with the examples we were both using because they seemed incomplete, selective, and anecdotal. I wanted something definitive. When I realized there was a general rule for roots of complex numbers, and that negative numbers were just a subset of complex numbers with the form re , I realized we could test if the rule forbidding negative numbers held when considering complex numbers, which it did not.

The last point I made certainly is no where near the first point I made on this thread. I've learned a bunch of things on this thread including the rule for complex roots which I may find handy in the future and the nature of mathematical proofs. Anyway, that kind of learning only comes from a give and take, so thanks (And thanks to Fyz too) for giving me a thread to discuss it.

See you around CR4,

Roger

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#44

### Re: 1=2

06/20/2007 1:21 PM

You've convinced me - Whitehead and Russell finally won the terminology argument, and reductio ad absurdum has acquired that narrow meaning in mathematical argument. Others (possibly mostly from the same establishment) argued that the phrase had a different and wider meaning, and so should not be used for contradictions of a proved statement. I stand corrected.

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#16

### Re: 1=2

06/18/2007 7:22 AM

Hello Roger

Agreed, it's a bit like saying that as e2.pi.i = 1 this means epi.i = 1 (by taking sqrt) while we know by coming from the "opposite direction" that epi.i = -1

Codey

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#5

### Re: 1=2

06/17/2007 2:27 AM

Whenever taking a root, you have to consider all possible roots. The roots of a^2 are both a & -a. You have to continue your calculations in parallel for both roots until one is proven invalid. A resultant of 1=2 invaladates the use of that root.

Anonymous Poster
#6

### Re: 1=2

06/17/2007 2:45 AM

Agreed - when expanding:

ab-a^2=b^2-a^2, you have to also consider the case of:

-a(-b+a)=(b-a)(b+a), which cannot be reduced and will result in 0=0 if you substitute any real value for a & b.

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#7

### Re: 1=2

06/17/2007 4:58 AM

Its wrong. ab-a2 = b2-a2 has no relation with a = a+b.............. fake

Anonymous Poster
#9

### Re: 1=2

06/17/2007 10:40 AM

You can not cancel (b-a) or divide by (b-a) both side as if b=a then b-a = 0 i.e why above said equation if invalid

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#11

### Re: 1=2

06/17/2007 9:39 PM

sqrt (-1) X sqrt (-1) = -1, not 1, so the conclusion is incorrect.

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#13

### Re: 1=2

06/18/2007 12:40 AM

sqrt (-1) isn't just i, it is also -i since

-i x -i = -1

in other words:

sqrt (-1) = ± i

which means that when you write

sqrt (-1) x sqrt (-1), you could be writing:

i x i = -1
i x -i = 1
i x -i = 1
-i x -i = -1

So the sqrt(-1) x sqrt(-1) = ±1,

just as the sqrt(1) = ±1

Anonymous Poster
#14

### Re: 1=2

06/18/2007 5:18 AM

As stated in the begining a=b

ab-a2 =0 and b2-a2 = 0

then what is the point of going further down

even if you go

ax0 = 0 and 0(b+a) = 0

Sasee

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#43

### Re: 1=2

06/19/2007 9:21 PM

Just a fun,

one paper slite into two piece of papers, thus 1=2;

one pregnant woman has a baby in hospital, thus 1=2;

a coupler build up a family; 1=2;

...

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#61

### Re: 1=2

06/23/2007 2:31 AM

Anyone for Riemann Surface Tennis ?

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#62

### Re: 1=2

06/23/2007 2:20 PM

Yeah! First match is tonight during prime time.

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#63

### Re: 1=2

06/23/2007 3:08 PM

Your way with words is cracking me up !

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#64

### Re: 1=2

06/23/2007 4:34 PM

Does this sentence remind you of pain?

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#65

### Re: 1=2

06/24/2007 2:55 AM

Life isn't handed to us on (or even under) a plate. Baptists take note.

Geller has probably twisted the spoon by now.

It also occurs to me that a bird in the hand is worth two in the bush.QED

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