Can 1 Ever Equal 2 Under the Peano Axioms?

[mar01]

given a=b

a²=ab

subtract b² on both sides

a²-b²=ab-b²
(a+b)(a-b)=b(a-b)
a+b=b
a+a=a
2a=a
2=1
simple

 

[emyt]

given a=b

a²=ab

subtract b² on both sides

a²-b²=ab-b²
(a+b)(a-b)=b(a-b)
a+b=b
a+a=a
2a=a
2=1
simple

division by zero alert!

 

[Sobeita]

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.

:)

 

[Bob_for_short]

No, 1 cannot be 2. Normally they say "2 in 1" or "3 in 1" but never vice versa.

 

[epkid08]

You're basically asking: "If one equals two, then can one equal two?"

Unfortunately, one will never equal two, so the underlying assumption that one can equal two is false.

 

[Mark44]

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 a^x = a^y, then e^{x lna} = e^{y lna}
This implies that x = y OR that a = 1.

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

 

[Sobeita]

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.

 

[EngWiPy]

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 $$x-1$$.
Second: $$x+1=0$$, then $$x=-1$$ => $$(-1)+1=0$$

 

[EngWiPy]

Not true.
Assuming a > p, if a^x = a^y, then e^{x lna} = e^{y 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.

 

[disregardthat]

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.

 

[Sobeita]

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

 

[LumenPlacidum]

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.

 

[Sobeita]

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

 

[kthouz]

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

using those identity you should get 0=0, since log1=0

 

[Count Iblis]

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]