Recent content by pankaz712

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    Polynomial divided by another polynomial

    If the polynomial x^4-16x^2-25x+10 is divided by another polynomial x^2-2x+k, the remainder comes out to be x+a find k and a. My approach: I tried dividing the first polynomial by the second and then equating the remainder to (x+a). It didn't work out. Are my calculations wrong or am i on the...
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    Proving Root n is Irrational: Perfect Square Affects Proof

    Yes...i believe u are right. I will surely check again and see. Please could u explain your method of proving it.
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    Proving Root n is Irrational: Perfect Square Affects Proof

    yes i see that not all composite numbers can be split into the products of only 2 primes. But the proof remains the same even if when the number can be split into " n" number of primes.
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    Proving Root n is Irrational: Perfect Square Affects Proof

    OK here is how i approached it: n can either be a prime number or a composite 1) if n is prime: a) assume root n is rational. therefore root n = p/q( where p, q are co-prime) then n= p^2/q^2, p^2= n *q^2 n divides p^2, therefore n divides p. Now for some integer k, p=nk...
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    Proving Root n is Irrational: Perfect Square Affects Proof

    How to prove that root n is irrational, if n is not a perfect square. Also, if n is a perfect square then how does it affect the proof.
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    Cube root of 6 is irrational. (please check if my proof is correct)

    thnx i understood ur method. I have just one more question: How to prove that root n is irrational, if n is not a perfect square. Also, if n is a perfect square then how does it affect the proof.
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    Cube root of 6 is irrational. (please check if my proof is correct)

    i did that because i thought Euclid's Lemma was true only for prime numbers.
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    Cube root of 6 is irrational. (please check if my proof is correct)

    Prove that cube root of 6 irrational. Solution: I am trying to prove by contradiction. Assume cube root 6 is rational. Then let cube root 6 = a/b ( a & b are co-prime and b not = 0) Cubing both sides : 6=a^3/b^3 a^3 = 6b^3 a^3 =...
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