- #1
issacnewton
- 1,000
- 29
Hi
I am trying to prove that 1 > 0. I am going to assume a contradiction .
Assume [tex] 1 \leqslant 0[/tex]
First consider 1=0, Let [itex]a \in \mathbb{R}[/itex] be arbitrary. So
[tex] a=1.a = 0.a = 0 [/tex] using field axioms. Actually 0.a=0 is not field axiom , but I have
proved it separately .
that means [tex] \forall \;\; a\in \mathbb{R} \Rightarrow a=0 [/tex]
since all numbers are equal to 0 , no inverse exists in the set R. So R is not a field. Hence
the contradiction. So
[tex] 1 \neq 0 [/tex]
that means we have to consider the second option
[tex] 1 < 0 [/tex]
Consider [itex] a > 0 \backepsilon a \in R [/itex]
Now I have already proved another theorem.
If q,b,c are in R and q < b , c < 0 then qc > bc
letting c =1 and q = 0 , b= a we can say that
(0)(1) > (a)(1) which means 0 > a or a < 0 but this is contradiction since
we assumed that a> 0 . So our assumption that 1 < 0 , is wrong.
Hence 1 > 0
is my proof right ?
I am trying to prove that 1 > 0. I am going to assume a contradiction .
Assume [tex] 1 \leqslant 0[/tex]
First consider 1=0, Let [itex]a \in \mathbb{R}[/itex] be arbitrary. So
[tex] a=1.a = 0.a = 0 [/tex] using field axioms. Actually 0.a=0 is not field axiom , but I have
proved it separately .
that means [tex] \forall \;\; a\in \mathbb{R} \Rightarrow a=0 [/tex]
since all numbers are equal to 0 , no inverse exists in the set R. So R is not a field. Hence
the contradiction. So
[tex] 1 \neq 0 [/tex]
that means we have to consider the second option
[tex] 1 < 0 [/tex]
Consider [itex] a > 0 \backepsilon a \in R [/itex]
Now I have already proved another theorem.
If q,b,c are in R and q < b , c < 0 then qc > bc
letting c =1 and q = 0 , b= a we can say that
(0)(1) > (a)(1) which means 0 > a or a < 0 but this is contradiction since
we assumed that a> 0 . So our assumption that 1 < 0 , is wrong.
Hence 1 > 0
is my proof right ?