Help with arithmetic tables

[*Jas*]

Arithmetic tables...?!?


Any help would be v much appreciated!

 

[rodigee]

Are you asking what those tables are? How to read them? How to answer that question in picture?

 

[CRGreathouse]

The tables work like ordinary multiplication tables -- choose the row & column of the numbers you're multiplying, then their intersection has the product.

Hint on the questions: the squares are on the diagonal of the multiplication table.

 

[John Creighto]

The tables work like ordinary multiplication tables -- choose the row & column of the numbers you're multiplying, then their intersection has the product.

Hint on the questions: the squares are on the diagonal of the multiplication table.

What if x is greater then 6 though?

x^2=3+n*7

Aside from trying every possible value of n I'm not sure how to prove n can't be an integer.

 

[rodigee]

What if x is greater then 6 though?

x^2=3+n*7

Aside from trying every possible value of n I'm not sure how to prove n can't be an integer.

How can x be greater than 6?

 

[DaveC426913]

The tables work like ordinary multiplication tables -- choose the row & column of the numbers you're multiplying, then their intersection has the product.

Hint on the questions: the squares are on the diagonal of the multiplication table.

Has math changed so much since I was a boy?


Since when does 6+6 = 5?
Since when does 6x6 = 1?

It looks like there's some sort of modulo going on.

 

[John Creighto]

Has math changed so much since I was a boy?


Since when does 6+6 = 5?
Since when does 6x6 = 1?

It looks like there's some sort of modulo going on.

Yes. The subscript on the Z means it's modulo 7 arithmetic.
http://en.wikipedia.org/wiki/Modular_arithmetic

That is why I wrote:
x^2=3+n*7

So x could be greater then 6 for a suitably large choice of n but I don't know if it can be an integer.

 

[John Creighto]

So I had some thoughts of n greater then 6. In general any number can be written in modula 7 via the division algorithm
x=r+nq

So if we square x we get:
(r+nq)^2=r^2+2rnq+n^2q^2
Which is equal to 3. Therefore:
r^2+2rnq+n^2q^2=3
rearranging we get:
q^2n^2+2rqn+(r^2-3)=0

Now, perhaps someone who knows something of group theory can tell me under what conditions the above polynomial of n, will have integer roots. If we know that condition are we easily able to choose an r and a q that will satisfy this condition?