# proof that 0 is not equal to 1

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• Jan 14th 2011, 10:48 PM
Noxide
proof that 0 is not equal to 1
I'm not very experienced when it comes to writing proofs, and I have come up with two proofs for 0 =/= 1, both by reductio ad absurdum. Is either one of them valid?

for any field F, 0 denotes the additive identity and 1 denotes the multiplicative identity
the real numbers are a set
sets are a collection of distinct elements
if 0 = 1 then 0 is not in the real numbers xor 1 is not in the real numbers
if 0 xor 1 are not in the real numbers then, the real numbers are no longer a field
but the real numbers are a field
therefore 0 =/= 1

a is a real number
suppose 0 = 1, then a + 0 = a and a + 1 = a
the real numbers are a field
there exists an element 0 for any a in a field F such that a + 0 = a
therefore 0 =/= 1
• Jan 14th 2011, 11:14 PM
dwsmith
Quote:

Originally Posted by Noxide
I'm not very experienced when it comes to writing proofs, and I have come up with two proofs for 0 =/= 1, both by reductio ad absurdum. Is either one of them valid?

for any field F, 0 denotes the additive identity and 1 denotes the multiplicative identity
the real numbers are a set
sets are a collection of distinct elements
if 0 = 1 then 0 is not in the real numbers xor 1 is not in the real numbers
if 0 xor 1 are not in the real numbers then, the real numbers are no longer a field
but the real numbers are a field
therefore 0 =/= 1

a is a real number
suppose 0 = 1, then a + 0 = a and a + 1 = a
the real numbers are a field
there exists an element 0 for any a in a field F such that a + 0 = a
therefore 0 =/= 1

Let $\displaystyle x=0$

$\displaystyle x(x-y)=0*(x-y)\Rightarrow x(x-y)=0 \ \ y\in\mathbb{Z}, \ \ y\neq 0$

$\displaystyle (x-y)=0\Rightarrow x=y$

We have reached a contradiction since $\displaystyle x = y\neq 0$
• Jan 14th 2011, 11:30 PM
dwsmith
Here is a proof from MIT on the matter.

Given any x,

$\displaystyle \displaystyle\int_0^1\frac{x^3}{y^2}e^{-x^2/y}dy=\left[xe^{-x^2/y}\right]_0^1=xe^{-x^2}$

$\displaystyle \displaystyle e^{-x^2}(1-2x)=\frac{d}{dx}(xe^{-x^2})$

$\displaystyle \displaystyle =\frac{d}{dx}\int_0^1\frac{x^3}{y^2}e^{-x^2/y}dy$

$\displaystyle \displaystyle =\int_0^1\frac{\partial}{\partial x}\left(\frac{x^3}{y^2}e^{-x^2/y}\right)dy$

$\displaystyle \displaystyle =\int_0^1 e^{-x^2/y}\left(\frac{3x^2}{y^2}-\frac{2x^4}{y^3}\right)dy$

Now, set x =0.

$\displaystyle \displaystyle e^{0}=\int_0^1 e^{0}\left(0\right)dy\Rightarrow 1=\int_0^1 0dy\Rightarrow 1=0$
• Jan 15th 2011, 02:14 AM
emakarov
Quote:

I'm not very experienced when it comes to writing proofs, and I have come up with two proofs for 0 =/= 1, both by reductio ad absurdum.
Proving simple things is often more complicated because one has to be very careful in choosing the means. Some things look very ordinary and therefore obvious, but they can't be used because they have not been proved. Also, proofs of simple things should use smaller steps, whereas in proving more complicated facts one is sometimes justified to omit a page of calculations noting "It is obvious that...". This is like a map's scale: smaller objects require larger scale.

So, I understand that you are proving 0 ≠ 1 in real numbers.

Quote:

if 0 = 1 then 0 is not in the real numbers xor 1 is not in the real numbers
It is not clear why 0 = 1 implies that one of them is not a real number. Also, in my experience, "xor" is never used to mean "either... or" in regular text; it is only used in a discussion of Boolean connectives.

Quote:

a is a real number
Which one?
Quote:

suppose 0 = 1, then a + 0 = a and a + 1 = a
the real numbers are a field
there exists an element 0 for any a in a field F such that a + 0 = a
There is no need to say, "There exists an element 0 for any a". First, 0 is the same for all a; second, one usually uses "there exists" when one cannot exhibit an object explicitly. Saying, "There exists a 0" is like saying, "There exists a city in the United States (you may have never heard of it), called New York, which is the largest city in the New York State."
Quote:

therefore 0 =/= 1
It's not clear how this follows (see the remark about small steps above).

Quote:

Originally Posted by dwsmith
Let $\displaystyle x=0$

$\displaystyle x(x-y)=0*(x-y)\Rightarrow x(x-y)=0 \ \ y\in\mathbb{Z}, \ \ y\neq 0$

$\displaystyle (x-y)=0\Rightarrow x=y$

We have reached a contradiction since $\displaystyle x = y\neq 0$

You probably mean, since 0 = 1, 0 = 0 * (x - y) = 1 * (x - y) = x - y, so x = y. However, it is not clear why there exists an y ≠ 0.

I did some search and it seems that 0 ≠ 1 is accepted as an axiom in (some) rings, fields, ordered fields and reals. In particular, the axioms of reals consist of three parts: $\displaystyle \mathbb{R}$ is a field, it has a total order compatible with operations, and the completeness axiom. As far as I can see, if the axiom 0 ≠ 1 is removed, all other axioms are true in the one-element set {0}.
• Jan 15th 2011, 08:33 AM
HallsofIvy
Both of these proofs are invalid for the same reason- they both prove only that in the field of real numbers, $\displaystyle 0\ne 1$. They do not apply to general fields.

Quote:

Originally Posted by Noxide
I'm not very experienced when it comes to writing proofs, and I have come up with two proofs for 0 =/= 1, both by reductio ad absurdum. Is either one of them valid?

for any field F, 0 denotes the additive identity and 1 denotes the multiplicative identity
the real numbers are a set
sets are a collection of distinct elements
if 0 = 1 then 0 is not in the real numbers xor 1 is not in the real numbers
if 0 xor 1 are not in the real numbers then, the real numbers are no longer a field
but the real numbers are a field

Why are you talking about the real numbers? This does not prove that 0 is not equal to 1 in a general field, only that it is not true in the field of real numbers (and you prove that by asserting that it is not true for the real numbers!). IF the theorem were "in all fields, $\displaystyle 0\ne 1$" then exhibiting one "counter example", the real numbers, would be sufficient. But that is not what you want to prove. You want to prove that there are no fields in which 0= 1.

[quote]therefore 0 =/= 1

Quote:

a is a real number
suppose 0 = 1, then a + 0 = a and a + 1 = a
the real numbers are a field
there exists an element 0 for any a in a field F such that a + 0 = a
therefore 0 =/= 1
Again, you are only saying that $\displaystyle 0\ne 1$ for the field of real numbers, not for a general field. No proof about general fields should mention the real numbers- from the basic definition of "field", fields would exist even if there were no "real numbers"!

"0" and "1" in a general field are NOT the real numbers 0 and 1- they are simply the additive and multiplicative identities.
And any proof that they are not the same, for a general field, must use the definition of "0" and "1".

"0" is the multiplicative identity: a+ 0= a for any a in the field.
"1" is the multiplicative identity: a(1)= a for any a in the field.

You can prove, using the distributive law, that 0(a)= 0 for any a in the field: a(b+ 0)= ab+ a(0) but b+ 0= b so a(b+ 0)= ab. Now we have ab= ab+ a(0) and, adding the additive inverse of ab to both sides, 0= a(0).

From that, if 0= 1, it follows that 1(a)= a= 0(a)= 0. That is, if 1= 0, every member of the field is equal to 0. Actually, some definitions of "field" allow fields having only one member, 0, in which case, it is true that "1= 0".

More commonly, definitions of "field" require more than one element, or simply assert, as an axiom, that "$\displaystyle 0\ne 1$".
• Jan 15th 2011, 09:13 AM
emakarov
Quote:

Originally Posted by HallsofIvy
Both of these proofs are invalid for the same reason- they both prove only that in the field of real numbers, $\displaystyle 0\ne 1$. They do not apply to general fields.

No, the proofs are invalid because they are just wrong (or presented not clearly enough), even for real numbers. It is not clear to me whether the OP wanted to prove 0 ≠ 1 for an arbitrary field or for reals only, but the real problem is the logic of the proofs.
• Jan 15th 2011, 09:54 AM
magus
Ignore This My Proof Doesn't Work
• Jan 15th 2011, 09:58 AM
emakarov
Quote:

Originally Posted by magus
$\displaystyle 0=1$, $\displaystyle 1\cdot 0 = 1 \cdot1 = 1$

but we know that by definition that $\displaystyle 1 \cdot 0=0$

This is our contradiction and therefore $\displaystyle 1 \neq 0$

So, where exactly is the contradiction?
• Jan 15th 2011, 10:06 AM
magus
I'm sorry it only works for the field of real numbers where $\displaystyle 0\cdot a=0$. I'll keep quiet from now on.
• Jan 15th 2011, 10:43 AM
emakarov
Quote:

Originally Posted by magus
I'm sorry it only works for the field of real numbers where $\displaystyle 0\cdot a=0$.

No, 0 * a = 0 is true in any ring (and field), as HallsofIvy showed above. The problem is that, even though 1 * 0 = 1 and 1 * 0 = 0 imply that 0 = 1, this is not necessarily a contradiction because the whole field can be {0}. In fact, 0 = 1 was assumed in the beginning of the proof.
• Jan 15th 2011, 11:27 AM
magus
First I apologize for hijacking this thread. I really thought I knew the answer to the question and was being helpful.

I see what you're talking about and you're right it was incorrect. I don't know what I was thinking.
• Jan 15th 2011, 03:05 PM
Noxide
No problem magus, that means at least two of us learned something from this thread.

With help from Ivy, I came up with the following proof:

F denotes a field
there exists F: 0 = 1 ==>
By Theorem 3.1.ii and M3, For every a in F a = a * 1 = a * 0 = 0
F must have at least two elements
QED

Should I write more, I'm really not sure if my proof is clear enough. I think it is, but i'm afraid of writing too much and looking like a fool for explaining obvious steps or writing too little and not having the proof be clear.
• Jan 15th 2011, 04:22 PM
Noxide
Quote:

Originally Posted by emakarov
There is no need to say, "There exists an element 0 for any a". First, 0 is the same for all a; second, one usually uses "there exists" when one cannot exhibit an object explicitly. Saying, "There exists a 0" is like saying, "There exists a city in the United States (you may have never heard of it), called New York, which is the largest city in the New York State."

From what I understand, 0 is just the name of the element, but if you're creating other fields 0 could represent a monstrously large expression so saying there exists a 0 might not be incorrect if i'm talking about a general field since i'm not explicitly stating what zero is... i'm not sure though

also, would the same principle apply to:

for every a in a field F, there exists an element (-a) ?

thanks
• Jan 16th 2011, 04:07 AM
emakarov
Quote:

F denotes a field
there exists F: 0 = 1 ==>
By Theorem 3.1.ii and M3, For every a in F a = a * 1 = a * 0 = 0
F must have at least two elements
QED
Since you are not saying where Theorem 3.1.ii came from, I assume it is from The Book. From Wikipedia:
Quote:

[Paul Erdős] had his own idiosyncratic vocabulary: he spoke of "The Book", an imaginary book in which God had written down the best and most elegant proofs for mathematical theorems. Lecturing in 1985 he said, "You don't have to believe in God, but you should believe in The Book."... When he saw a particularly beautiful mathematical proof he would exclaim, "This one's from The Book!".
Could you tell me how to get a copy? :)

I guess, you are proving that in field with more than one element, 0 ≠ 1. In this case I agree.

Quote:

for every a in a field F, there exists an element (-a)
When this is formulated for the first time, say, as an axiom, one has to say "there exists some element, which we will denote by -a" because we aren't necessarily given an easy way of identifying this element. Once we fix the notation -a, there is no longer a reason to say, "since there exists a $\displaystyle -a$ such that $\displaystyle a + (-a) = 0$..."; one can just say, "since a + (-a) = 0...". In any case, though, this is an issue of style; inserting an extra "there exists" does not make a statement false.
• Feb 19th 2015, 07:22 PM
mekun
Re: proof that 0 is not equal to 1
My professor asked us to prove this and this is exactly how I proved it.
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