Just join in for the fun.
Why is 0/0 not equal to 1 like 2/2 = 1 or y/y = 1 ?
The answer is indeterminate form (L'Hospital Rule)
Why 0^0 not equal to 1 like 2^0 = 1 or a^0 = 1 ?
I have googled it but I cannot find any answer for it.
Anyone knows of any maths theory that will give an answer for 0^0.
Thanks.
Originally posted by Lee012lee:Just join in for the fun.
Why is 0/0 not equal to 1 like 2/2 = 1 or y/y = 1 ?
The answer is indeterminate form (L'Hospital Rule)
Why 0^0 not equal to 1 like 2^0 = 1 or a^0 = 1 ?
I have googled it but I cannot find any answer for it.
Anyone knows of any maths theory that will give an answer for 0^0.
Thanks.
0 to any power except 0 is equal to 0.
By mathematical induction, 0 to the power of 0 is also equal to 0.
Originally posted by skythewood:0 to any power except 0 is equal to 0.
By mathematical induction, 0 to the power of 0 is also equal to 0.
might as well say...
any number except 0 to the power of 0 is equal to1
By mathematical induction, 0 to the power of 0 is also equal to1
-.-"
Originally posted by greengoblin:Prove or disprove that 111...111 (n times) where n is a positive integer > 1 has an even number of positive divisors.
hey...i want to try. haha.but dont know whether correct, because the solution i thought of seems a bit stupid to me-.-
first you factorise out the 111...111(n times) into its prime factors.
then you take out a single digit existing in its prime factor....eg if the prime factors end up to be 3^2 x 5^7 x 11^6 then you take out any one digit you like. for instance then i take out....the digit 3 => 3 x (3 x 5^7 x 11^6)
then, to find no. of divisors you simply take the (power + 1) then multiply all the powers together
so if the prime factors is of the form (k1^1) [(k1^n1)(k2^n2)....(Kx^nx)] (this is after u take out 1 digit already, assuming k1^1), then your number of divisors is 2(n1+1)(n2+1)...(nx + 1)= even number, since there is a multiplier of 2 at the start already.
Originally posted by Lee012lee:Just join in for the fun.
Why is 0/0 not equal to 1 like 2/2 = 1 or y/y = 1 ?
The answer is indeterminate form (L'Hospital Rule)
Why 0^0 not equal to 1 like 2^0 = 1 or a^0 = 1 ?
I have googled it but I cannot find any answer for it.
Anyone knows of any maths theory that will give an answer for 0^0.
Thanks.
ask yourself this question. i share 0 apple with 0 person.
Originally posted by freedom4ever:ask yourself this question. i share 0 apple with 0 person.
The question of 0/0 looks simple but the answer is not so straightforward.
The answer for 0/0 is not 0 or 1.
2 apples shared by 2 persons, so the answer is 1 apple per person
0 apple shared by 0 person, the answer is NOT 0 apple per person
The answer to the question requires the knowledge of limit theory and L' Hospital Rule. The answer for 0/0 is indeterminate form ie the answer will depend on the given function.
Thanks for the useful link to the answer for the question of 0^0.
Originally posted by absol: hey...i want to try. haha.but dont know whether correct, because the solution i thought of seems a bit stupid to me-.-first you factorise out the 111...111(n times) into its prime factors.
then you take out a single digit existing in its prime factor....eg if the prime factors end up to be 3^2 x 5^7 x 11^6 then you take out any one digit you like. for instance then i take out....the digit 3 => 3 x (3 x 5^7 x 11^6)
then, to find no. of divisors you simply take the (power + 1) then multiply all the powers together
so if the prime factors is of the form (k1^1) [(k1^n1)(k2^n2)....(Kx^nx)] (this is after u take out 1 digit already, assuming k1^1), then your number of divisors is 2(n1+1)(n2+1)...(nx + 1)= even number, since there is a multiplier of 2 at the start already.
Your method will work but you will need to tweat a little. For the general case, as pointed out by Rudin, you just need to prove that 11...111 can never be a perfect square and hence using your method above, it can never have an odd number of divisors :) Good try!
An extention of the problem will be to consider kkk...kkk where k is an integer between 1 to 9 inclusive. I leave it as a open problem for the interested solver.
My answers
1. 11^11, dun anyhow cock 11!^11! ( i never say can use !) or infinite
2. One pound of feathers......
Originally posted by SBS n SMRT:My answers
1. 11^11, dun anyhow cock 11!^11! ( i never say can use !) or infinite
2. One pound of feathers......
You never say cannot use ! also