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Apologies. If you substitute in (+ 3) you only prove part of the theorem, namely that for all elements there exists a unique such that . Let be any other element, so there exists a unique such that . We need to show that .Now
Add
to the first equation and to the second equation. By using a combination of the commutative and associative axioms for addition, you should getBy uniqueness,
as required.
This is a nice proof. Thank you!
Hi guys,
Suppose following axiomatization of real numbers:
(+ 1)
(. 1)
(+ .)
(< 1) forall
forall , one of the following holds: , ,(sup) Let
be non-empty set, s.t. it has upper bound. ThenNow following theorem holds: There is precisely one element (which will be denoted
), which is solution of equation .Now proof is straightforward and I will not finish it. My problem is that
is formal symbol of our first-order theory (i.e. it is non-logical constant) and no axiom define any property that should obbey. However the first step of the theorem says something like . My question is why is this correct? I don't see any inference rule which interlinks and axiom (+ 3).any ideas why is that correct? thanks
Thanks guys. I have nice proof now (Im not an author):
From 1* we have kc = a+b < c+c. So k < 2 and k must be 1.
Next from 2* we have that l > 1 (otherwise a = 0). 2a + b = lb < 3b. So l < 3 and l must be 2.
From 2* now we have b = 2a. Finally from a+b = c we have a + 2a = c.
hi jozou
i have found an equation which links k,l and m,but i am not sure if it is true:
if you can find any numbers that satisfy this equation than you can probably find a,b and c by solving a system of equations given by the first three equations you posted.
Can you please expand, how you got
?Find all integers of the form 1 ≤ a < b < c, s.t.:
1* ∃k a+b = kc
2* ∃l a+c = lb
3* ∃m b+c = ma
Attempt: Well, I have answer that solution is of the form a, 2a, 3a. I can prove that if b = 2a, then c = 3a:
From 2* we have c = a(2l - 1)
Then from 1* and previous we have 3a = ka(2l - 1). So k = 1 or k = 3:
If k = 3, then
1 = 2l - 1, l = 1, a +c = 1.b = 2a what contradicts assumption that a < c.
So k = 1. Then a + b = 3a = 1.c
But I can't prove that b=2a. Any hints? Thanks!
hi jouzo
If a > 1, then 4c ≤ a.b.c = a+b+c. So 3c ≤ a+b, what contradicts a ≤ b ≤ c.
Don't understand where the 4c bit came from. Can you expand this please?
Bob
If a > 1, then a ≥ 2. Next b ≥ a, so b ≥ 2. Hence a.b ≥ 4.
So a+b+c = a.b.c ≥ 4c.
One guy found really nice proof:
Let a ≤ b ≤ c be positive integers, s.t. equation a+b+c = a.b.c holds.
If a > 1, then 4c ≤ a.b.c = a+b+c. So 3c ≤ a+b, what contradicts a ≤ b ≤ c.
So a= 1. Now we have to solve 1+b+c = b.c. Then b.(c-1) = 1+b+c - b = 1 + c, (b-1).(c-1) = 1 + c - (c-1) = 2.
So (b-1).(c-1) = 2. Hence the only solution satistying assumptions is b = 2, c = 3.
Hi guys,
find all solutions of equation a + b + c = a . b . c, where a ≤ b ≤ c are positive integers.
Well, by trial and error method I found out that 1+2+3 = 1.2.3
Probably there is no other solution, but how can I proove that? I tried case analysis, but unsuccessfully. Any idea? Thanks!
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