Can 1 Ever Equal 2 Under the Peano Axioms?

[navneet1990]

can 1 = 2 ??

is this possible
1 = 2
??

 

[waht]

Consider the equation,

x + 1 = x + 2

And you are asked to solve for x.

By looking at it, you can say there is no solutions.

Subtract x from both sides, and you get

1 = 2

But what if you could subtitute 1 = 2 in one eqation

you get x + 1 = x + 1

or x + 2 = x + 2

That makes more sense now.

 

[CRGreathouse]

Sure, if the underlying assumptions are inconsistent.

 

[navneet1990]

i didnt get it
for any value of x how can x+1 = X+2
the assumption itself is wrong
right??

 

[radou]

This really has nothing to do with math.

 

[arildno]

Sure 1=2 in some number systems.
In a number system that denies the axiom 0\neq{1}, for example, 1=2 will be a true statement.

 

[StatusX]

1 and 2 are just symbols. By their standard definition, they correspond to two distinct real numbers, and so the statement '1=2' is false. If you want to use these symbols in a different way, then the statement '1=2' can be true, false, or meaningless, depending on how you define them.

 

[loom91]

1 is not equal to 2 in any of the standard formulations of natural numbers (Peano axioms, set theory etc). However you can very easily define two symbols 1 and 2 in a number system such that they are equal.

 

[navneet1990]

cant we use complex numbers to prove it
like :

i = i

root -1 = root -1

hence ,

root -1/ root 1 = root -1 / root 1

hence,
whole root [-1/1] = whole root [ -1/1]

hence,
whole root [-1/1] = whole root [ 1/-1]

[ -5/4 can also be written as 5/-4...cant it? i mean -1/1 is the same as 1/-1 right??]

hence,
root -1 / root 1 = root 1 / root -1

hence,

i / 1 = 1 / i

i square = 1

hence,
-1 = 1

adding 3/2 on both sides,

3/2 + ( -1 ) = 3/2 + 1

3/2 - 1 = 3/2 + 1

2/ 2 = 4/2

therefore,

1 = 2

can this be possible

 

[J77]

hence,
root -1 / root 1 = root 1 / root -1

hence,

i / 1 = 1 / i

Nope.

You can't take roots on both the top and bottom like that.

i=e^{i\pi/2}

1/i=e^{-i\pi/2}

With these 1=2 things, there's always a mistake/trick.

 

[navneet1990]

um...
i kinda understood a little
but i didnt understand
the

i = e raised to (-i Pie/ 2)
what is that

 

[chroot]

He's expressing complex numbers as complex exponentials.

i=e^{i\pi/2}

is the value on the complex unit circle corresponding to an angle of pi/2. It is equal to i.

His point is that 1/i and i/1 are quite different numbers, on opposite sides of the unit circle, so your "proof" contains an error.

- Warren

 

[navneet1990]

ohk
thank you

 

[radou]

...[ -5/4 can also be written as 5/-4...cant it? i mean -1/1 is the same as 1/-1 right??]...

A rational number is, by definition, a number of the form \frac{p}{q}, where p \in Z and q \in N. So, you can't write -5/4 as 5/-4.

 

[arildno]

A rational number is, by definition, a number of the form \frac{p}{q}, where p \in Z and q \in N. So, you can't write -5/4 as 5/-4.

Eeeh, wherever do you have this limitation from??
Not saying you might not be right, but I really don't see the necessity of this limitation.

 

[jnorman]

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.
:-)

 

[chroot]

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.
:-)

You should tell your father to take more math classes. You cannot divide both sides by x-1, when x=1, because that is equivalent to division by zero. Division by zero is not a "legal" mathematical operation.

- Warren

 

[mattmns]

A rational number is, by definition, a number of the form \frac{p}{q}, where p \in Z and q \in N. So, you can't write -5/4 as 5/-4.

That is NOT the definition of a rational number

http://mathworld.wolfram.com/RationalNumber.html

A rational number is a number p/q where p and q are Integers and q \neq 0

So, you can write (-5)/4 as 5/(-4)

 

[CRGreathouse]

A rational number is, by definition, a number of the form \frac{p}{q}, where p \in Z and q \in N. So, you can't write -5/4 as 5/-4.

How quaint.

 

[HallsofIvy]

A rational number is, by definition, a number which can be written in the form \frac{p}{q} where p \in Z and q \in N. Whether or not the number is written that way is irrelevant.

Yes, you can write -5/4 as 5/-4 just as you could write it as -1.25.

 

[tony873004]

1 quart = 2 pints

 

[radou]

It's a standard definition. Btw, does it appear logical to divide with a negative number as well as take Wolfram definitions sooo seriously ..Or let's state it this way: there is no need for q to be an integer. It is enough for q to be a natural number.

 

[shmoe]

It's a standard definition.

Allowing any non-zero integer in the denominator is much more standad from books I've seen. This is what happens in the more general case of the field of fractions of an integral domain where you don't necessarily have any sort of order.

Btw, does it appear logical to divide with a negative number

Yes it does. Why wouldn't it be?

as well as take Wolfram definitions sooo seriously

Wolfram is generally pretty good at reflecting the 'usual' definitions. This one fits with my experience. You could define them as Halls says, as things that *can* be written a/b where a,b are integers and b>0, but any definition that tries to say 5/(-4) is not rational, and is not the same as (-5)/4 is non-standard and different from everyone else's.

 

[arildno]

It's a standard definition. Btw, does it appear logical to divide with a negative number as well as take Wolfram definitions sooo seriously ..Or let's state it this way: there is no need for q to be an integer. It is enough for q to be a natural number.

So, you didn't have any arguments after all.
Of course 1/(-4) has a neat interpretation:
It is the number that multiplied with (-4) equals 1.
The number (-1)/4 is, by definition equal to (-1)*(1/4).

 

[radou]

Right, but let's put it this way. Let's define the set of rational numbers as Q=\left\{\frac{p}{q}:p \in Z , q \in N\right\}. If we compare two different rational numbers, then we have p_{1}>p_{2} \Rightarrow \frac{p_{1}}{q}>\frac{p_{2}}{q}. Now, let's define the set of rational numbers as Q=\left\{\frac{p}{q}:p, q \in Z , q \neq 0 \right\}. Then we have p_{1}>p_{2} \Rightarrow \frac{p_{1}}{q}>\frac{p_{2}}{q} if q > 0 and p_{1}>p_{2} \Rightarrow \frac{p_{1}}{q}<\frac{p_{2}}{q} if q < 0. So, in the first case, it's easier to compare two rational numbers, which may make the first definition more convenient. Sorry if I'm tiresome, ( ) but it's the defiition that I found in almost all my math textbooks (mathematical analysis, elementary math, etc.), so I'm convinced there's a reason for it.

 

[HallsofIvy]

Right, but let's put it this way. Let's define the set of rational numbers as Q=\left\{\frac{p}{q}:p \in Z , q \in N\right\}. If we compare two different rational numbers, then we have p_{1}>p_{2} \Rightarrow \frac{p_{1}}{q}>\frac{p_{2}}{q}. Now, let's define the set of rational numbers as Q=\left\{\frac{p}{q}:p, q \in Z , q \neq 0 \right\}. Then we have p_{1}>p_{2} \Rightarrow \frac{p_{1}}{q}>\frac{p_{2}}{q} if q > 0 and p_{1}>p_{2} \Rightarrow \frac{p_{1}}{q}<\frac{p_{2}}{q} if q < 0. So, in the first case, it's easier to compare two rational numbers, which may make the first definition more convenient. Sorry if I'm tiresome, ( ) but it's the defiition that I found in almost all my math textbooks (mathematical analysis, elementary math, etc.), so I'm convinced there's a reason for it.

Nobody's arguing that it's not! This a all started when you said

A rational number is, by definition, a number of the form , where and . So, you can't write -5/4 as 5/-4.

That is just flat wrong and neither of the definitions you cite say that. A rational number is not a number written in that form- it is a number that is equal to something in that form!
5/(-4) or, for that matter, -1.25, is equal to (-5)/4 and so is a rational number.

 

[Hubert]

So what is the fault in the "proof"? I assume it has something to do with the fact that both i and 1/i (or -i) are solutions to the equation x^2 -1 = 0, just as both 2 and -2 are solutions to the equation x^2 + 4 = 0.

 

[StatusX]

It is true that we can't define a function f(z) by "f(z) = the solution x to x²-z=0", because, as you point out, there are always two solutions to this equation, each the negative of the other. To define a "square root" function, we have to pick only one of these solutions for each z. For positive real z, both solutions will be real, and we usually adopt the convention of taking the positive solution as "the" square root.

This is arbitrary though, and when we try to extend to complex numbers, this arbitrariness becomes important, because there is no notion of "positive" for complex numbers, nor even of "greater than." But say we have somehow chosen a root for each complex number, and denote √(z) by whichever solution to x²-z=0 we have chosen for z. Furthermore, assume √(z) coincides with the positive root for the positive real numbers and that √(-1)=i (as opposed to -i).

Now going back to the faulty proof, we can easily isolated where the problem is by checking the validity of each line. The line -1/1=1/-1 is true (and has nothing to do with the problem with the proof, despite the attention it's received in this thread), and if we have √(z) as above, √(-1/1)=√(1/-1) is true, since we are evaluating the function at the same point. But the next line, i/1=1/i is false. The problem must be in asserting:

√(1/-1)=√(1)/√(-1)

we know √(a/b)=√(a)/√(b) holds for all real numbers a and b, but here we see it cannot hold for all complex numbers. In fact, this "faulty" proof is ironically a perfectly good proof by contradiction of the previous statement. And if you go back and pick √(-1)=-i instead, you'll reach a similar contradiction, showing that √(a/b)=√(a)/√(b) cannot be true in general no matter how we pick the roots for each z.

 

[jnorman]

warren - i am aware of the 'divide by zero' step - hence the " :-) " at the end of my response...

 

[Hubert]

So what is the fault in the "proof"? I assume it has something to do with the fact that both i and 1/i (or -i) are solutions to the equation x^2 -1 = 0, just as both 2 and -2 are solutions to the equation x^2 + 4 = 0.

Since this thread was bumped up recently, just for clarity I meant x^2 + 1=0 . I think that StatusX responded to me as if I said that anyway, but my mistake was bugging me .

 

[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]