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<div>Quite often in mathematics we find ourselves wanting to prove a statement that we <span style="font-style: italic;">think</span> is true for every natural number $n$.</div>
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<div>For example, you may have met the formula $\frac{1}{6}n(n+1)(2n+1)$ for the sum $$ \sum_{i=1}^n i^2=1^2+2^2+\ldots+n^2. $$ We can try some values of $n$, and see that the formula seems to be right: \begin{eqnarray} n=1: & & \sum_{i=1}^n i^2 = 1^2 = 1;\\ & & \frac{1}{6}n(n+1)(2n+1) = \frac{1}{6}\times 1\times 2\times 3 = 1\\ \\ n=2: & & \sum_{i=1}^n i^2 = 1^2 + 2^2 =
1+4 = 5;\\ & & \frac{1}{6}n(n+1)(2n+1) = \frac{1}{6}\times 2\times 3\times 5 = 5\\ \\ n=3: & & \sum_{i=1}^n i^2 = 1^2 + 2^2 + 3^2 = 1+4+9 = 14;\\ & & \frac{1}{6}n(n+1)(2n+1) = \frac{1}{6}\times 3\times 4\times 7 = 14.\\ \end{eqnarray} But we want to <span style="font-style: italic;">prove</span> that this is true for <span style="font-style: italic;">all</span> positive
integers $n$, and it's going to be impossible if we try to do this by putting in all possible values! Instead, we're going to have to be a bit more cunning$\ldots$</div>
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<div>Have you ever played that game with dominoes where you line them up on end and then, by knocking over the first one, knock over the whole lot? You can think of proof by induction as the mathematical equivalent (although it does involve infinitely many dominoes!).</div>
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<div>Suppose that we have a statement $P(n)$, and that we want to show that it's true for all $n$. So in our example above, $P(n)$ is: $$"\sum_{i=1}^n i^2 = \frac{1}{6}n(n+1) (2n+1)".$$ Think of each $P(n)$ as a domino. If we can show that $P(1)$ is true (that is, we can knock over the first domino)and that if $P(n)$ is true then so is $P(n+1)$ (knocking over one domino means the next one will
also fall over), do you agree that we've then shown that $P(n)$ is true for all $n\geq 1$ (because all of the dominoes will fall over)?</div>
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<div>We'll know that $P(1)$ is true,</div>
<div>so we'll know that $P(1+1)=P(2)$ is true,</div>
<div>so we'll know that $P(2+1)=P(3)$ is true,</div>
<div>so we'll know that $P(3+1)=P(4)$ is true,</div>
<div>$\ldots$</div>
<div>You get the idea.</div>
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<div>Let's go back to our example from above, about sums of squares, and use induction to prove the result. We know that $P(1)$ is true (we did that before!). Now let's see what would happen <span style="font-style: italic;">if</span> we knew that $P(n)$ is true. Then we'd know that $\sum_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1)$. So what can we say about $\sum_{i=1}^{n+1} i^2$? Well,
\begin{eqnarray} \sum_{i=1}^{n+1} i^2 & = & (n+1)^2 + \sum_{i=1}^n i^2\\ & = & (n+1)^2 + \frac{1}{6}n(n+1)(2n+1)\quad\textrm{(by the inductive hypothesis)}\\ & = & \frac{1}{6}(n+1)\left[6(n+1)+n(2n+1)\right]\\ & = & \frac{1}{6}(n+1)\left[6n+6+2n^2+n\right]\\ & = & \frac{1}{6}(n+1)\left[2n^2+7n+6\right]\\ & = & \frac{1}{6}(n+1)(n+2)(2n+3).
\end{eqnarray} This looks familiar; what <span style="font-weight: bold;">is</span> $P(n+1)$? Well, let's substitute $n+1$ in place of $n$ in our statement of $P(n)$. So we're aiming for $$ \sum_{i=1}^{n+1} i^2 = \frac{1}{6}(n+1)(n+2)(2(n+1)+1)=\frac{1}{6}(n+1)(n+2)(2n+3), $$ which is exactly what we've just got! (Working out what you're aiming for can often give you an idea of how to manipulate
the algebra.) So we've shown that <span style="font-style: italic;">if</span> $P(n)$ is true, <span style="font-style: italic;">then</span> $P(n+1)$ is true. Since we also know that $P(1)$ is true, we know that $P(2)$ is true, so $P(3)$ is true, so $P(4)$ is true, so $\ldots$ In other words, we've shown that $P(n)$ is true for all $n\geq 1$, by mathematical induction.</div>
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<div><span style="font-style: italic; font-weight: bold;">Warning:</span> When encountering induction for the first time, people usually remember to show "if $P(n)$ is true then $P(n+1)$ is true'', because that's what induction is really about, but a common mistake is to forget to show that $P(1)$ is true. Proof by induction is a two-stage process, even if one stage is usually very easy. The
dominoes won't fall over unless you knock over the first one!</div>
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<div>Don't forget that your first domino doesn't have to be $P(1)$. It could be $P(2)$, or $P(19)$, or $P(1000000)$. For example, we can use induction to show $3^n&gt; n^3$ for $n\geq 4$ (see the exercises below)</div>
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<div>One last thing: induction is only a method of <span style="font-style: italic;">proof</span> . For example, if you're trying to sum a list of numbers and have a guess for the answer, then you may be able to use induction to prove it. But you can't use induction to find the answer in the first place. Also, there are often other methods of proof: I've given some examples below of things that
you might like to try to prove by induction, but several of them can be proved at least as easily by other methods (indeed, you've probably seen some proved by other methods).</div>
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<div>Don't forget that if you get stuck with these, you can ask for help on Ask NRICH. You can also consult the Hints if you need to.</div>
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<li>Prove by induction that $$ \sum_{i=1}^n i^3 = \frac{1}{4}n^2(n+1)^2\textrm{ for }n\geq 1. $$</li>
<li>Use induction to show that $$ \sum_{i=1}^n r^{i-1}=\frac{r^n - 1}{r-1}\textrm{ for }n\geq 1\textrm{ and }r\neq 1. $$</li>
<li>Use induction to show that $3^n&gt; n^3$ for $n\geq 4$. (Note that you have to start at $n=4$ as the result isn't true for $n=3$!)</li>
<li>Use induction to show that $$ \left(\begin{array}{c}n+1\\r\end{array}\right) = \left(\begin{array}{c}n\\r\end{array}\right) + \left(\begin{array}{c}n\\r-1\end{array}\right)\textrm{ for }n\geq 1 $$ (where $\left(\begin{array}{c}n\\r\end{array}\right)$ is the binomial coefficient). (You will need to use induction on $n$ and keep $r$ fixed). Remark: This is rather fiddly; starting with the
right-hand side will make things easier. This example is an excellent lesson in why you should not always use induction: thinking about counting things will help you prove the result much more easily. Can you see how?</li>
<li>Use induction to show that $4^n+6n$ is divisible by 6 for $n\geq 1$. [No this is not a misprint. What part of the argument works? What part fails?]</li>
<li>Use induction to show that $4^n + 6n -1$ is divisible by 9 for $n\geq 1$.</li>
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<div><span style="font-style: italic;">Vicky finished a degree in Maths at Cambridge last summer and is now doing a fourth year course studying Combinatorics, Number Theory and Algebra, still at Trinity College, Cambridge.</span></div>
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Note 1:<br></br>
"By the inductive hypothesis'' is just a fancy way of saying "using P(n) since we're assuming it at the moment''.<br></br></mdoxml>
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<mdoxml version="1.0">(3) Prove: $3^n> n^3$ for $n\geq 4$.<br></br>
Hint: $3k^3 - (k+1)^3 = (k-1)^3 + k(k^2-6)$<br></br>
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(4) The left-hand side is the number of ways of choosing $r$ balls
from $n+1$. Suppose one ball is coloured blue (and the others
aren't). Now explain why the right-hand side is the number of ways
of picking $r$ balls including the blue one plus the number of ways
of picking $r$ balls excluding the blue one.<br></br>
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(5) OK, this isn't true. But the inductive step works. So what's
gone wrong? It's not true for $n=1$. This is why it's absolutely
vital that you check the starting point!<br></br>
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(6) The inductive hypothesis is that $4^k + 6k - 1$ is divisible by
9. That is, $4^k+6k-1=9m$ for some integer $m$. Now use this to get
an expression for $4^k$ that you can substitute into
$4^{k+1}+6(k+1)-1$. Alternatively, what is $4(4^k+6k-1)-18k+9$?
This is more elegant, but perhaps harder to spot without
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Mathematical Induction
This article gives an introduction to mathematical induction, a powerful method of mathematical proof.
Advanced Algebra
Summation of series
Using, Applying and Reasoning about Mathematics
Mathematical induction