Prove that (-1)•a= -a
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Prove that (-1)•a= -a

[From: ] [author: ] [Date: 11-11-29] [Hit: ]
Im going to assume youre in high school, so sorry if Im wrong and what Im saying sounds condescending.When we get to things this basic, we should point out what our axioms are, that is, what we fundamentally believe to be true to the point that we do not need to prove them.......
Im really confused on my homework. I don't know if it's really simple and I just can see it. I need a two column method where you show the proof and the reasons.

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I'm not sure exactly what level of maths you're doing. Proving something this fundamental makes me think you're doing university level mathematics (since high school doesn't concern itself with proofs like this), but asking for a "two column method" sounds like high school level. I'm going to assume you're in high school, so sorry if I'm wrong and what I'm saying sounds condescending. :-)

When we get to things this basic, we should point out what our axioms are, that is, what we fundamentally believe to be true to the point that we do not need to prove them. The following are axioms that are taught in a real analysis course that apply to the real numbers:

1) (a + b) + c = a + (b + c) ... (associativity of +)
2) a + b = b + a ... (commutativity of +)
3) There exists an element 0 in R such that 0 + a = a for any a in R ... (identity of +)
4) For each a in R, there exists an -a in R such that -a + a = 0 ... (inverses in +)
5) a(bc) = (ab)c ... (associativity of *)
6) ab = ba ... (commutativity of *)
7) There exists an element 1 in R such that 1a = a for any a in R ... (identity of *)
8) For each a in R, except a = 0, there exists an a^-1 in R such that (a^-1)a = 1 ... (inverses in *)
9) a(b + c) = ab + ac ... (distributive law)
10) - 14) Not important right now

This long list of statements are fundamental enough to be offered without proof. It is from these that we can prove (-1)a = -a. First we must prove that 0x = 0 for any x. We know that:

0 + 0 = 0 ... (since 0 is the + identity)
x(0 + 0) = x0
x0 + x0 = x0 ... (distributive law)
-(x0) + (x0 + x0) = -(x0) + x0 ... (x0 has an additive inverse)
(-(x0) + x0) + x0 = -(x0) + x0 ... (associativity of +)
0 + x0 = 0 ... (since -(x0) and x0 are additive inverses, they sum to 0)
x0 = 0 ... (additive identity)
0x = 0 ... (commutativity of *)

Now, we know that:

-1 + 1 = 0 ... (additive inverses)
(-1 + 1)a = 0a = 0 ... (we just proved this)
(-1)a + 1a = 0 ... (distributive law)
(-1)a + a = 0 ... (multiplicative identity)
((-1)a + a) + (-a) = 0 + (-a) ... (note that -a is the additive inverse of a, but (-1)a is -1 times a)
(-1)a + (a + (-a)) = 0 + (-a) ... (associativity)
(-1)a + (-a + a) = 0 + (-a) ... (commutativity)
(-1)a + 0 = 0 + (-a) ... (inverses in +)
0 + (-1)a = 0 + (-a) ... (commutativity)
(-1)a = -a ... (identity of +)

So, using only the axioms, we obtain that (-1)a = -a.
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