Mathematical Induction: Prove P(n)=(n^2)-n divisible by 2
- Thread starter portia-h
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Re: Mathematical Induction
First, show n=1 is true. Then we assume 2 is a factor of \(\displaystyle k^{2}-k\)
equivalently, \(\displaystyle k^{2}-k=2p\) for some integer p.
We want to show that \(\displaystyle P_{k+1}\) is true. Namely, 2 is a factor of:
\(\displaystyle (k+1)^{2}-(k+1)\)
\(\displaystyle =k^{2}+2k+1-k-1\)
Rearrange terms:
\(\displaystyle =\underbrace{(k^{2}-k)}_{\text{2p}}+2k\)
Now, can you finish?.
First, show n=1 is true. Then we assume 2 is a factor of \(\displaystyle k^{2}-k\)
equivalently, \(\displaystyle k^{2}-k=2p\) for some integer p.
We want to show that \(\displaystyle P_{k+1}\) is true. Namely, 2 is a factor of:
\(\displaystyle (k+1)^{2}-(k+1)\)
\(\displaystyle =k^{2}+2k+1-k-1\)
Rearrange terms:
\(\displaystyle =\underbrace{(k^{2}-k)}_{\text{2p}}+2k\)
Now, can you finish?.