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I just started to learn induction in my first year course. I'm having a difficult time grasping the concept. I believe I understand the basics but could someone summarize simple induction and strong induction and explain what the differences are? The video I'm watching explains that if $P(k)$ is true then $P(k+1)$ is true for simple induction, and for strong induction if $P(i)$ is true for all $i$ less than equal to $k$ then $P(k+1)$ is true. I don't really know what that means.

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  • $\begingroup$ For further reading regarding mathematical induction read this. $\endgroup$ – user 170039 Aug 12 '16 at 4:37
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With simple induction you use "if $p(k)$ is true then $p(k+1)$ is true" while in strong induction you use "if $p(i)$ is true for all $i$ less than or equal to $k$ then $p(k+1)$ is true", where $p(k)$ is some statement depending on the positive integer $k$.

They are NOT "identical" but they are equivalent.

It is easy to see that if simple induction is true then strong induction is true: if you know that statement $p(i)$ is true for all $i$ less than or equal to $k$, then you know that it is true, in particular, for $i=k$ and can use simple induction.

It is harder to prove, but still true, that if strong induction is true, then simple induction is true. That is what we mean by "equivalent".

Here we have a question. It is not why we still have "weak" induction - it's why we still have "strong" induction when this is not actually any stronger.

My opinion is that the reason this distinction remains is that it serves a pedagogical purpose. The first proofs by induction that we teach are usually things like $\forall n\left[\sum_{i=0}^n i= \frac{n(n+1)}{2}\right]$. The proofs of these naturally suggest "weak" induction, which students learn as a pattern to mimic.

Later, we teach more difficult proofs where that pattern no longer works. To give a name to the difference, we call the new pattern "strong induction" so that we can distinguish between the methods when presenting a proof in lecture. Then we can tell a student "try using strong induction", which is more helpful than just "try using induction".

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    $\begingroup$ Another important reason for the distinction is that when working in other well-ordered sets, they need no longer be equivalent (strong induction always works, simple induction might require more base cases to still work). $\endgroup$ – Tobias Kildetoft Oct 7 '13 at 8:01
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    $\begingroup$ I dislike the isolation of weak induction. I’d rather see induction taught first in terms of (non-existence) of a least counterexample. Strong induction comes naturally that way, and weak induction is obviously just a special case; moreover, since least ultimately generalizes to well-founded relations in general, you also get structural induction. $\endgroup$ – Brian M. Scott Oct 7 '13 at 8:09
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    $\begingroup$ I don't get how it is "harder to prove" that strong induction implies weak. That direction is completely straightforward, since the base case and induction step for a "weak" induction proof directly imply the induction step of strong induction. $\endgroup$ – Henning Makholm Oct 7 '13 at 9:34
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    $\begingroup$ @Tobias: Yes, so if you have done the step for weak induction, you have proved $P(n)$ with strictly fewer assumptions that long induction allows you to use, and you can reuse your induction step unchanged. In contrast, if you want to go in the other direction you have, in general, to change what the property you prove is. $\endgroup$ – Henning Makholm Oct 7 '13 at 9:43
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    $\begingroup$ Don't you mean it's easier to see that strong induction implies simple induction, and it's harder to prove that simple induction implies strong induction? If you assume $(\forall i \le k): p(i) \implies p(k+1)$ is true, then it's straightforward to show that $p(k)\implies p(k+1)$ is true, like you said, taking the case $i=k$. The harder proof is showing the reverse direction (simple to strong). $\endgroup$ – chharvey Oct 1 '16 at 3:28
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One example of the use of strong induction is in the inductive proof that any prime $p\not \equiv 3 \pmod 4$ is the sum of squares of two integers. We have $2=1^2+1^2.$ For prime $p\equiv 1 \pmod 4,$ at some point in the proof we need to employ the inductive hypothesis that every prime $q$ such that $p>q\not \equiv 3 \pmod 4$ is the sum of two squares, because (after a fair bit of other work) such a $q$ appears in the algebra but we don't know which one it is.

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