where is the integer part of .

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- Sep 12th 2008, 02:06 AMjames_bondSolve the equation
where is the integer part of .

- Sep 12th 2008, 07:22 AMSoroban
Hello, james_bond!

I haven't solved it yet, but here's an approach.

. . Maybe it will inspire someone . . .

Quote:

.where is the integer part of

We have: .

Since the right side must be an integer,

. . That is: .

Substitute into [1]: .

. . which simplifies to: .

And that's it for now . . .

- Sep 12th 2008, 08:12 AMwisterville
Hello,

Soroban made a mistake since x is not necessarily an integer. Instead, as he pointed out, it is true that (7x-1)/3 is an integer. Thus, x=(3k+1)/7 for some integer k.

You have an inequality .

In this case, you get (4x+1)/5 - 1 < (7x-1)/3 <= (4x+1)/5 which means -7/23<x<=8/23.

Now, substituting x=(3k+1)/7, you get k=-1, 0.

Bye. - Sep 24th 2008, 08:42 AMbkarpuzRemarks on Soroban's post
In addition to the comments by

**wisterville**, I have the followings.

At the right-hand side of [1], we**do not**know that is an integer, what we know is is an integer exactly.

Letting , we indeed see that but we have .

We clearly have for all .

Then, using the equation you gave with , we get

.

You can solve these inequalities by seperating into two parts. (Wink)