Exploring the Mathematical Error: Why Does 2 Equal 3 in This Scenario?

  • Thread starter PrincePhoenix
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In summary, the conversation revolves around a supposed proof that 2 equals 3, where the participants discuss the steps taken to arrive at this conclusion and point out the flaws in the reasoning. The main mistake is assuming what is being proven, and the conversation highlights the importance of starting with true statements in a proof.
  • #1
PrincePhoenix
Gold Member
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2
Why is 2 = 3 over here?

Suppose 2=3

Then subtract 5/2 from both sides.
2 - 5/2 = 3 - 5/2
4-5/2 = 6-5/2
-1/2 = 1/2

Take square root
(-1/2)2 = (1/2)2
1/4 = 1/4

What's wrong? why is it proved that 2 = 3?
 
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  • #2
You didn't prove anything. Note that when you square both sides you effectively multiply the sides by different numbers.
 
  • #3
Maybe it is because I am not fully awake yet, but I do not follow your arithmetic.

2 - 5/4 = 3/4 not -1/2
3 - 5/4 = 7/4 not 1/2
 
  • #4
Doc Al said:
You didn't prove anything. Note that when you square both sides you effectively multiply the sides by different numbers.
So you mean that taking square was wrong? I thought I did everything on both sides of the equation. I mean which step is incorrect?
 
  • #5
pbandjay said:
Maybe it is because I am not fully awake yet, but I do not follow your arithmetic.

2 - 5/4 = 3/4 not -1/2
3 - 5/4 = 7/4 not 1/2
I corrected the mistake. Check again.
 
  • #6
When you square an equation, you run the risk of creating extraneous roots. For example, [itex]-2^2 = 2^2 = 4[/itex] but this does not prove that [itex]-2 = 2[/itex]. Also, another mistake you made is assuming what you're trying to prove. To have a correct proof you would need to begin with the equaltiy [itex]1/4 = 1/4[/itex] and proceed to show that [itex]2 = 3[/itex] (which you can't do).

Edit: Just so it's clear, the biggest fallacy in your "proof" was the assumption that [itex]2 = 3[/itex]. You can't assume what you're trying to prove.
 
  • #7
To absolutely clear, here's why you can't assume what you're proving . . .

Let [itex]a,b \in \mathbb{R}[/itex] and suppose that [itex]a = b[/itex]. Then clearly we have that [itex]0*a = 0*b = 0[/itex]. Hence, if [itex]a[/itex] and [itex]b[/itex] are any two real numbers, then they are equal to each other (using your same logic). A correct proof would need to start at the equality [itex]0 = 0[/itex] and work from there (which is impossible).
 
  • #8
Thanks for the answers. I didn't want to PROVE anything here. This was just shown to me by a friend and I just wanted to know what was wrong.
 

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