To do this with only the field axioms I found that it can be proven like this: I'm using associativity and commutativity axioms freely without showing explicitly. Other than that, each step should be the application of one field axiom. Lemma 1: x + y = x => y = 0 where x,y in R. pf. x + y = x => (-x) + x + y = (-x) + x => 0 + y = 0 => y = 0. QED Lemma 2: x0 = 0 where x in R. pf. x + x0 = x(1) + x0 = x(1 + 0) = x(1) = x. Hence by lemma 1, x0 = 0. QED Lemma 3: -(-x) = x where x is in R. pf. x + (-x) = 0. Plugging -x in for x, we have (-x) + (-(-x)) = 0 => x + (-x) + (-(-x)) = 0 + x => 0 + (-(-x)) = x => -(-x) = x. QED Lemma 4 : -(xy) = (-x)y where x, y in R. pf. xy + (-(xy)) = 0 => (-x)y + xy + (-(xy)) = (-x)y + 0 => (-x + x)y + (-(xy)) = (-x)y => 0y + (-(xy)) = (-x)y and by lemma 2, 0y = 0. Hence, we have 0 + (-(xy)) = (-x)y => -(xy) = (-x)y. QED Theorem 1: (-1)x = -x for x in R. pf. (-1)x + (-((-1)x)) = 0 => (-1)x + ((-(-1))x) = 0 by lemma 4 and (-(-1))x = (1)x by lemma 3 and (1)x = x. Hence, (-1)x + ((-(-1))x) = 0 => (-1)x + x = 0 = > (-1)x + x + (-x) = 0 + (-x) => (-1)x + 0 = -x => (-1)x = -x. QED Corollary 1: (-1)(-1) = 1. pf. (-1)(-1) = -(-1) by Theorem 1 and -(-1) = 1 by lemma 3. Hence (-1)(-1) = -(-1) = 1. QED Let me know if you find mistakes or have a quicker route only using field axioms.
Yeah it's beautiful and there is no mistake....To approach to the prove you pick really suitable lemma's using the field axioms. Thank you for your comment 🥰 BTW I don't have yet any route using field axioms for this question.
No, See since it is the theorem by which we are going to prove. Here if we consider a=-1 then -a will belongs to positive real number which is true because of trichotomy property. But according that theorem we can't conclude about (-1)(-1)= 1 cuz we don't know about -(-1) is equals to what till yet ...So that's why we use another theorem there which define -(-1) will be equals to what so that's not beg the question.
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To do this with only the field axioms I found that it can be proven like this:
I'm using associativity and commutativity axioms freely without showing explicitly. Other than that, each step should be the application of one field axiom.
Lemma 1: x + y = x => y = 0 where x,y in R.
pf. x + y = x => (-x) + x + y = (-x) + x => 0 + y = 0 => y = 0. QED
Lemma 2: x0 = 0 where x in R.
pf. x + x0 = x(1) + x0 = x(1 + 0) = x(1) = x. Hence by lemma 1, x0 = 0. QED
Lemma 3: -(-x) = x where x is in R.
pf. x + (-x) = 0. Plugging -x in for x, we have (-x) + (-(-x)) = 0 => x + (-x) + (-(-x)) = 0 + x => 0 + (-(-x)) = x => -(-x) = x. QED
Lemma 4 : -(xy) = (-x)y where x, y in R.
pf. xy + (-(xy)) = 0 => (-x)y + xy + (-(xy)) = (-x)y + 0 => (-x + x)y + (-(xy)) = (-x)y => 0y + (-(xy)) = (-x)y and by lemma 2, 0y = 0. Hence, we have 0 + (-(xy)) = (-x)y => -(xy) = (-x)y. QED
Theorem 1: (-1)x = -x for x in R.
pf. (-1)x + (-((-1)x)) = 0 => (-1)x + ((-(-1))x) = 0 by lemma 4 and (-(-1))x = (1)x by lemma 3 and (1)x = x. Hence, (-1)x + ((-(-1))x) = 0 => (-1)x + x = 0 = > (-1)x + x + (-x) = 0 + (-x) => (-1)x + 0 = -x => (-1)x = -x. QED
Corollary 1: (-1)(-1) = 1.
pf. (-1)(-1) = -(-1) by Theorem 1 and -(-1) = 1 by lemma 3. Hence (-1)(-1) = -(-1) = 1. QED
Let me know if you find mistakes or have a quicker route only using field axioms.
Yeah it's beautiful and there is no mistake....To approach to the prove you pick really suitable lemma's using the field axioms.
Thank you for your comment 🥰
BTW I don't have yet any route using field axioms for this question.
Awesome thank you
Pls make videos on Sir Srinivasa Ramanujan's work🙏🙏
Yes offcourse I will....
Stay tuned with us..🤝
Is 5:05 not begging the question? Can you explain this step a little more?
No, See since it is the theorem by which we are going to prove. Here if we consider a=-1 then -a will belongs to positive real number which is true because of trichotomy property. But according that theorem we can't conclude about (-1)(-1)= 1 cuz we don't know about -(-1) is equals to what till yet ...So that's why we use another theorem there which define -(-1) will be equals to what so that's not beg the question.