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I am not sure if Im on the right track here but if any one could help out I would greatly appreciate it.

Prove if $x_1,...,x_n$ are natural numbers with $n\geq2$ then $x_1x_2...x_n$ is odd iff $x_i$ is odd for all $i$, $1\leq i\leq n$

Formulate an induction and prove it.

Suppose $x_1x_2\dots x_n=x_{n^2+1}$ and $x_1=x_2=1$

Let p(n):=$x_1x_2\dots x_n=x_{n^2+1}$

Base case:

Let $n=1$ LHS= $x_1=1$, RHS=$x_{(1)^2+1}=x_2=1$.

So LHS=RHS=1,

Thus, the base case holds.

Induction hypothesis:

Assume $x_1x_2\dots x_n=x_{n^2+1}$

NTS: $x_1x_2\dots x_n x_{n+1}=x_{(n+1)^2+1}$

Inductive Step:

By the induction hypothesis we can simplify $x_1x_2\dots x_n x_{n+1}=x_{(n+1)^2+1}$ to $x_{n^2+1} x_{n+1}=x_{(n+1)^2+1}$

Im not sure how to continue from here and I believe its because the induction I formulated is incorrect but I am having trouble thinking of what adjustments I need to make.

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  • $\begingroup$ Suppose otherwise. Then at least one of the $ x_i$ are even and can be written as $2k$. But then the product is even as well. In the other direction, suppose the product is even. Since two is prime, it must divide some $x_i$. $\endgroup$ – JMoravitz Jun 26 '16 at 1:07
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I think you have your wires crossed; particularly it seems like you're thinking about how the sum of the first $n$ odd numbers gives $n^2$. This is woefully incorrect. Here is how you should frame it:

Suppose that $x_1,\ldots, x_n\ge 2$, then you want to show that $x_1\cdots x_n$ is odd if and only if $x_i$ is odd for all $i$. In terms of an induction argument, this would go like this:

Base case (this is trivial)

 

Induction hypothesis $y = x_1\ldots x_n$ is odd if and only if $x_i$ is odd for all $1\le i \le n$.

Then you want to show that this if and only if statement holds true for each $x_1,\ldots, x_{n+1}$, i.e. you want to show that

$z =x_1\cdots x_{n+1}$ is odd if and only if $x_i$ is odd for all $1\le i \le n+1$.

But $z = x_1\cdots x_n\cdot x_{n+1} = y\cdot x_{n+1}$. Can you take it from here?

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