Invalid proof

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In mathematics, there are a variety of spurious proofs of obvious contradictions. Although the proofs are flawed, the errors, usually by design, are comparatively subtle. These fallacies are normally regarded as mere curiosities, but can be used to show the importance of rigor in mathematics.

Most of these proofs depend on some variation of the same error. The error is to take a function f that is not one-to-one, to observe that f(x) = f(y) for some x and y, and to (erroneously) conclude that therefore x = y. Division by zero is a special case of this; the function f is x → x × 0, and the erroneous step is to start with x × 0 = y × 0 and to conclude that therefore x = y. Similarly, the argument below that purports to demonstrate 5=4 makes this same error with the function f(x) = x². The erroneous step starts with the correct assertion that for certain x and y, x² = y², and then makes the incorrect deduction that x = y.

Contents 1 Examples in algebra 1.1 Proof that 1 = −1 1.1.1 Alternative proof 1 1.1.2 Alternative proof 2 1.1.3 Alternative proof 3 1.2 Proof that 2 = 1 1.2.1 Alternative proof 1 1.2.2 Alternative proof 2 1.3 Proof that 0 = 1 1.3.1 Alternative proof 1.4 Proof that -2 = 1 1.5 Proof that 4 = 5 1.6 Proof that 1 < 0 1.7 Proof that ∞ = 1/4 1.8 Proof that i² = 1 1.9 Proof that x = 0 for any real x 1.10 Proof that x = y for any real x, y 1.10.1 Alternative proof 2 Examples in geometry 2.1 Proof that any angle is zero 2.2 Proof that any parallelogram has infinite area 2.3 Proof that any triangle is isosceles 3 Conclusion 4 See also 5 External links

[edit] Examples in algebra

[edit] Proof that 1 = −1

• We start with

  − 1 = − 1

• Then we convert these into vulgar fractions

  

• Applying square roots on both sides gives

  

• Which is equal to

  

• If we now clear fractions by multiplying both sides by and then , we have

  

• But any number's square root squared gives the original number, so

  1 = − 1

Q.E.D.

This proof is invalid since it applies the following principle for square roots incorrectly:

This principle is only correct when y is a positive number. In the "proof" above, this is not the case. Thus, the proof is invalid.

[edit] Alternative proof 1

Q.E.D.

If we interpret this over the complex numbers, it's almost true, but on the left side of the equation there are two numbers, not just one.

[edit] Alternative proof 2

Q.E.D.

The fallacy is the last step where the square root can be, and in this case is, negative.

[edit] Alternative proof 3

Say x=180°.

Therefore,

− 1 = 1

Q.E.D.

Again, the fallacy is in the second step, where the square root has positive and negative roots.

[edit] Proof that 2 = 1

• Let a and b be equal non-zero quantities

  a = b

• Multiply through by a

  a² = ab

• Subtract b²

  a² − b² = ab − b²

• Factor both sides

  (a − b)(a + b) = b(a − b)

• Divide out (a − b)

  a + b = b

• Observing that a = b

  b + b = b

• Combine like terms on the left

  2b = b

• Divide by the non-zero b

  2 = 1

Q.E.D.

The fallacy is in line 5: the progression from line 4 to line 5 involves division by (a − b), which is zero since a equals b. Since division by zero is undefined, the argument is invalid.

[edit] Alternative proof 1

• Let x and y be equal, non-zero quantities

  x = y

• Add x to both sides

  2x = x + y

• Take 2y from both sides

  2x − 2y = x − y

• Factor out a two on the left side

  2(x − y) = x − y

• Divide out (x − y)

  2 = 1

Q.E.D.

The fallacy here is the same as above in that by dividing by (x − y), you are dividing by zero and as such, this argument is invalid. Also by this argument:

[edit] Alternative proof 2

• By the common intuitive meaning of multiplication we can see that

  

• It can also be seen that for a non-zero x

  (x terms)

• Now we multiply through by x

  (x terms)

• Then we take the derivative with respect to x

  (x terms)

• Now we see that the right hand side is x which gives us

  2x = x

• Finally, dividing by our non-zero x we have

  2 = 1

Q.E.D.

The fallacy lies in line two. Our definition of x assumed that x was an integer; this equation is not meaningful for non-integer real numbers. Functions are only differentiable on a continuous space such as the reals, not on integers. For any particular integer x, you get a true equation. But to differentiate both sides you need an equation of functions, not an equation of integers. The right-hand function "with x terms" is not a meaningful function on the reals (how can you have x terms?) and thus not differentiable.

Also, when taking the derivative in line 4 the derivative is taken with respect to each of the terms individually, but not with respect to the numbers of terms. This is erroneous, as the number of terms is x, the variable of differentiation. The chain rule is incorrectly not applied on the right-hand side of the equation.

[edit] Proof that 0 = 1

• Start with the addition of an infinite succession of zeros

  

• Then recognize that 0 = 1 − 1

  

• Applying the associative law of addition results in

  

• Of course − 1 + 1 = 0

  

• And the addition of an infinite string of zeros can be discarded leaving

  0 = 1

Q.E.D.

The error here is that the associative law cannot be applied freely to an infinite sum unless the sum would converge without any parentheses. Here that sum is 1 − 1 + 1 − 1 + · · ·, a classic divergent series. In this particular argument, the second line gives the sequence of partial sums 0, 0, 0, ... (which converges to 0) while the third line gives the sequence of partial sums 1, 1, 1, ... (which converges to 1), so these expressions need not be equal. This can be seen as a counterexample to generalizing Fubini's theorem and Tonelli's theorem to infinite integrals (sums) over measurable functions taking negative values.

[edit] Alternative proof

• Begin with the evaluation of the indefinite integral

  

• Through integration by parts, let

  and dv = dx

• Thus,

  and v = x

• Hence, by integration by parts

  

  

  0 = 1

Q.E.D.

The error in this proof lies in an improper use of the integration by parts technique. Upon use of the formula, a constant, C, must be added to the right-hand side of the equation. This is due to the derivation of the integration by parts formula; the derivation involves the integration of an equation and so a constant must be added. In most uses of the integration by parts technique, this initial addition of C is ignored until the end when C is added a second time. However, in this case, the constant must be added immediately because the remaining two integrals cancel each other out.

In other words, the second to last line is correct (1 added to any antiderivative of 1/x is still an antiderivative of 1/x); but the last line is not. You cannot "cancel" because they are not necessarily equal. There are infinitely many antiderivatives of a function, all differing by a constant. In this case, the antiderivatives on both sides differ by 1.

This problem can be avoided if we use definite integrals (i.e. use bounds). Then in the second to last line, 1 would be evaluated between some bounds, which would always evaluate to 1 - 1 = 0. The remaining definite integrals on both sides would indeed be equal.

[edit] Proof that -2 = 1

Let's start with a simple problem.

• Solve the equation
• Taking the cube:

• Replacing the expression within parenthesis by the initial equation, and cancelling common terms:

• Take the cube:

• Solution: x = 2. Replacing in the original equation, we get:

• So:

Q.E.D.

[edit] Proof that 4 = 5

• Start with the identity

  − 20 = − 20

• Express both sides in slightly different, yet equivalent ways

  25 − 45 = 16 − 36

• Factor both sides

  

• Add the same thing to both sides

  

• Now factor both sides again

  

• Take the square root of both sides

  

• Cancel the common term

  5 = 4

Q.E.D.

The error in the proof comes from the fact that x² = y² does not imply that x = y. Except for the fact that the initial step of -20 = -20 implies that x squared = x squared, thus x = x. The arithmetic up until this point is correct, and in fact

It is also important to note that if we subtract the term 9/2 from 4, we end up with -1/2. If we then square the term, we arrive at a positive 1/4th. The next logical mathematical step is to take the square root of both sides. If we do this, we can see that 1/2 is equal to 1/2. The original problem of -20=-20 does in fact result in a correct identity if the problem is worked out in the proper way.

[edit] Proof that 1 < 0

• Let us first suppose that

  0 < x < 1

• Now we will take the logarithm on both sides. As long as x > 0, we can do this because logarithms are monotonically increasing. Observing that the logarithm of 1 is 0, we get

  ln(x) < 0

• Dividing by ln (x) gives

  1 < 0

Q.E.D.

The violation is found in the last step, the division. This step is invalid because ln(x) is negative for 0 < x < 1. While multiplication or division by a positive number preserves the inequality, multiplication or division by a negative number reverses the inequality, resulting in the correct expression 1 > 0.

[edit] Proof that ∞ = 1/4

• Since an infinitely large plane has the coordinates of (-∞,∞) × (-∞,∞), this means that

  

• Which can be simplified into

  

• And finally

  

• Now combine the ∞'s:

  

• This itself then simplifies into

  

• And finally, to find the value of ∞ itself,

  

• This can be checked by starting with the equation given in step 1,

  

• Substitute in the above value of ∞ to see if it really works:

  

• Which is then simplified to get

  

• And that then simplifies into

  

Q.E.D.

This proof's fallacy is using ∞ (infinity) to represent a finite value – in reality infinity is thought of as a direction as opposed to a destination. One of the more unusual aspects of this type of invalid proof is that it can be "checked," unlike many of the above proofs, particularly the ones which rely on division by zero.

[edit] Proof that i² = 1

• Start with the identity

  

• Square both sides

  

• Multiply the square root functions

  

• The square root of 1 is 1

  i² = 1

Q.E.D.

The error in the proof comes from the fact that if and only if a and b are values greater than or equal to zero. Correct multiplication of the radical would give . This would lead to i² = − 1. Taking the square root of both sides gives the correct identity .

[edit] Proof that x = 0 for any real x

• Let x equal y such that x and y are any two non-zero quantities

  x = y

• Multiply by x

  x² = xy

• Subtract y²

  x² − y² = xy − y²

• Factor both sides

  (x − y)(x + y) = y(x − y)

• Divide by (x - y)

  x + y = y

• Subtract y from both sides

  x + y − y = y − y

• As + y − y cancels out and y − y is zero,

  x = 0

Q.E.D.

The fallacy occurs when dividing by (x − y). The two quantities x and y are set equal at the beginning of the proof, so (x − y) = 0, and therefore cannot be divided by.

[edit] Proof that x = y for any real x, y

• We start with:

  x − y = c

• Multiply by x - y:

  (x − y)(x − y) = (x − y)c

• Write out the multiplication:

  x² − 2xy + y² = xc − yc

• Rearranging all, we get:

  x² − xy − xc = xy − y² − yc

• Factorize both members:

  x(x − y − c) = y(x − y − c)

• Cancel the common factor:

  x = y

Q.E.D.

The catch is that since x − y = c, x − y − c = 0, and as a result we have performed an illegal division by zero.

[edit] Alternative proof

• Let x and y be any two numbers
• Then let
• Let
• Let's compute:

• Replacing , we get:

• Let's compute
• Replacing :

• So:

• Replacing :

Q.E.D.

[edit] Examples in geometry

[edit] Proof that any angle is zero

Construct a rectangle ABCD. Now identify a point E such that CD=CE and the angle DCE is a non-zero angle. Take the perpendicular bisector of AD, crossing at F, and the perpendicular bisector of AE, crossing at G. Label where the two perpendicular bisectors intersect as H and join this point to A, B, C, D, and E.

Now, AH=DH because FH is a perpendicular bisector; similarly BH=CH. AH=EH because GH is a perpendicular bisector, so DH=EH. And by construction BA=CD=CE. So the triangles ABH, DCH and ECH are congruent, and so the angles ABH, DCH and ECH are equal.

But if the angles DCH and ECH are equal then the angle DCE must be zero.

Q.E.D.

The error in the proof comes in the diagram and the final point. An accurate diagram would show that the triangle ECH is a reflection of the triangle DCH in the line CH rather than being on the same side, and so while the angles DCH and ECH are equal in magnitude, there is no justification for subtracting one from the other; to find the angle DCE you need to subtract the angles DCH and ECH from the angle of a full circle (2π or 360°).

[edit] Proof that any parallelogram has infinite area

Take a parallelogram ABCD. Rule an infinite number of lines equal and parallel to CD along AD's length until ABCD is completely full of these lines. As these lines all equal CD, the total area of these lines (and thus the parallelogram) is ∞ × (CD), thus infinity.

Q.E.D.

The fallacy here is that infinity is assumed to be a dimension. Infinity is not a dimension, therefore the only dimension is CD. Parallelograms have two dimensions, invalidating this proof.

[edit] Proof that any triangle is isosceles

It is sufficient to prove that any two sides of a triangle are congruent.

Refer to the diagrams here.

Given a triangle △ABC, proof that AB = AC:

1. Draw a line bisecting ∠A
2. Call the midpoint of line segment BC, D
3. Draw the perpendicular bisector of segment BC, which contains D
4. If these two lines are parallel, AB = AC, by some other theorem; otherwise they intersect at point P
5. Draw line PE perpendicular to AB, line PF perpendicular to AC
6. Draw lines PB and PC
7. By AAS, △EAP ≅ △FAP (AP = AP; ∠PAF ≅ ∠PAE since AP bisects ∠A; ∠AEP ≅ ∠AFP are both right angles)
8. By HL, △PDB ≅ △PDC (∠PDB,∠PDC are right angles; PD = PD; BD = CD because PD bisects BC)
9. By SAS, △EPB ≅ △EPC (EP = FP since △EAP ≅ △FAP; BP = CP since △PDB ≅ △PDC; ∠EPB ≅ ∠FPC since they are vertical angles)
10. Thus, AE ≅ AF, EB ≅ FC, and AB = AE + EB = AF + FC = AC

Q.E.D.

As a corollary, one can show that all triangles are equilateral, by showing that AB = BC and AC = BC in the same way.

All but the last step of the proof is indeed correct (those three triangles are indeed congruent). The error in the proof is the assumption in the diagram that the point P is inside the triangle. In fact, whenever AB ≠ AC, P lies outside the triangle. Furthermore, it can be further shown that, if AB is shorter than AC, then E will lie outside of AB, while F will lie within AC (and vice versa). (Any diagram drawn with sufficiently accurate instruments will verify the above two facts.) Because of this, AB is actually AE - EB, whereas AC is still AF + FC; and thus the lengths are not necessarily the same.

[edit] Conclusion

Most of these proofs rely on erroneously taking the square root without considering that it can be both positive and negative.

Some of these arguments do constitute valid proofs, but not of the claimed assertions. For example, there is no a priori reason why division by zero should be defined (it's not a field axiom, for example, though 1 ≠ 0, from which 2 ≠ 1 follows, is an axiom), and the "proof" that 2 = 1 is, in fact, simply a demonstration that division by zero cannot be defined in general. A proof that division by zero could be defined would demonstrate a contradiction and show that the axiomatic system we are working under is logically inconsistent.

On the other hand it is possible to construct useful mathematical systems where 1 is equivalent to 2. Mathematics in domains modulo 1 are one example. In such domains

[edit] See also

• Paradox

[edit] External links

• More invalid proofs from AhaJokes.com
• "Proof" that 3=1 by Joe Watkins