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Homework Help: Plzz check the answers to these difficult questions?

  1. Apr 13, 2006 #1
    A book of mine has the following questions with no answers.So I am not sure if I am right or if the proceure is correct and the best.So,plz help-

    1.For a given real number a>0,define[tex] A_n=(1^a + 2^a + 3^a +....+n^a)^n[/tex]
    and[tex] B_n=n^n(n!)^a[/tex] for all n=1,2,.... Then
    a. [tex]A_n<B_n[/tex]for all n>1
    b.there exists an integer n>1 such that [tex]A_n<B_n[/tex]
    c.[tex]A_n>B_n[/tex] for all n>1
    d.there exists integers n and m both larger than one such that [tex]A_n>B_n[/tex] and [tex]A_m<B_m[/tex]

    As there's no specification about the value of a in the options given,I considered a special case taking a=1 and then put n=2 and n=3.I found [tex]A_n>B_n[/tex] and [tex]A_n<B_n[/tex] respectively.So I think the answer is d.
    2.Is 1 a prime number?

    3.Let C denote the set of all complex numbers.Define A and B by
    A={(z,w):z,w [tex]\in [/tex]C and mod z=mod w}
    B={(z,w):z,w [tex]\in [/tex]C and [tex]z^2=w^2[/tex]}

    a.A=B b.A[tex]\sqsubseteq[/tex]B and A not equal to B
    c.B[tex]\sqsubseteq[/tex]A and B not equal to A
    d.none of the above.

    I think z^2=w^2 means mod z=mod w but the reverse is not true,so is the answer b?

    4.If positive numbers a,b,c,d are such that 1/a,1/b,1/c,1/d are in A.P then we always have
    a.a+d[tex]\geq[/tex]b+c b.a+b[tex]\geq[/tex]c+d
    c.a+c[tex]\geq[/tex]b+d d.none of the above

    1/a +1/d =1/b +1/c or,(a+d)/ad =(b+c)/bc. Now 1/ad<1/bc or,ad>bc.so,a+d[tex]\geq[/tex]b+c i.e.,(a)?

    thanking you in advance.And how do you put "not equal to" and "modulus" in latex?
    Last edited: Apr 13, 2006
  2. jcsd
  3. Apr 14, 2006 #2

    Hi! Here are some of my thoughts with regard to your questions...

    Q2) 1 is not a prime number. A prime number is defined as a number which is divisible only by itself and one. It can then be taken to mean that every prime number has 2 distinct factors. Hence, 1 is not prime.

    Q3) Your reasoning that "[tex]z^2=w^2[/tex] means [tex]\mid z \mid =\ \mid w \mid [/tex] but the reverse is not true" is valid. However, doesn't this mean that all elements in B are in A but not all elements in A are in B? So, the answer is...

    Q4) (a) is indeed the correct option, but I do not understand one of the steps in your working. Why do you immediately conclude that [tex]\frac{1}{ad} < \frac{1}{bc} [/tex]? Can you clarify? Thanks.
    Last edited: Apr 15, 2006
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