Mathematical Fallacies -- _1 Equals 2

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1=2: A Proof using Beginning Algebra

The Fallacious Proof:

Step 1: Let a=b.

Step 2: Then ,

Step 3: ,

Step 4: ,

Step 5: ,

Step 6: and .

Step 7: This can be written as ,

Step 8: and cancelling the from both sides gives 1=2.

See if you can figure out in which step the fallacy lies. When you think you've figured it out, click on thatstep and the computer will tell you whether you are correct or not, and will give an additional explanation

of why that step is or isn't valid.

See how many tries it takes you to correctly identify the fallacious step!

This page last updated: May 26, 1998

Original Web Site Creator / Mathematical Content Developer: Philip Spencer

Current Network Coordinator and Contact Person: Joel Chan - [email protected]


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1=2: A Proof using Complex Numbers

This supposed proof uses complex numbers. If you're not familiar with them, there's a brief introduction

to them given below.

The Fallacious Proof:

Step 1: -1/1 = 1/-1

Step 2: Taking the square root of both sides:

Step 3: Simplifying:

Step 4: In other words, i /1 = 1/ i.

Step 5: Therefore, i / 2 = 1 / (2i),

Step 6: i /2 + 3/(2i) = 1/(2i) + 3/(2i),

Step 7: i (i /2 + 3/(2i) ) = i ( 1/(2i) + 3/(2i) ),

Step 8: ,

Step 9: (-1)/2 + 3/2 = 1/2 + 3/2,

Step 10: and this shows that 1=2.

See if you can figure out in which step the fallacy lies. When you think you've figured it out, click on thatstep and the computer will tell you whether you are correct or not, and will give an additional explanation

of why that step is or isn't valid.

See how many tries it takes you to correctly identify the fallacious step!

Complex Numbers

This proof uses complex numbers. For those who are unfamiliar with them, we give a brief sketch here.

The complex numbers are a set of objects that includes not only the familiar real numbers but also an

additional object called "i". Addition and multiplication are defined on this larger set in such a way that

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i^2 = -1. So, although -1 does not have any square root within the context of real numbers, but it does

have a square root within the larger context of complex numbers.

One thing that often confuses people is the question "what is this strange object i"? There are different

ways to define it, all essentially equivalent. One convenient way is to define complex numbers to be pairs

(a,b) of real numbers. Addition is defined by the rule (a,b) + (c,d ) = (a+c, b+d ), and multiplication is

defined by the strange-looking rule (a,b)(c,d ) = (ac-bd , ad +bc).

With this definition, a complex number of the form (a,0) is identical in all its arithmetic properties to the

real number a, and is therefore just a different way of representing the same abstract concept that the real

number a does. So it is legitimate to think of the complex number (a,0) and the real number a as being

"the same thing", and with this understanding, the complex numbers are a set of objects that contains the

real numbers along with other things.

(This is exactly the same thing that one does when one first learns about rational numbers (fractions). A

fraction is a pair of integers (a,b), commonly written as a / b. There are rules for addition: a / b + c / d =

(ad +bc)/(bd ) and for multiplication: (a / b)(c / d ) = (ac)/(bd ). With these rules, a fraction of the form a /1

behaves identically to the integer a, so we consider the rational number a /1 and the integer a to be "the

same thing", and with this understanding, the rational numbers are a set of objects that contains the

integers along with other things).

Having defined complex numbers as pairs of real numbers, we define i to be the pair (0,1). If you apply

the above rule of multiplication, you find that (0,1)(0,1) = (-1,0) (which, remember, is really the same

thing as the real number -1). In other words, i^2 = -1.

Finally, where did the above rule for multiplication come from? It came from the following reasoning:

We want to have in our set an object called i whose square is -1. We also want to have the usual real

numbers in our set. We want a way to add and multiply elements in our set, so as well as i and the usualreal numbers we also need to have all sums and products involving i and real numbers; in other words, for

each pair (a,b) of real numbers, there needs to be a corresponding element a+bi in our set.

So, since to every pair of real numbers (a,b) there needs to correspond an element in our set, why don't we

just let our set be the set of all pairs of real numbers. Now, what should our rule for multiplication be? We

want addition and multiplication to satisfy the familiar properties of commutativity, distributivity, and

associativity. That means (a + bi) + (c + di) should equal (a+c) + (b+d )i, and (a+bi)(c+di) should equal ac

+ bci + adi + bdi^2 = (ac-bd ) + (bc+ad )i.

This reasoning says that there is only one possible way to define addition and multiplication and still haveany hope at all of them satisfying the properties we want. That's why one makes the definition above.

Then, having made that definition, one proves that this definition does in fact have all the properties we

want. That's a separate exercise.





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Congratulations; you have correctly identified the

fallacious step!

This innocent-looking step is in fact quite wrong. The problem is that there is no rule that guarantees

, except in the case in which a and b are both positive.

If this surprises you, think about the question

Why should equal ?

If you were to try to convince someone of this, you'd have to start with the definition of what a "square

root" is: it's a number whose square is the number you started with. So, from first principles, all that has to

be true is that squared is a, squared is b, and squared is a / b.

So, when you square , you will get a / b, and when you square , you will also get a / b. That's

all that the definition of square root tells you.

Now, the only way two numbers x and y can have the same square is if x = +/- y. So, what is true is that

,

but in general there's no reason it has to be rather than , unless a and b are

both positive: for then (because by convention we take the positive square root) everything in the above

equation is positive, and that's why we obtain . But remember it's only because everything

is positive that we obtain it!

In our case, it is true that , but is not . The fallacy comes

from using the latter instead of the former.

In fact, the whole proof really boils down to the fact that (-1)(-1) = 1, so , but

(not 1). The proof tried to claim that these two were equal (but in a more disguised

way where it was harder to spot the mistake).

This fallacy is a good illustration of the dangers of taking a rule from one context and just assuming itholds in another. When you first learned about square roots you had never encountered complex numbers,

so the only objects that had sqare roots were positive numbers. In this case, is always true,

and you were probably taught it as a "rule". But it is only a mathematical truth in that original context,

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and fails to remain true after you extend the definition of "square root" to allow the square roots of

negative and complex numbers.

To see the explanations for the other steps, finding out exactly why they're correct (or, in some cases,

finding out which other steps have slight mistakes in them), go back to the list of steps in the proof . To try

your hand at finding the fallacy in a different problem, go back to the Classic Fallacies index page.

This page last updated: May 26, 1998 Original Web Site Creator / Mathematical Content Developer: Philip Spencer




2 7/18/2015 2:12 PM




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Congratulations; you have correctly identified the

fallacious step!

To "cancel" a quantity from both sides of an equation is to divide both sides of the equation by it. So, in

this step of the proof we are attempting to divide both sides of the equation by .

However, division only makes sense when the number you are dividing by is non-zero. In this proof,

, because we assumed in step 1 that a=b!

Therefore, it is not legitimate to divide both sides of the equation by , because that would be

division by zero, which does not make any sense (as explained below).

In essence, this proof boils down to saying "1 times 0 equals 2 times 0, therefore 1 equals 2". The fallacy

is that, just because two numbers give you the same answer (zero) after you multiply them each by zero,

doesn't necessarily mean that the two numbers are the same, because anything when multiplied by zero

gives zero.

This is also the reason division by zero does not make sense: there isn't just one unambiguously

determined number q such that , so there isn't any number that we can uniquely and

unambiguously define the quotient 0/0 to be.

If you tried to divide 1 (or some other non-zero number) by 0, you'd run into a different problem: in this

case, there is no number q at all such that , so there is nothing that we can define the quotient 1/0

to be.

That's why division by zero is undefined (not just because it's a rule somebody decided on!)

To see the explanations for the other steps, finding out exactly why they're correct (or, in some cases,

finding out which other steps have slight mistakes in them), go back to the list of steps in the proof . To try

your hand at finding the fallacy in a different problem, go back to the Classic Fallacies index page.






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Mathematical Fallacies -- _1 Equals 2