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Here's my work so far:

  • base case: $n=1$
    $2^1 = 2$ , $2>1$
  • induction hypothesis: $p(k) = 2^k > k$ $\quad \forall$ $k>0$
  • induction step:
    $p(k+1) = 2^k+1 > k$ $\quad \forall$ $(k+1)>0$
    $2^k+1 = 2^k\cdot 2 = 2^k + 2^k$
    $2^k + 2^k > k+1$
    $2^k + 2$
    $2^k + 2 > k+1$
Tita
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HeinricheBerry
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  • Recommend reading http://math.stackexchange.com/questions/1253956/proof-writing-how-to-write-a-clear-induction-proof on how to write a clear proof by induction. – JMoravitz Nov 14 '16 at 00:01

2 Answers2

2

Here is what your induction step should look like. $$2^{k+1}=2\cdot 2^k>2\cdot k=k+k\geq k+1$$ The first inequality comes from the induction hypothesis.

Laars Helenius
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Assume that $2^k > k$. In other words, the upper bound for $k$ is $2^k$. If the induction is true, then: $$2^{k + 1} > k + 1$$ $$2*2^k > k + 1$$ Let's assume the worst. What if $k$ is at its upper bound? The following must still hold: $$2*2^k > 2^k + 1$$ $$2^k > 1$$ $$k > 0$$ Since this is true for all cases, this proves the induction for all positive integers.

Kaynex
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