2011/029) can x^y be rational for both irrational x and y

we know that sum and product of irrational x and y can be rational

(a + sqrt(b)) + (a- sqrt(b))= 2a

and product = a^2 - b^2

so taking a and b rational such that sqrt(b) is irrational we can get easily


but for power

say x = sqrt(2) and y = sqrt(2)


is x^y rational

we do noy know

but if it is then we are through

and if it is not then (x^y)^y = 2 whhich is rational

hence x^y can be rational for both irrational x and y

2011/028) to prove that sqrt(2) is irrational

before proving it we need to prove a lemma

lemma: if 3 / (a^2+ b^2) then 3 | a and 3| b

proof:

we have a mod 3 = 0 or 1 or 2

so a^2 mod 3 = 0 or 1
and b^2 mod 3 = 0 or 1

so a^2 + b^2 mod 3 = 0 if a mod 3 = 0 and b mod 3 = 0
or 1 or 2 otherwise

hence lemma is proved

now for the proof of sqrt(2) is irrational

let sqrt(2) = a/b where GCD(a,b) = 1 ( we can always reduce it in this form)

so a^2 = 2b^2 or a^2 + b^2 = 3b^2

so 3 / (a^2+b^2) or hence as per above lemma 3 | a and 3 | b

so GCD(a,b) is not 1 so a contradiction

so sqrt(2) is irrational

2011/027) obtain the sum of

1/(x+1) + 2/(x^2+1) + 4/(x^4+1) + (2^2n/(x^2n + 1)

we realise that 1/(x+1) - 1/(x-1) = -2 /(x^2-1)

so add and subtract 1/(x-1) to get

1/(x-1) + (1/(x+1) - 1/(x- 1) + 2/(x^2+1) + 4/(x^4+1) +.... (2^2n/(x^2n + 1)))
= 1/(x-1) + (-2/(x^2-1)+ 2/(x^2+1) + 4/(x^4+1) + (2^2n/(x^2n + 1)))
= 1/(x-1) + (-4/(x^4-1)+ ... + (2^2n/(x^2n + 1)))

applying repeatedly we get 2^(4n)/(x^4n+1) + 1/(x-1)

2011/026) Factor by^2-b^2y-ay^2+a^2y+ab^2-ba^2

rewrite in decreasing power of any one say y and then proceed

by^2-b^2y-ay^2+a^2y+ab^2-ba^2

= by^2-ay^2- b^2y+a^2y+ab^2-ba^2
= y^2(b-a) +y (a^2-b^2) + ab(b-a)
= (b-a) (y^2 + y(a-b)(a+b) + ab)
= (b-a) (y^-y(a+b) + ab)
= (b-a) (y-a)(y-b)

thus it becomes easy

2011/025) solve x = a/(b+c) = b/(c+a) = c/(a+b)

x cannot be zero as if x= 0 then a = b =c so a/(b+c) = b/(c+a) = c/(a+b) are indeterminate form

so we take reciprocals

1/x = (b+c)/ a= (c+a)/b = (a+b)/c
add 1 on all 2 sides

1/x +1 = (a+b+c)/a = (a+b+c)/b = (a+b+c)/ c

so a+b+c = 0 = 1/x + 1 or a= b = c

a+b+c = 0 = 1/x + 1 => x= -1 => b+ c = - a,if a = 1 then b= k and c = -1 - k

this satisfies all 1/x = (b+c)/ a= (c+a)/b = (a+b)/c
add 1 on all 2 sides


if a=b= c then we get x = 1/2 so solution (x=1/2, a= k. b= k,c =k ) for any k

if x = -1 then (x= -1, a = 1 , b= k, c = - 1 -k) or any permutation or any multiple of the same.

2011/024) if p and p^2 + 8 are both primes then prove p^3+4 is a prime

proof:
p is either 2 or 3 or of the form (6n+/-1)

if p is of the form (6n+/-1)

then p^2 + 8 = 36n^2-12n + 9 = 3 ( 12n^2 - 4n + 3) divisible by 3 and ( 12n^2 - 4n + 3) > 1

so not a prime

if p = 2 then p^2 + 8 is even and not a prime

so only p =3 is left and p^2+ 8 = 17 is a prime and p =3 is the only number satisfies the criteria and p^3+ 4 = 31 is a prime as well

2011/023) the number of real solutions of the equation

sin e^x = 5^x + 1/5^x

is

a) 0
b) 1
c) 2
d) many

ans:

we know sin e^x is between -1 and + 1

and
5^x + 1/5^x is cannot lie between - 2 and 2

so zero solutions

2011/022) To show that 1 = 1/2+1/4+ 1/8 + 1/16+ ….

We can prove from the RHS that

1/2+1/4+ 1/8 + 1/16+ .. is a GP with a= 1/2 and r = 1/2 and the GP converges to

1/2/(1-/12) = 1/2/(1/2) = 1

but beauty lies in expanding from LHS

1 = 1/2 + 1/2
= 1/2 + 1/4 + 1/4 ( take the last term and make into 2 halves)
= 1/2 + 1/4 + 1/8 + 1/8 and so on to get the result

2011/021) Pythagorean triplet with Gaussian Integer

Here we discuss the Pyhagorean triplet using Gaussian integers

As I have discussed in the previous blog entry about Gaussian integers that Gaussian integers are complex number of the form α = u + vi, where u and v are ordinary integers d i is the square root of minus one.

A primitive Pythagorean triple is one in which a and b are coprimes

That is integers (a,b,c) form a primitive Pythagorean triple where a^2+b^2= c^2 and

a and b are coprime

as a and b are co primes both cannot be even and only one can be odd. Both cannot be odd as

a mod 2 = 1 and b mod 2 = 1 shall give a^2 + b^2 mod 4 = 2 which is not a perfect square.

Now c^2 = (a^2+ b^2) = (a+ib)(a-ib)

Now both factors a + ib and a-ib cannot share any common factor that is they most be coprime to each other

So both of them must be perfect squares
Let
(a+ib) = (u+iv)^2 (and so (a-ib) = (u-iv)^2)

squaring RHS we get

a = u^2-v^2
b= 2uv

and c = (u+iv)(u-iv) = (u^2+v^2)

so we get

(u^2-v^2, 2uv, u^2+v^2) as Pythagorean triplet