how come 1 = 2 ?! just proved it plz tell me whts wrong!?

Question

LET

2 2 2 2
{ x - x } ={ x - x } dont u agree?? thus:

x (x - x) = ( x - x) ( x +x) difference bet. 2 squares

x = x + x (after removing (x-x) frm both sides)
1 = 2!!

Answer 1

Its very simple
log1 = 0
log2= log(1+1)=log1+log1=0+0=0
log1=log2
therefore1=2
hence proved

Answer 2

You removed (x-x) from both sided by dividing both sides by (x-x). But x-x is 2-2 is 0, so you divided by 0.

Answer 3

Your problem is in removing (x-x) from both sides. Since x-x=0 this would result in a division by zero, which is undefined.

Answer 4

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.
Contents
[hide]

* 1 Examples
o 1.1 Proof that 1 equals −1
o 1.2 Proof that 1 is less than 0
o 1.3 Proof that 2 equals 1
o 1.4 Proof that a equals b
o 1.5 Proof that 0 equals 1
+ 1.5.1 Alternative proof
o 1.6 Another proof that 2 equals 1
o 1.7 Proof that 4 equals 5
o 1.8 Proof that any angle is zero
o 1.9 Proof that ∞ = 1/4
* 2 Conclusion
* 3 See also
* 4 External links

[edit]

Examples
[edit]

Proof that 1 equals −1

* We start with

− 1 = − 1

* Then we convert these into vulgar fractions

\frac{1}{-1} = \frac{-1}{1}

* Applying square roots on both sides gives

\sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}}

* Which is equal to

\frac{\sqrt{1}}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}}

* If we now clear fractions by multiplying both sides by \sqrt{-1} and then \sqrt{1}, we have

\sqrt{1}\sqrt{1} = \sqrt{-1}\sqrt{-1}

* 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:

\sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}}

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]

Proof that 1 is less than 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 wrong because the number we are dividing by is negative as can clearly be seen in the previous line. A multiplication with or division by a negative number flips the inequality sign; in other words, we should obtain 1 > 0, which is indeed correct.
[edit]

Proof that 2 equals 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.

A variation:

* 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.
[edit]

Proof that a equals b

* We start with:

a − b = c

* Now, square both sides:

a2 − 2ab + b2 = c2

* Since a − b = c, substitute:

a2 − 2ab + b2 = (a − b)c

* Write out the multiplication:

a2 − 2ab + b2 = ac − bc

* Rearranging all, we get:

a2 − ab − ac = ab − b2 − bc

* Factorize both members:

a(a − b − c) = b(a − b − c)

* Cancel the common factor:

a = b

Q.E.D.

The catch is that since a − b = c, a − b − c = 0, and as a result we have performed an illegal division by zero.
[edit]

Proof that 0 equals 1

* Start with the addition of an infinite succession of zeros

0 = 0 + 0 + 0 + \cdots

* Then recognize that 0 = 1 − 1

0 = (1 - 1) + (1 - 1) + (1 - 1) + \cdots

* Applying the associative law of addition results in

0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + \cdots

* Of course − 1 + 1 = 0

0 = 1 + 0 + 0 + 0 + \cdots

* 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. 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 it is unclear in what sense these expressions can be considered equal.
[edit]

Alternative proof

* Begin with the evaluation of the integral

\int \frac{1}{x} dx

* Through integration by parts, let

u=\frac{1}{x}

* Thus,

du=-\frac{1}{x^2}dx

* Hence, by integration by parts

\int \frac{1}{x} dx=\frac{x}{x} - \int \left ( - \frac{1}{x^2} \right ) x dx
\int \frac{1}{x} dx=1 + \int \frac{1}{x} dx
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.
[edit]

Another proof that 2 equals 1

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

4 \times 3 = 3 + 3 + 3 + 3

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

x = 1 + 1 + \cdots + 1 (x terms)

* Now we multiply through by x

x^2 = x + x + \cdots + x (x terms)

* Then we take the derivative with respect to x

2x = 1 + 1 + \cdots + 1 (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 error: 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 x + x + \cdots + x "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 righthandside of the equation.
[edit]

Proof that 4 equals 5

* Start with the identity

− 20 = − 20

* Express both sides in slightly different, yet equivalent ways

25 − 45 = 16 − 36

* Factor both sides

5^2 - 5 \times 9 = 4^2 - 4 \times 9

* Add the same thing to both sides

5^2 - 5 \times 9 + \frac{81}{4} = 4^2 - 4 \times 9 + \frac{81}{4}

* Now factor both sides again

\left(5 - \frac{9}{2}\right)^2 = \left(4 - \frac{9}{2}\right)^2

* Square root both sides

5 - \frac{9}{2} = 4 - \frac{9}{2}

* Cancel the common factor

5 = 4

Q.E.D.

The error in the proof comes from the fact that x2 = y2 does not imply that x = y. The arithmetic up until this point is correct, and in fact

-\left(5 - \frac{9}{2}\right) = 4 - \frac{9}{2}. It is also important to note that if we subract 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 any angle is zero
Diagram for 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 ∞ = 1/4

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

\frac{}{}\infin = [\infin - (-\infin)]^2

* Which can be simplified into

\frac{}{}\infin = (2\infin)^2

* And finally

\frac{}{}\infin = 4\infin^2

* Now combine the ∞'s:

\frac{\infin}{\infin^2} = 4

* This itself then simplifies into

\frac{1}{\infin} = 4

* We can get ∞ out of the denominator by doing

\frac{}{}1 = 4\infin

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

\frac{1}{4} = \infin

* This can be checked by starting with the simplified equation given in step 3,

\frac{}{}\infin = 4\infin^2

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

\frac{1}{4} = 4(\frac{1}{4})^2

* Which is then simplified to get

\frac{1}{4} = 4(\frac{1}{16})

* And that then simplifies into

\frac{1}{4} = \frac{1}{4}

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]

Conclusion

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 0.5 + 0.5 = 0 = 1 = 2\dots .




Proof that 2 = 1
let a = b

a² = ab
Multiply both sides by a
a² + a² - 2ab = ab + a² - 2ab
Add (a² - 2ab) to both sides
2(a² - ab) = a² - ab
Factor the left, and collect like terms on the right
2 = 1
Divide both sides by (a² - ab)

This proof can also be used as a proof by contradiction that a can never be equal to b.

Answer 5

x - x = 0. You can't divide both sides by zero.

Answer 6

You're using the same variable for two different values. All your x's don't equal eachother. Your math is flaw, moron.

Answer 7

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.
Contents
[hide]

* 1 Examples
o 1.1 Proof that 1 equals −1
o 1.2 Proof that 1 is less than 0
o 1.3 Proof that 2 equals 1
o 1.4 Proof that a equals b
o 1.5 Proof that 0 equals 1
+ 1.5.1 Alternative proof
o 1.6 Another proof that 2 equals 1
o 1.7 Proof that 4 equals 5
o 1.8 Proof that any angle is zero
o 1.9 Proof that ∞ = 1/4
* 2 Conclusion
* 3 See also
* 4 External links

[edit]

Examples
[edit]

Proof that 1 equals −1

* We start with

− 1 = − 1

* Then we convert these into vulgar fractions

\frac{1}{-1} = \frac{-1}{1}

* Applying square roots on both sides gives

\sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}}

* Which is equal to

\frac{\sqrt{1}}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}}

* If we now clear fractions by multiplying both sides by \sqrt{-1} and then \sqrt{1}, we have

\sqrt{1}\sqrt{1} = \sqrt{-1}\sqrt{-1}

* 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:

\sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}}

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]

Proof that 1 is less than 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 wrong because the number we are dividing by is negative as can clearly be seen in the previous line. A multiplication with or division by a negative number flips the inequality sign; in other words, we should obtain 1 > 0, which is indeed correct.
[edit]

Proof that 2 equals 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.

A variation:

* 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.
[edit]

Proof that a equals b

* We start with:

a − b = c

* Now, square both sides:

a2 − 2ab + b2 = c2

* Since a − b = c, substitute:

a2 − 2ab + b2 = (a − b)c

* Write out the multiplication:

a2 − 2ab + b2 = ac − bc

* Rearranging all, we get:

a2 − ab − ac = ab − b2 − bc

* Factorize both members:

a(a − b − c) = b(a − b − c)

* Cancel the common factor:

a = b

Q.E.D.

The catch is that since a − b = c, a − b − c = 0, and as a result we have performed an illegal division by zero.
[edit]

Proof that 0 equals 1

* Start with the addition of an infinite succession of zeros

0 = 0 + 0 + 0 + \cdots

* Then recognize that 0 = 1 − 1

0 = (1 - 1) + (1 - 1) + (1 - 1) + \cdots

* Applying the associative law of addition results in

0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + \cdots

* Of course − 1 + 1 = 0

0 = 1 + 0 + 0 + 0 + \cdots

* 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. 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 it is unclear in what sense these expressions can be considered equal.
[edit]

Alternative proof

* Begin with the evaluation of the integral

\int \frac{1}{x} dx

* Through integration by parts, let

u=\frac{1}{x}

* Thus,

du=-\frac{1}{x^2}dx

* Hence, by integration by parts

\int \frac{1}{x} dx=\frac{x}{x} - \int \left ( - \frac{1}{x^2} \right ) x dx
\int \frac{1}{x} dx=1 + \int \frac{1}{x} dx
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.
[edit]

Another proof that 2 equals 1

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

4 \times 3 = 3 + 3 + 3 + 3

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

x = 1 + 1 + \cdots + 1 (x terms)

* Now we multiply through by x

x^2 = x + x + \cdots + x (x terms)

* Then we take the derivative with respect to x

2x = 1 + 1 + \cdots + 1 (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 error: 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 x + x + \cdots + x "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 righthandside of the equation.
[edit]

Proof that 4 equals 5

* Start with the identity

− 20 = − 20

* Express both sides in slightly different, yet equivalent ways

25 − 45 = 16 − 36

* Factor both sides

5^2 - 5 \times 9 = 4^2 - 4 \times 9

* Add the same thing to both sides

5^2 - 5 \times 9 + \frac{81}{4} = 4^2 - 4 \times 9 + \frac{81}{4}

* Now factor both sides again

\left(5 - \frac{9}{2}\right)^2 = \left(4 - \frac{9}{2}\right)^2

* Square root both sides

5 - \frac{9}{2} = 4 - \frac{9}{2}

* Cancel the common factor

5 = 4

Q.E.D.

The error in the proof comes from the fact that x2 = y2 does not imply that x = y. The arithmetic up until this point is correct, and in fact

-\left(5 - \frac{9}{2}\right) = 4 - \frac{9}{2}. It is also important to note that if we subract 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 any angle is zero
Diagram for 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 ∞ = 1/4

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

\frac{}{}\infin = [\infin - (-\infin)]^2

* Which can be simplified into

\frac{}{}\infin = (2\infin)^2

* And finally

\frac{}{}\infin = 4\infin^2

* Now combine the ∞'s:

\frac{\infin}{\infin^2} = 4

* This itself then simplifies into

\frac{1}{\infin} = 4

* We can get ∞ out of the denominator by doing

\frac{}{}1 = 4\infin

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

\frac{1}{4} = \infin

* This can be checked by starting with the simplified equation given in step 3,

\frac{}{}\infin = 4\infin^2

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

\frac{1}{4} = 4(\frac{1}{4})^2

* Which is then simplified to get

\frac{1}{4} = 4(\frac{1}{16})

* And that then simplifies into

\frac{1}{4} = \frac{1}{4}

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]

Conclusion

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 0.5 + 0.5 = 0 = 1 = 2\dots .




Proof that 2 = 1
let a = b

a² = ab
Multiply both sides by a
a² + a² - 2ab = ab + a² - 2ab
Add (a² - 2ab) to both sides
2(a² - ab) = a² - ab
Factor the left, and collect like terms on the right
2 = 1
Divide both sides by (a² - ab)

This proof can also be used as a proof by contradiction that a can never be equal to b.

Answer 8

oh come on now, there is someone out there that could prove that you don't know your a*s from a hole in the ground as well. does that make it so?

Answer 9

Wouldn't it be better to say:

Let a=b
a-b=a-b
a(a-b)=(a-b)(a+b)
a=a+b
so, if a=1,
1=2

Answer 10

wrong