Difference between revisions of "Proof that 2=1"
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+ | ==Proof== | ||
+ | |||
1) <math>a = b</math>. Given. | 1) <math>a = b</math>. Given. | ||
+ | |||
2) <math>a^2 = ab</math>. Multiply both sides by a. | 2) <math>a^2 = ab</math>. Multiply both sides by a. | ||
+ | |||
3) <math>a^2-b^2 = ab-b^2</math>. Subtract <math>b^2</math> from both sides. | 3) <math>a^2-b^2 = ab-b^2</math>. Subtract <math>b^2</math> from both sides. | ||
+ | |||
4) <math>(a+b)(a-b) = b(a-b)</math>. Factor both sides. | 4) <math>(a+b)(a-b) = b(a-b)</math>. Factor both sides. | ||
+ | |||
5) <math>(a+b) = b</math>. Divide both sides by <math>(a-b)</math> | 5) <math>(a+b) = b</math>. Divide both sides by <math>(a-b)</math> | ||
+ | |||
6) <math>a+a = a</math>. Substitute <math>a</math> for <math>b</math>. | 6) <math>a+a = a</math>. Substitute <math>a</math> for <math>b</math>. | ||
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7) <math>2a = a</math>. Addition. | 7) <math>2a = a</math>. Addition. | ||
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8) <math>2 = 1</math>. Divide both sides by <math>a</math>. | 8) <math>2 = 1</math>. Divide both sides by <math>a</math>. | ||
+ | |||
+ | ==Error== | ||
+ | Usually, if a proof proves a statement that is clearly false, the proof has probably divided by zero in some way. | ||
+ | |||
+ | In this case, the quantity of <math>a-b</math> is <math>0</math> as <math>a = b</math>, since one cannot divide by zero, the proof is incorrect from that point on. | ||
+ | |||
+ | <b>Thus, this proof is false.</b> | ||
+ | |||
+ | ==Note:== | ||
+ | If this proof were somehow true all of mathematics would collapse. Simple arithmetic would yield infinite answers. This is why one cannot divide by zero. | ||
+ | |||
+ | ==Alternate Proof== | ||
+ | Consider the continued fraction <math>3-\frac{2}{3-\frac{2}{3-\frac{2}{3- \cdots}}}.</math> If you set this equal to a number <math>x</math>, note that | ||
+ | <math>3-\frac{2}{x}=x</math> due to the fact that the fraction is infinitely continued. But this equation for <math>x</math> has two solutions, <math>x=1</math> and <math>x=2.</math> Since both <math>2</math> and <math>1</math> are equal to the same continued fraction, we have proved that <math>2=1.</math> QED. | ||
+ | |||
+ | ==Error== | ||
+ | |||
+ | The proof translate the continued fraction into a quadratic, which has multiple solutions. Therefore, <math>2 \neq 1</math>. |
Latest revision as of 19:22, 27 January 2022
Contents
Proof
1) . Given.
2) . Multiply both sides by a.
3) . Subtract from both sides.
4) . Factor both sides.
5) . Divide both sides by
6) . Substitute for .
7) . Addition.
8) . Divide both sides by .
Error
Usually, if a proof proves a statement that is clearly false, the proof has probably divided by zero in some way.
In this case, the quantity of is as , since one cannot divide by zero, the proof is incorrect from that point on.
Thus, this proof is false.
Note:
If this proof were somehow true all of mathematics would collapse. Simple arithmetic would yield infinite answers. This is why one cannot divide by zero.
Alternate Proof
Consider the continued fraction If you set this equal to a number , note that due to the fact that the fraction is infinitely continued. But this equation for has two solutions, and Since both and are equal to the same continued fraction, we have proved that QED.
Error
The proof translate the continued fraction into a quadratic, which has multiple solutions. Therefore, .