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Having trouble proving stuff

  1. Oct 10, 2013 #1
    Prove for all ## x \geq -1 ## and n being a natural number, ## (1+x)^n \geq 1 + nx ## I've done this using induction

    hence deduce from this that for any real number # a \ geq 2 ##, ## a^n > n ## for all n belonging to the natural numbers.

    I'm stuck at the deduce part

    I'm not sure what deduce means, it seems fairly obvious so instead I tried to prove it, by again, induction:

    a = 2,
    2^n > n which is true a n >= 1
    assume true for a = k

    then when a = k+1

    (k+1)^n > n

    I don't see how I can prove it from here, it seems obvious as k >= 2 and n >=1 this will always hold true, and thus a^n > n but I can't seem to prove it. Does anyone have any tips on proving as I'm fairly new at this

    thank you
     
  2. jcsd
  3. Oct 10, 2013 #2

    pasmith

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    Homework Helper

    It means "show that this follows from what you have just proved".

    In this case, you've shown that if [itex]x \geq -1[/itex] then [itex](1 + x)^n \geq 1 + nx[/itex].

    Now see what happens if you set [itex]a = 1 + x[/itex].
     
  4. Oct 10, 2013 #3
    how would that work exactly? it states that a >= 2 if we let a = 1 + x then a >= 0 as x >= -1?
     
    Last edited: Oct 10, 2013
  5. Oct 10, 2013 #4
    if I ignore my confusion above I ge tthis:

    a^n >= 1 + nx

    as x >= -1 a^n >= 1 - n

    as n is a natural number n >=1
    hence a^n > n
     
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