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here's a fun proof that 7 =14

 
 
shug23
 
Reply Sun 14 Jun, 2020 06:31 am
2+2 =2^2
3+3+3 =3^2
4+4+4+4=4^2
by induction x+x+x+x....(x times)=x^2
Take derivatives of both sides of last statement:
1+1+1+1....(xtimes)=2X, but 1+1+1+1....x times equals x
therefore X=2X for all X, or specifically,
7=14

do you see the flaw ? Smile
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shug23
 
  1  
Reply Sun 14 Jun, 2020 07:24 am
@shug23,
most proofs that result in this kind of nonsense usually sneak in a division by zero.......not the case here, which makes it a bit unique
0 Replies
 
Olivier5
 
  2  
Reply Sun 14 Jun, 2020 12:47 pm
@shug23,
The derivation of x+x+x (x times) is NOT 1+1+1+1....(x times). Because "x times" is NOT derivated.
Sturgis
 
  0  
Reply Sun 14 Jun, 2020 01:25 pm
Not to mention, 14=5
0 Replies
 
izzythepush
 
  1  
Reply Sun 14 Jun, 2020 01:34 pm
I don’t see the fun.
0 Replies
 
shug23
 
  1  
Reply Sun 14 Jun, 2020 03:00 pm
@Olivier5,
I'm going to say this is close enough.....we are not dealing with continuous functions and hence not differentiable
Olivier5
 
  1  
Reply Sun 14 Jun, 2020 03:08 pm
@shug23,
I don't think so.

X + x + x ... (x times) is but an odd way to write "x times x", usually shortened x * x or x^2.

Y = x^2 is a continuous function.
0 Replies
 
shug23
 
  1  
Reply Sun 14 Jun, 2020 05:42 pm
we are dealing with integers on the left side of the equation... I agree f(x)=x^2 is continuous and differentiable.....but consider g(x)= x+x+x+x..............x times. We know g(x)= x^2 is true if x is an integer....is it true if x=, say, pi ? is it true if x = -4.8 ?

For those that remember the definition of a derivative (lim as h goes to zero of (g(x+h)-g(x))/h, try to take the derivative of g(x) and it will fall apart.

However, if you think g(x) is differentiable, then there is nothing wrong with the proof and indeed, 7 does in fact equal 14
Olivier5
 
  1  
Reply Mon 15 Jun, 2020 12:12 am
@shug23,
Okay, but your problem statement does not explicitly limit x to integers.
shug23
 
  1  
Reply Mon 15 Jun, 2020 05:38 am
@Olivier5,
Correct, I worded the proof to mislead .... As an aside, I can also 'prove' all triangles are isosceles if interested
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