I'm asked to indicate which natural numbers $n$ each of the below inequality is true, and then I am required to prove this via induction, but I'm wondering what that means... Strong induction?
$n^2\leq n!$
$n^2\leq 2^n$
$2^n\leq n!$
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1There is little conceptual difference between strong and "weak" induction. Anything that can be proved with one can be proved with the other, though often strong induction is more natural. For the above inequalities, ordinary induction works fine. – André Nicolas Aug 26 '13 at 15:46
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Prove the inequality $n! \geq 2^n$ by induction – GNUSupporter 8964民主女神 地下教會 May 10 '18 at 16:52
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I think the question means to use standard induction. You don't need strong induction to demonstrate these inequalities just standard induction.
For the first inequality we wish to find for which natural numbers $n$ $n^{2}\le n!$. First note that the inequality holds for $n=1$ and $n=4$ by checking directly. I claim that the inequality holds for $n=1$ and for all $n\ge4$. We proceed by induction starting from $n=4$. At $n=4$ we have $4^{2}=16\le24=4!$. Suppose for $n\ge4$ the inequality holds. Then for $n+1$ we have:
$(n+1)^{2}=n^{2}+2n+1\le n!+2n+1\le n!+n!+n!=3n!\le(n+1)n!=(n+1)!$.
Note that $2n\le n!$ for $n\ge3$ since $2$ and $n$ appear in the product $n!=n(n-1)\cdot...\cdot2\cdot1$ and any numbers between $2$ and $n$ (if there are any) are greater than 1.
A similar process can be applied to check the other inequalities.
user71352
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Each of these inequalities is true for $n > N$ for some $N \in \mathbb{N}$. You need to find that $N$ for each, demonstrate that the inequality does not hold for all $k \in \mathbb{N}$ with $k < n$, and then, by induction (Possibly strong, but that may not be necessary for some of them) prove that $\forall n > N$, the inequality holds. That is, use $N$ as your base case, and then induct as normally.
qaphla
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