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| − | ==Proof==
| + | #REDIRECT [[Fallacy#2=1]] |
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| − | 1) <math>a = b</math>. Given.
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| − | 2) <math>a^2 = ab</math>. Multiply both sides by a.
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| − | 3) <math>a^2-b^2 = ab-b^2</math>. Subtract <math>b^2</math> from both sides.
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| − | 4) <math>(a+b)(a-b) = b(a-b)</math>. Factor both sides.
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| − | 5) <math>(a+b) = b</math>. Divide both sides by <math>(a-b)</math>
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| − | 6) <math>a+a = a</math>. Substitute <math>a</math> for <math>b</math>.
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| − | 7) <math>2a = a</math>. Addition.
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| − | 8) <math>2 = 1</math>. Divide both sides by <math>a</math>.
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| − | ==Error==
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| − | Usually, if a proof proves a statement that is clearly false, the proof has probably divided by zero in some way.
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| − | 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.
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| − | <b>Thus, this proof is false.</b>
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| − | ==Note:==
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| − | 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.
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| − | ==Alternate Proof==
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| − | 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
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| − | <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.
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| − | ==Error==
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| − | The proof translate the continued fraction into a quadratic, which has multiple solutions. Therefore, <math>2 \neq 1</math>.
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