IIT JEE , AIEEE Forum - Mathematics , Algebra - Open Challenge to all Experts and Math-Wiz kids..

GoNik

WELL THATS VERRY NICE RAHUL..U MISSED ANOTHER ONE...BUT GUD I RATED U...the only solutions are x = 12, y = ±36...u and spidey both got atleast 2 each and seem to b working real hard....gud keep it up..

spideyunlimited

Q5. If x is a positive rational number, show that x^x is irrational unless x is an integer.

If x is rational, it can be expressed as x = p/q where p,q are integers and pq > 0

Now x^x = (p/q)^(p/q)
This can only be a rational number if the denominator of the exponent is 1.
This would mean that the number x = p/1
which indicates that x is an integer.

GoNik

thats nice @spidey...verry gud..u got 3 answers..rated u ...the answer given is as follws..u made it simpler..
_________
This question is amenable to a reductio ad absurdum proof.

Assume x is rational but not an integer; that is, x can be written as a/b, irreducible, with b > 1. Assume (a/b)a/b = c/d is irreducible.

Raising each side to the power b we get (a/b)^a = (c/d)^b.
Hence (a^a)*( d^b) = (c^b)*(b^a).

Now we use the Fundamental Theorem of Arithmetic, which states that every integer greater than 1 can be written uniquely as a product of finitely many prime numbers.

Since b > 1, it will have at least one prime factor, p > 1. Consider the number of times p occurs in the prime factorization of each of the above terms. Letting b = (p^r)*u, where p and u are relatively prime, then:

* b^a: ra times
* a^a: 0 times, since a and b are relatively prime
* d^b: sb times, where s is the number of times p occurs in d (since p occurs on the right-hand side, it must also occur on the left-hand side, so it must be a factor of d)
* c^b: 0 times, since c and d are relatively prime

By the Fundamental Theorem of Arithmetic, ra = sb. But since a and b are relatively prime, b must divide r.
That is, r >= b >= p^r, which is absurd for p > 1.

This completes the reductio ad absurdum proof.
Hence (a/b)^a/b is irrational.

Of course, if b = 1, then x is an integer for any integer a, and x^x is rational.

Therefore, if x is a positive rational number, xx is irrational unless x is an integer.
***********
Remark:
The above result, in conjunction with the Gelfond-Schneider Theorem, can be used to show that the positive real root of x^x = 2 is a transcendental number.

Clearly x^x = 2 does not have an integer solution. Hence, by the above result, x cannot be rational.

Now, using the Gelfond-Schneider Theorem, if x is algebraic and irrational, x^x is transcendental, and so cannot be equal to 2.

Therefore the positive real root of x^x = 2 is transcendental. The root is approximately equal to 1.559610469462369349970388768765

GoNik

in question no. 2 ........x*y means x multiplied to y....equivalent to xy

spideyunlimited

I'm working on 2, g2g.

GoNik

SOLUTION TO QUESTION NO. 3

1/x + 1/y = ?1 --------(1)
x^3 + y^3 = 4 --------(2)

(1) implies x + y = ?xy
(2) implies (x + y)^3 ? 3xy(x + y) = 4
Hence ?(xy)^3 + 3(xy)2 ? 4 = 0

By inspection, xy = ?1 is a solution of this cubic equation.
Factorizing, we have (xy + 1)(xy ? 2)2 = 0.
Hence xy = ?1, x + y = 1, or xy = 2, x + y = ?2.

If xy = ?1 and x + y = 1, then x, y are roots of the quadratic equation
u^2 ? u ? 1 = 0.
(Consider the sum and product of the roots of

(u ? A)(u ? B) = u^2 ? (A + B)u + AB = 0.)

Hence u = [1 ± root(5)]/2.

If xy = 2 and x + y = ?2, then x, y are roots of u^2 + 2u + 2 = 0.
This has complex roots: u = ?1 ± i.

Therefore the real solutions are x = [1 ± root(5)]/2, y = [1 -+ root (5)]/2.
____________________
CHEERS!!!!!!!!!!!!!!!!!

GoNik

SOLUTION TO QUESTION NO. 6

The fallacy lies in ignoring the fact that the number of x's being added is not constant. Not only is x changing; the number of x's is also changing.

Another potential objection is that, since the function is defined only when x is an integer, it is not continuous, and therefore not differentiable. A function is said to be differentiable at a point if its derivative exists at that point.

For example, if x = 2.4, we write f(x) = x + x + 0.4x. If x = 2.5, we write f(x) = x + x + 0.5x, and so on.

Now, having restored continuity, we can again pose the question: why is the derivative of this function at x = 2.4 not equal to 1 + 1 + 0.4? The reason, as indicated above, is that we are ignoring the fact that the number of x's being added is also changing.

To make this even clearer, consider that the above extension is equivalent to the following definition: f(x) at x = 2.4 is defined as 2.4x, at x = 2.5 it is defined as 2.5x, and so on. In other words, at x = a, f(x) = a·a.

The derivative at x = a is defined as the limit, as h tends to zero, of
[f(a+h) ? f(a)] / h.
The fallacy lies in writing f(a+h) as a· (a+h). (From which f'(a) = a.)
The correct formulation is:
f(a+h) = (a+h)· (a+h).
Expanding, we find
f'(a) = 2a, as expected!!!!!!
CHEEERS!!!!!!!!!!!!!

GoNik

SOLUTION TO QUESTION NO. 17

We first note that the number (2^58 + 1)/5 is in fact an integer.
This may be seen by writing

2^58 = 2^2 · (2^4)^14 congruent to 4 .(1^14) congruent to 4(modulo 5)

Hence (2^58) + 1 congruent to 0 (mod 5).

(Alternatively, consider that 5 = 4 + 1 is a factor of (4^29) + 1 = (2^58) + 1, since (x + 1) is a factor of (x^n + 1) when n is odd.)

Now we factorize 2^58 + 1.

Note that (2^29 + 1)^2 = (2^58 + 1) + 2^30, and so 2^58 + 1 can be written as a difference of two squares:
2^58 + 1 = (2^29 + 1)^2 ? (2^15)^2 = (2^29 + 2^15 + 1)(2^29 ? 2^15 + 1).
Clearly, both factors are greater than 5, and so 2^58 + 1 = 5ab, where a and b are integers greater than 1.

Therefore, the number (2^58 + 1)/5 is composite.

CHEERS!!!!!!!!!!!!!!

GoNik

i gave only those solutions that have been already solved by spidey and rahul.....this is jus bcoz i promised to giv it....4 those who cudnt understand them(i hope theres noone, who dint understand spidey and rahul's work coz they made it look quite simpl than these solns).....cheers to them....

GoNik

and yes not only students shud try it but i think it can b tried by any expert.......ques 2 and finding digits abcd etc etc.questions are gud....plzzzz

rahul1993

ans 1)

i am not sure of the answer

I could only find h and g

first we count no of 5 as no of 5= no of zeroes at the end

here we get 8 fives

thus h=0

now we find the last digit 2³⁰ as 37! contains 30 twoes.

it comes out to be 4 hence g=4

h=0 , g=4

PLS CORRECT IF WRONG

ajit123

Question No. 2

I Think that answer to this question can be any 2 numbers within the conditions that are maintained in the question because S and P are related as no. x and y by the property: x+y and x*y of which the individual values are provided to them

ajit123

Answer : 7

Is the no. Prime. because it is neither divisible by 3,7,9,11,. But for this i used calculation without calculator and paper and pen. Is it valid.

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