Can 1 Ever Equal 2 Under the Peano Axioms?

  • Thread starter navneet1990
  • Start date
In summary, the conversation discusses the possibility of 1 being equal to 2. The participants consider equations and proofs using complex numbers, rational numbers, and division by zero. Ultimately, they conclude that 1 is not equal to 2 in any standard mathematical formulation.
  • #36


Sobeita said:
Okay, how about this one?

1^1 = 1; // Exponential identity.
1^2 = 1; // A power of 1 equals 1.
1^1 = 1^2. // Substitution of like terms.
1 = 2. // Exponents are equal if the bases are equal.

:)
Not true.
Assuming a > p, if ax = ay, then ex lna = ey lna
This implies that x = y OR that a = 1.

If a = 1 as in your "proof" then x and y can be unequal.
 
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  • #37


Ah, that makes more sense. Thank you. :) I've seen some 1=2 problems before, but most of them had a definite error (like (a+b)/(a-b) when a=b) rather than a subtle one like this.
 
  • #38


jnorman said:
i remember my father showing me a proof once that 1=2, but can't quite recall it. but, if you start with the equation:
x^2 -1 = 0, you can factor x^2 - 1 into (x+1)(x-1)=0
then divide both sides by x-1, and get x+1=0.
for a value of x=1, you have shown that 2=0.
:-)

First: you can not divide by [tex]x-1[/tex].
Second: [tex]x+1=0[/tex], then [tex]x=-1[/tex] => [tex](-1)+1=0[/tex]
 
  • #39


Mark44 said:
Not true.
Assuming a > p, if ax = ay, then ex lna = ey lna
This implies that x = y OR that a = 1.

If a = 1 as in your "proof" then x and y can be unequal.

Very nice proof.
 
  • #40


A proof for that 1=2 is not necessarily incorrect, we do not know whether arithmetic is consistent or not. To disprove such a "proof" one must find the error in all cases.
 
  • #41


If 1=2, then 2*2*2*2*2^infinity = 1. It would basically destroy everything we know. XD
 
  • #42


Sobeita said:
Okay, how about this one?

1^1 = 1; // Exponential identity.
1^2 = 1; // A power of 1 equals 1.
1^1 = 1^2. // Substitution of like terms.
1 = 2. // Exponents are equal if the bases are equal.

:)

I really like this one. It hides the division by zero very well.
 
  • #43


I made that one up, because I knew the powers of one could be pretty dodgy - I assumed there was some sort of rule (x=y or a=1) that could cover it, but I didn't remember what it was.

I just did a search and came up with this page:
http://en.wikipedia.org/wiki/Invalid_proof

bece7a4ff69a1ca64899e8693b697dc3.png


ee85bd408b1e82c63883c9dfcb6f9f91.png
 
  • #44


LumenPlacidum said:
I really like this one. It hides the division by zero very well.

using those identity you should get 0=0, since log1=0
 
  • #45


navneet1990 said:
is this possible
1 = 2
??


Yes, if the Peano Axioms are inconsistent.

http://en.wikipedia.org/wiki/Peano_axioms

When the Peano axioms were first proposed, Bertrand Russell and others agreed that these axioms implicitly defined what we mean by a "natural number". Henri Poincaré was more cautious, saying they only defined natural numbers if they were consistent; if there is a proof that starts from just these axioms and derives a contradiction such as 0 = 1, then the axioms are inconsistent, and don't define anything. In 1900, David Hilbert posed the problem of proving their consistency using only finitistic methods as the second of his twenty-three problems.[12] In 1931, Kurt Gödel proved his second incompleteness theorem, which shows that such a consistency proof cannot be formalized within Peano arithmetic itself.[13]
 

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