21

Assume you want to prove an identity such as

$$\sum_{k=m+1}^{n}A(k,m)-B(k,m)=S(m)+T(n,m)\qquad\text{for } n,m\in \mathbb{Z},n,m\geq 0.$$

Added: I applied mathematical induction on $m,n$ to prove it. I am unsure because up to now I have seen it applied to properties depending on a single variable only.

Question: does application of two inductive arguments, one on $m$ and the other on $n$, guarantee the validity of such a proof?

Américo Tavares
  • 38,602

2 Answers2

27

Here are some induction principles for two variables:

  • $P(0,0)$
  • $\forall x,y. P(x,y) \Rightarrow P(x+1,y)$

  • $\forall x,y. P(x,y) \Rightarrow P(x,y+1)$

  • $\forall x,y. P(x,y)$

and

  • $P(0,0)$
  • $\forall x,y. P(x,0) \Rightarrow P(x+1,0)$

  • $\forall x,y. P(x+1,y) \Rightarrow P(x,y+1)$

  • $\forall x,y. P(x,y)$

  • @muad: in the 2nd example is it $\forall x,y. P(x+1,y) \Rightarrow P(x,y+1)$ or $\forall x,y. P(x,y) \Rightarrow P(x,y+1)$? – Américo Tavares Oct 24 '10 at 11:04
  • @Américo Tavares, the second one is covering the x-axis and shifting diagonally back to cover everything above it. So it is $\forall x,y. P(x+1,y) \Rightarrow P(x,y+1)$. –  Oct 24 '10 at 11:11
  • Nice! You can extend this to transfinite ordinals by taking

    $$\forall z\Big(\big(\forall x,y\big[x<y\Rightarrow P(x,z)\big]\big)\Rightarrow P(y,z)\Big),$$

    and

    $$\forall z\Big(\big(\forall x,y\big[x<y\Rightarrow P(z,x)\big]\big)\Rightarrow P(z,y)\Big).$$

     – Alec Rhea May 08 '17 at 13:58
19

Suppose you are trying to prove a family of statements $P(x, y)$. This is the same as proving the family of statements $F(x)$, where $F(x) = \forall y : P(x, y)$. Each statement $F(x)$ can be proven by induction on $y$ (for fixed $x$), and then you can prove $P(x, y)$ by induction on $x$. You might want to try proving

$${n+1 \choose k+1} = {n \choose k+1} + {n \choose k}$$

this way.

But actually you can be much trickier than this. Sometimes it suffices to induct on $x + y$, for example.

Qiaochu Yuan
  • 419,620