1. ## 1>0 proof

I am trying to write this proof for class. I am trying to prove by contradiction so I try to prove that 0 is greater than or equal to 1.
"Case 1" gives 1>0 which is a contradiction.
"Case 2" is 1=0, which I can't disprove. Anyone have a proof for that? I've tried a=1 and playing with the algebra (this seems like the right track, though I'm not sure), but I can't get anywhere from there.
I have a test tomorrow, so either a proof that 1 doesn't equal 0 or a big hint would be greatly appreciated.

2. Originally Posted by reilyj
I am trying to write this proof for class. I am trying to prove by contradiction so I try to prove that 0 is greater than or equal to 1.
"Case 1" gives 1>0 which is a contradiction.
"Case 2" is 1=0, which I can't disprove. Anyone have a proof for that? I've tried a=1 and playing with the algebra (this seems like the right track, though I'm not sure), but I can't get anywhere from there.
I have a test tomorrow, so either a proof that 1 doesn't equal 0 or a big hint would be greatly appreciated.
to prove something like this, you need to rely on axioms. what axioms are you using?

you can use the axiomatic definitions of 1 and 0 respectively to prove they are not equal

and if 1 < 0 that would mean 1 is negative. what do the axioms regarding "1" have to say against this?

3. i can use the associative laws, the communitative laws, and the distributive law.

4. Originally Posted by reilyj
i can use the associative laws, the communitative laws, and the distributive law.
those won't help too much here.

usually the axioms regarding 1 and 0 are:

1*x = x for all x

x + 0 = x for all x

respectively

5. yeah I can use those

6. Originally Posted by reilyj
yeah I can use those
ok, so use those.

disproving 1 < 0 is easy.

assume 1 is negative. then 1*1 = 1 by the afore mentioned axiom. but that means 1^2 is negative, clearly a contradiction, as a negative number times a negative one should yield a positive number.

now, try to use the axioms to disprove 1 = 0

7. cool, thanks for the help buddy I got it. Sorry about the double post. catch you later.

8. i seen this question a few times, you can do a contradiction...
Suppose 1<=0, case 1. 1=o, then 0<=0, which holds, now consider 1not=0, if you are using Kirdwood books is theorem 1-7e) that states anot+=0 then a^2>0, so 1^2<=0, contradiction to the theorem....

9. And both of those proofs involve axioms reilyj did not mention:

If a> 0 and b> 0 then ab> 0

If a> 0 and b> 0 then a+ b> 0

If a and b are two numbers then one and only one of these must hold:
i) a= b
ii) a> b
iii) b> a