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Difference between revisions of "Proof that 2=1"

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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.
  
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5) <math>(a+b) = b</math>. Add both sides by <math>(a-b)</math>
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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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8) <math>2 = 1</math>.  Divide both sides by <math>a</math>.
 
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>.

Latest revision as of 19:22, 27 January 2022

Proof

1) $a = b$. Given.

2) $a^2 = ab$. Multiply both sides by a.

3) $a^2-b^2 = ab-b^2$. Subtract $b^2$ from both sides.

4) $(a+b)(a-b) = b(a-b)$. Factor both sides.

5) $(a+b) = b$. Divide both sides by $(a-b)$

6) $a+a = a$. Substitute $a$ for $b$.

7) $2a = a$. Addition.

8) $2 = 1$. Divide both sides by $a$.

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 $a-b$ is $0$ as $a = b$, 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 $3-\frac{2}{3-\frac{2}{3-\frac{2}{3- \cdots}}}.$ If you set this equal to a number $x$, note that $3-\frac{2}{x}=x$ due to the fact that the fraction is infinitely continued. But this equation for $x$ has two solutions, $x=1$ and $x=2.$ Since both $2$ and $1$ are equal to the same continued fraction, we have proved that $2=1.$ QED.

Error

The proof translate the continued fraction into a quadratic, which has multiple solutions. Therefore, $2 \neq 1$.

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