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Can you find the illegal step?

Started by Rick Lyons 3 weeks ago10 replieslatest reply 3 weeks ago154 views

Can you find the illegal step in the following silly proof? Given: a=b

$$ a^2=ab$$

$$a^2-b^2 = ab-b^2$$

$$(a+b)(a-b)=b(a-b)$$

$$a+b = b$$

$$b+b=b$$

$$2b=b$$


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Reply by CharlieRaderAugust 27, 2021
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Reply by philipoakleyAugust 27, 2021

This is mathematics, not real world, isn't it? 

The error, if there is one ;-), is the last step where 'b' should be subtracted from both sides, resulting in the true answer 'b=0; => 0=a', !

See also: dy/dx => $inf * 0, doesn't it? 

(Hence, Are there more infinitesimals than infinities?)


Thinking outside the box ;-)

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Reply by Rick LyonsAugust 27, 2021

OK OK, the above problem was pretty easy. Can you find the illegal step in the following?

Theorem: 4=5

Proof:

$$16-36=25-45$$

$$4^2–9*4=5^2–9*5$$

$$4^2–9*4+81/4=5^2–9*5+81/4$$

$$ (4-9/2)^2=(5-9/2)^2$$

$$4-9/2=5-9/2$$

$$4=5$$

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Reply by PlatybelAugust 27, 2021

Thanks for the mental challenge.  It is tricky. The illegal step is taking only one solution from each side when taking the square root.  There are two factors on each side of the equation.

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Reply by adiduaAugust 27, 2021

If $x^2 = y^2$, taking square root on both sides gives $x = +/- y$. In this case accepting the solution $x = -y$ would not have caused a contradiction.


Alternatively, 

$x^2 = y^2$, implies $(x+y)(x-y)=0$, so either $x=y$ or $x=-y$ or both. Given the specific values of $x$ and $y$ in this problem, we know that $x=y$ is not true, so we must have $x=-y$.

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Reply by CharlieRaderAugust 27, 2021

Going from step 3 to step 4, you divided by (a-b). Division by 0 is not meaningful.

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Reply by SarabethAugust 27, 2021

Like CharlieRader said, division by 0 is the truly "illegal" step, but to my mind, the worst decision is not factoring out common terms until step 3. The error between steps three and four becomes obvious if the first transformation is to reduce the initial equation to a = b, which also makes the multiplication by zero in step three apparent. 

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Reply by cortegaAugust 27, 2021

3rd step into 4th is only correct if a!=b, but the condition was a=b,

therefore proof is inconsistent

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Reply by neiroberAugust 28, 2021

If you let b = 1, then you get the "classic" homework result:

2 = 1

From this follows the other classic result:

1 = 0. 

Neil

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Reply by adiduaAugust 29, 2021

As a corollary, 2 = 0.