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+0  
 
+2
283
6
avatar+919 

Here's a cool 0=2 proof I found.

 

\(a=1, b=2\)

\(2a=b, 2a^2b=ab^2\)

\(2b=b^2, b^2-2b=0\)

\(b(b-2)=0\)

\(b=0, 2\)

So \(0=2\)...

 

Prove it wrong!

 Mar 18, 2019
 #1
avatar+893 
+1

Wait, but you already defined b to be 2. Then how can it be 0 or 2?

 Mar 18, 2019
 #2
avatar+6043 
+1

No.  0 is just an extraneous root of the quadratic you synthesized.

 Mar 18, 2019
 #3
avatar+919 
+1

Why?

CoolStuffYT  Mar 19, 2019
 #4
avatar+6043 
+1

consider the equation

 

\((x-2)(x-3)=0\\ x = 2, 3\\ \text{Would you then say }2=3?\)

Rom  Mar 19, 2019
 #6
avatar+919 
+1

Good point, thanks

CoolStuffYT  Mar 21, 2019
 #5
avatar+104472 
+1

Hi CoolStuffYT

Rom is right but it is good that you think about 'puzzles' like this.

You need to question the maths that is put in front of you.

 

Here is another example like your one

 

x=-3,   y=9

 

I can see that 9=(-3)^2 so I could say these fits the equation   y=x^2

but there are two solutions to y=x^2   when y=9    x could be  -3 or 3  

They are just 2 different solutions to the equation that i synthesized.       laugh

 Mar 20, 2019

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