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I'm quite stuck on wrapping my head around this question on linear algebra. I've just started studying it so I don't know too much about fields and such, but the only knowledge required for this question is the axioms for number systems.

The question I am doing is as follows:

"Define a paddock to be a set in which all the axioms for addition and multiplication hold, but instead of $0\ne 1$, $0=1$. Find an example of a paddock, and prove that every paddock has just one element. Clearly the definition of a paddock is of no use outside of this exercise."

The sets that instantly came to mind are $\{0\}$ and $\{1\}$, but possibly also the set $\{0,1\}$.

$\{0\}$ would be the only logical set to pick if I wanted all the axioms for addition to hold (keeping in mind the axiom that states that for each $x \in S$, there exists a number $-x \in S$).

However, would the set $\{{0\}}$ not be equivalent to the set $\{1\}$, because $0=1$? And if $0=1$ then wouldn't $1=2$, meaning $\{0\}=\{1\}=\{2\}=\{n\}$?

Additionally, going back to my previous guesses, wouldn't the set $\{0,1\}$ be equivalent to $\{0,0\}$ and $\{1,1\}$? In which case $\{0,0\}$ would be a valid set, and hence a set containing an infinite amount of zeros or ones or even an infinite amount of any numbers would be valid too, as in the end everything is each other and also 0? I feel like I'm getting myself tangled into a huge confused loop.

In any case, I'm unsure how to actually prove any of this - the fact that every paddock has just one element (even though I feel like a set with an infinite number of elements are allowed, but I could be completely wrong), or even what exactly the elements in the possible sets are.

Any kind of help or pointers are appreciated! Thank you in advance!

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    In a set, we don't list elements more than once. The set $\{ 0, 0 \}$ is just the set $\{ 0 \}$. Moreover, you should find that the sets $\{ 0 \}$ and $\{ 1 \}$ are isomorphic with respect to this paddock structure. – 2017-01-18
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    "And if 0=1 then wouldn't 1=2". If you define 0 to be the addition identity and 1 to be the multiplication identity then you can't say that 0=1 then 0+1=1+1 then 1=2, that requires treating 1 as if it has a property that relates to addition but that isn't so. – 2017-01-18
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    "The sets that instantly came to mind are {0} and {1}, but possibly also the set {0,1}." If 0 =1 then those three sets are all the same thing. – 2017-01-18
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    @hugh I get what you are saying but I'm not sure I agree. If 0 is defined as the additive identity and 1 is defined as the multiplicative identity. And 2 is diffined as 1+1 that is fine. Then if 0 = 1 we have 2 = 1 + 1 = 0 + 0 = 0. That's fair. 2, 1, and 0 are not different things in this paddock. They are all just different symbols for the same element. We need to prove there can't be any other element. – 2017-01-18
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    You are thinking that that "infinite" elements of the same thing are infinite elements. If they are all equal then they are only one thing. If I came up with than infinite 3 = 6/2 = 9/3 = 12/4 =.... = 2+1 = 1+2 = the number of letters in "CAT" = .... I don't have an infinite set. I have a set with only *one* thing in it: 3. so if 0 = 1 = 2 then {0} and {1} and {0,0} and {1,2} are not merely "equivalent" they *ARE* the same thing. All you have is {single element} where single element is additive identity, multiplicative identity, and any other possible expressions of elements. – 2017-01-18
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    Is $a *0 = 0$ an axiom or a provable proposition. If you know that you are done because $a = a*1 = a*0 = 0=1$. Whatever you want to call it there is only one element. "all" elements equal it, so it is the only element there is. – 2017-01-18

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You are trying to hard thinking about the symbols $0$ and $1$ being god-given things.

You have a multiplicative identity and call it $1$. We have an additive inverse and we call it $0$. But in this case they are the same thing. We'll call it $X$. We'll call it $X_m$ when it is acting like the multiplicative inverse and we'll call it $X_a$ when it is acting like the additive inverse. $X_m = X_a = X$

And we know by axioms that for any element $a$ that $a*X = a$ and $a + X = a$.

We need to prove that there is only one element. If $a$ is any element at all then $a = X$. That's it. $X$ is all there is.

So suppose there is an element $a$. Then $a*X_m =a$ (that's the multiplicative identity).

Then $a+a = a*X + a*X = a(X+X)$ (that's the distributive property)

Then $a+a = a(X + X_a) = a*X = a$ (that's the additive identity: $k+X_a = k$ for any $k$)

so $a + a + (-a) = a + (-a)$ (every number has an additive inverse)

so $a + X_a = X_a$ (that the additive inverse: every number, $a$ has a corresponding number $-a$ so that $ a + (-a) = X_a$).

so $a + X_a = a = X_a$ (that's the additive identity: $a + X_a = a$).

So $a = X_a = X$. All elements equal $X$ and so $X$ is the only element there is.

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    All of your explanations and comments made a lot of sense and I have a much clearer understanding of what's going on. Thank you so much! – 2017-01-19
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In any ring with at least two elements one has $0\ne 1$

If $0=1$, then take an arbitrary element $x$ and you have $x\cdot 0=x\cdot 1$, so $x=0$.

So what you call "paddock" is the Zero ring

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    Isn't this just repeating what needs to be shown? – 2017-01-18
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    This proves that a paddock must have one element. And if you have one element, there is only one way of defining addition and multiplication on it (think about how the addition/multiplication table would look like). – 2017-01-18
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Under the "axioms for multiplication and addition", i.e., the usual axioms for a ring, if $0 = 1$, then for any element $x$, we have $x = 1x = 0x = 0$. So a ring in which $0 = 1$ (i.e., a "paddock" according to the terminology of the question) has exactly one element.

The "paddock"/"field" joke is (I concede) an entertaining reminder that the axioms for a field always include $0 \neq 1$.

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Suppose we have a paddock $(S, + , \cdot, 0_S, 1_S)$. Now consider two $x,y \in S$.

Then we have $x \cdot 0_S = y \cdot 0_S = 0_S$, since $0_S$ is the zero element. But since $0_S = 1_S$, we also have that $x \cdot 0_S = x \cdot 1_S = x$ and $y \cdot 0_S = y \cdot 1_S = y$. And this means that $x = y$. By transitivity of equality, we get that all elements in $S$ are equal. In other words, $S$ is a singleton set.