# Prove 0 < 1 (1 Viewer)

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#### Jamin2112

1. The problem statement, all variables and given/known data

Prove 0 < 1 using the properties of an ordered field.

2. Relevant equations

All the field properties

3. The attempt at a solution

Suppose 0=1 or 0 < 1.

If 0=1, then adding a to both sides of the equation yields

(a) + 0 = (a) + 1

Because 0 is the identity element for addition,
a = a + 1

.....

Now can I just say "this a contradiction" and move on?

#### Pagan Harpoon

That is only a contradiction if it relies on other properties of the space with which you're working. For example, you might be in a space where every element is 0, then 1 and a (if they are in the space) are also 0, so there is no contradiction there. If we are looking at real numbers here, then probaly in the early stages of defining numbers and addition, you would assume that 1 is not the identity element of addition. That means that indeed a is not =a+1 for any a and you have your contradiction. However, this approach relies on the ASSUMPTION that 0 is not =1 so it is circular logic.

I think a better way to do it is this... assume that there is an element a that is not =0. Then, multiply the equation by a and you get 0=a which is a more explicit contradiction.

-edit-

Mmm, yes. Perhaps you can just say that they're not euqal because if they are, the only number in the "field" would be 0 and then addition and multiplication would be the same. So you wouldn't have two operations, so you wouldn't have a field.

-another edit-

I suppose you could also argue that my assumption that there is an element a not =0 implicitly assumes that 1 is not equal to 0 so it is also circular logic.

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#### Jamin2112

That is only a contradiction if it relies on other properties of the space with which you're working. For example, you might be in a space where every element is 0, then 1 and a (if they are in the space) are also 0, so there is no contradiction there. If we are looking at real numbers here, then probaly in the early stages of defining numbers and addition, you would assume that 1 is not the identity element of addition. That means that indeed a is not =a+1 for any a and you have your contradiction. However, this approach relies on the ASSUMPTION that 0 is not =1 so it is circular logic.

I think a better way to do it is this... assume that there is an element a that is not =0. Then, multiply the equation by a and you get 0=a which is a more explicit contradiction.
Okay. My packet doesn't explicitly state that 1 is the identity element for addition, but I'll just pretend it does.

#### fluxions

by trichotomy either 0 < 1, 0 = 1, or 0 > 1. the definition of a field rules out the possibility that 0 = 1, so that leaves 0 < 1 or 0 > 1. can you think of a way to rule out the case 0 > 1?

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#### Fredrik

Staff Emeritus
Gold Member
The definition of "field" usually includes the requirement that 1≠0. If your book does, it's sufficient to prove that 1<0 implies 0<1. (This is a contradiction, so the assumption must be false). If your book doesn't include that requirement, you should also prove that 1=0 implies that x=0 for all x. This means that 1≠0 in all non-trivial fields.

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