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## Homework Statement

"Prove that Axiom 10[, the existence of identity in a linear space,] can be deduced from the other axioms."

## Homework Equations

Now, I know that these axiom numberings are fairly arbitrary, but I'll put them in anyways for easy reference. I'm listing them word for word in case there's a subtlety I'm missing.

"Let

*V*denote a nonempty set of objects, called

*elements*. The set

*V*is called a linear space if it satisfies the following ten axioms...:

Axiom 1. Closure Under Addition.

*For every pair of elements x and y in V there corresponds a unique element in V called the sum of x and y, denoted by x + y.*

Axiom 2. Closure Under Multiplication by Real Numbers.

Axiom 3. Commutative Law.

*For all x and y in V, we have x + y = y + x.*

Axiom 4. Associative Law.

*For all x, y, and z in V, we have (x + y) + z = x + (y + z).*

Axiom 5. Existence of Zero Element.

*There is an element in V, denoted by O, such that*

x+ O = x for all x in V.

x+ O = x for all x in V.

Axiom 6. Existence of Negatives.

*For every x in V, the element (-1)x has the property*

x + (-1)x = O

x + (-1)x = O

Axiom 7. Associative Law.

*For every x in V and all real numbers a and b, we have*

a(bx) = (ab)x.

a(bx) = (ab)x.

Axiom 8. Distributive Law for Addition in V.

*For all x and y in V and all real a, we have*

a(x + y) = ax + ay

a(x + y) = ax + ay

Axiom 9. Distributive Law for Addition of Numbers.

*For all x in V and all real a and b, we have (a + b)x = ax + bx.*

Axiom 10. Existence of Identity.

*For every x in V, we have 1x = x.*"

Afterwards, it states that when the scalars are real, such a linear space is sometimes called a real linear space while if the scalars are complex numbers, then the space is called a complex linear space. It also states that linear spaces are sometimes reffered to as linear vector spaces or vector spaces. The book states that whenever it refers to a linear space without designation, the space could be either real or complex.

Also, the uniqueness of the zero element and negative elements are proved as theorems later on. And while the book never explicitly says so, from sample proofs of theorems from the axioms I can see that substitution is another, though "hidden," assumption.

## The Attempt at a Solution

I am quite stuck on this one. It's an axiom so shouldn't it be impossible to prove it from the other axioms? I think I can prove it if I assume the standard rules of multiplication of real numbers. They are real scalars after all, so shouldn't that be fine (after I do the same assuming the scalars are complex)? However, I have a problem with that since the statement explicitly said to prove it from the other axioms of a linear space, so I would assume that's all I can use.

Anyways, this is as far as I've gotten: By Axiom 2, there is an element in V, say z, such that z = 1x. By Axiom 6, z + (-1)z = O. Thus, 1x +(-1)1x = O. By Axioms 9 and 6: 1x +(-1)1x = (1 + (-1)1)x = Ox. Therefore, 1x + (-1)1x = Ox = O = x + (-1)x (last part by Axiom 5).

Now, if I could somehow prove 1 + -1 = O, 1x + (-1)x = O, or x + (-1)1x = O, then I could finish the proof by adding x (or 1x) to both sides of 1x + (-1)1x = x + (-1)x getting 1x + (-1)1x + x = x +(-1)x + x = 1x + O = x + O = 1x = x. (1 + -1 = O would allow me to show 1x + (-1)x = Ox = O). Of course, one problem is the lack of a unique zero element, but I suppose I would have to prove that too along the course of my proof? Or do I need to stick strictly to the Axioms? I'm probably missing something simple here, if this can be proved at all.

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