lol... also look into why 1 is > than 0

1 is not actually defined to be greater than zero, it can be proven. The real numbers are an ordered field (technically, a complete ordered field, but completeness is not needed in this proof). If you are unfamiliar with the field axioms and the definition of a total ordering, you should look those up first. In fact, since there are different ways of defining a total ordering, I'll just give you the definition I'll be using:

If S is a set, define a subset P (called the positive set) by the following three conditions:

1.) Elements of P are closed under addition

2.) Elements of P are closed under multiplication

3.) If x is an element of S, then one and only one of the following are true: a is in P, -a is in P, a = 0

The third condition is called the law of trichotomy. -a is defined to be the additive inverse of a and 0 is defined to be the additive identity. We define the order relation > by

a > b if and only if a - b is an element of P.

Now, before proving this, I have to start with a lemma. I need to prove that if a is a non-zero real number, then a^2 > 0 (meaning it's in the positive set).

Suppose a is a non-zero real number. By the law of trichotomy, either a > 0 or a < 0. If a > 0, then by closure of multiplication in the positive set, a^2 > 0. If a < 0, then -a > 0. So,

a^2 = aa = (-a)(-a) = (-a)^2, which is in the positive set, again by closure. Thus, a^2 > 0 in all cases where a is not equal to zero.

Note that I didn't justify the step aa = (-a)(-a). This can be proven as well, but this is taking long enough as it is, so I'll let you figure it out (it's not that hard if you use the distributivity axiom). Now, onto the proof.

We first want to show that 1 is not equal to zero (bear in mind that 1 is defined to be the multiplicative identity). We'll do this by assuming the opposite and arriving at a contradiction. Suppose that 1 = 0. If a is a non-zero real number, we have a = a*1 = a*0 = 0, which is a contradiction (because we assumed a was non-zero). Note that the step a*0 = 0 was also unjustified. Again, this isn't difficult to prove using the field axioms, but I'd rather not get any more off track.

So, since 1 is not equal to zero, then the law of trichotomy says that either 1 > 0 or 1 < 0. Now, by the definition of multiplicative identity, 1^2 = 1*1 = 1. However, our lemma says that given any non-zero real number a, a^2 > 0. Thus, 1 > 0. hee hee hee