Difference between revisions of "HW 1:" - Rhea

Revision as of 17:14, 14 January 2010

T. Qu: Does anyone know how to do #50 in section 4.1?

It is asking us to prove that for any number n (lets say 4) you can pick n+1 integers (5) that are either less than or equal to 2n (8) at least one of the chosen integers can divide one of the other chosen integers in the set.

So if n=4 then the integers that are available to choose from are less than or equal to 2n=8

1, 2, 3, 4, 5, 6, 7, 8

We then choose any 5 (n+1) of these integers and we will see that it is not possible to choose 5 without having at least one of the integers able to divide another.

This is not an answer to the problem. I am just trying to make it more clear what it is asking for anyone who found the statement of the problem confusing.

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Have you done #40?

Here is what I am thinking.

Even thought x and y are positive numbers, but x-1 and y-1 are may not positive numbers.

ex: x=1 is positive number but x-1 is not. so, x-1 and y-1 can't be inductive hypothesis.

I'm not sure it is right.

any idea?