Freind of mine showed me this 3= 4?

  • Context: High School 
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Discussion Overview

The discussion revolves around a mathematical claim that suggests the equation 3 = 4 can be derived from manipulating algebraic expressions. Participants explore the validity of these manipulations and the implications of dividing by zero.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant presents an algebraic manipulation leading to the conclusion that 4 = 3, questioning what might be missing in their reasoning.
  • Another participant identifies that the manipulation involves dividing by zero, which is not valid in mathematics.
  • A third participant clarifies that the equation 4(a+b-c) = 3(a+b-c) can only hold true if (a+b-c) = 0, emphasizing that dividing by this expression leads to an indeterminate form.
  • Further, a participant points out that the assumption a+b = c leads to a contradiction when manipulated similarly, reinforcing the idea that 4 cannot equal 3.
  • One participant expresses gratitude for the clarification, indicating they are still learning basic mathematics.
  • A later reply acknowledges that everyone starts learning from the basics, fostering a supportive environment.

Areas of Agreement / Disagreement

Participants generally agree that the manipulations leading to 4 = 3 are flawed due to the division by zero. However, the discussion remains focused on the exploration of these flawed arguments rather than reaching a consensus on any alternative conclusions.

Contextual Notes

Limitations include the reliance on assumptions such as a+b = c and the implications of dividing by zero, which are not universally accepted in mathematical reasoning.

Who May Find This Useful

Individuals interested in foundational mathematics, algebraic manipulation, and the principles surrounding division by zero may find this discussion insightful.

phEight
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what am i missing here? this seems to make sense...

a + b = c



4a - 3a + 4b - 3b = 4c - 3c



4a + 4b - 4c = 3a + 3b - 3c



4(a+b-c) = 3(a+b-c)



4 = 3
 
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You divided by zero.

You can do it even more simply:

[tex]a=b[/tex]
[tex]4a = 3a+b[/tex]
[tex]4a - 4b = 3a - 3b[/tex]
[tex]4(a-b) = 3(a-b)[/tex]
[tex]4=3[/tex]

it's equivalent to saying

[tex]3 \cdot 0 = 4 \cdot 0 \Longrightarrow 3 = 4[/tex]

which is silly
 
4(a+b-c) = 3(a+b-c)

This statement can be true only if (a+b-c) = 0

If you try to divide both sides by ( a+b-c) then it becomes 0/0 i.e. indeterminate form of the function. hence 4 is never equal to 3, at least not for a mathematician.
 
It's even worse. We KNOW [itex]a+b-c=0[/itex]. Note his assumption, [itex]a+b = c[/itex]. Just like in my example, where I assume [itex]a=b \Longrightarrow a-b = 0[/itex].
 
phEight said:
what am i missing here? this seems to make sense...

a + b = c



4a - 3a + 4b - 3b = 4c - 3c



4a + 4b - 4c = 3a + 3b - 3c

4(a+b-c) = 3(a+b-c)
4*0 = 3*0
ERROR -> 4*0/0 = 3*0/0
4 = 3
 
Thanks for giving me a more "in depth" look at what is most probably basic mathematics to most of you. I'm still learning. :)
 
We all start somewhere.
 

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