rpenner
13th August 2011 - 05:32 PM
QUOTE (wzfa+Aug 13 2011, 01:45 AM)
所说的那些数都大于0
幂是n=2,式子x2^2-x1^2=c',x1+L=x2,让L成为有理数
c'=(x1+L)^2-x1^2=x1^2+2*x1*L+L^2-x1^2 =2*x1*L+L^2.
可见当c'为有理数时候x1必须是有理数
i*(x2^2-x1^2)=c'也有当c'为有理数时候x1必须是有理数这个性质.i是任意有理数
可见c'为无理数时x1为无理数
n=3式子为x2^3-x1^3=c,其导数是n=2的c',x1的个数一样多,因此c'和c的个数一样多,当c为任意无理数,可知道x1为无理数,由于c的个数和c'相同，那么x1就可以遍历所有(任意)无理数。这样任意有理数c就只能对应有理数x1了
[Ed: This is an annotated translation/interpretation ]
Working in the natural numbers greater than zero.
When the power, n, is 2 we state:
x₂² − x₁² = c'
x₁ + L = x₂ (L is rational)
[Ed: Note we immediately have x₂² − x₁² = (x₂ + x₁)(x₂ − x₁) = (x₂ + x₁) L = c' ]
c' = (x₁ + L)² − x₁² = x₁² + 2 x₁ L + L² − x₁² = 2 x₁ L + L²
[Ed: Note that c' = 2 x₁ L + L² = (2 x₁ + L) L = (2 x₁ + x₂ − x₁) L = (x₂ + x₁) L ]
So when x₁ is rational, so is c'.
[Ed. It is better to say that when any two of x₁ , x₂ and L are rational, then the remaining third variable and c' are rational. And when any two of x₁ , x₂ and L are integers, then both the third variable and c' are integers and L , if nor zero, evenly divides c'. When both x₁ and x₂ are natural numbers greater than zero, both L and c' are integers, but they both might be zero or negative.]
If y is another rational number, then (x₂² − x₁²)y is also rational.
[Ed. It is bad practice to use e or i as variables due to the use of these symbols in analysis and complex numbers. Frequently i will be used as an index variable in contexts where no confusion can arise.]
This shows that c' is an irrational number when x₁ is. [Ed. This is not a worthy statement since the first equation puts x₂ and x₁ as peers and the late-comer L is distinguished as rational. It is a trivial statement.]
The case when n=3 gives the formula x₂³ − x₁³ = c which allows us to derive similar statements about the rationality of c as we did c'
[Ed: c = x₂³ − x₁³ = ( x₂² + x₁ x₂ + x₁²)(x₂ − x₁) = ( x₂² + x₁ x₂ + x₁²)L ]
------------
Does it???
x₁ = (−3 + √57)/6
L = 1
x₂ = x₁ + L = ( 3 + √57)/6
c' = x₂² − x₁² = [ ( 3 + √57)/6]² − [(−3 + √57)/6]² = [ ( 11 + √57)/6] − [(11 − √57)/6] = √57/3
c = x₂³ − x₁³ = [( 3 + √57)/6]³ − [(−3 + √57)/6]³ = [(45 + 7√57)/18] − [(−45 + + 7√57)/18] = 90/18 = 5
And this is not the only counter example.
Pick c and L rational (don't pick L = 0).
Then
c/L is rational
c/L = x₂² + x₁ x₂ + x₁² = (x₁ + L)² + (x₁ + L)x₁ + x₁² = L² + 3 L x₁ + 3 x₁²
so the two solutions are:
x₁ = (±√3 √(4 c L − L⁴) − 3 L²)/(6 L)
which gives
x₂ = x₁ + 6 L² / (6 L) = (±√3 √(4 c L − L⁴) + 3 L²)/(6 L)
x₂³ = c/2 ± (c √3√(4 c L − L⁴))/(18 L²) ± (L √3√(4 c L − L⁴))/9
x₁³ = −c/2 ± (c √3√(4 c L − L⁴))/(18 L²) ± (L √3√(4 c L − L⁴))/9
x₂³ − x₁³ = c
wzfa
27th August 2011 - 08:27 AM
you are right, I am wrong, thank you for your good english and chinese
kowalskil
27th August 2011 - 03:33 PM
QUOTE (wzfa+Aug 27 2011, 08:27 AM)
you are right, I am wrong, thank you for your good english and chinese
Not very clear. Hahaha.
.
printingray
29th August 2011 - 04:13 PM
nice important post like it thanks for it.keep it up.
[Moderator: No spam, please.]
wzfa
8th September 2011 - 05:56 AM
I have another simple idea than the one above, but the n is much less than the Fermat Little Theorem. I have to check some days, to avoid loosing face again
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