Dividing by zero

The post that was quoted here has been removed

I'm probably going to regret attempting to talk about this problem ( I anticipate that you will poke fun at my arrogant mathematical ineptitude), but what do you consider reduced. Is it the fact that no common factor can be extracted between the numerator and denominator that makes it irreducible? It can be rewritten as a mixed number. 1 + (7*N +1)/(14*N + 3), and the mixed number is still irreducible. Thoughts?

Originally posted by joe shmo
I'm probably going to regret attempting to talk about this problem ( I anticipate that you will poke fun at my arrogant mathematical ineptitude), but what do you consider reduced. Is it the fact that no common factor can be extracted between the numerator and denominator that makes it irreducible? It can be rewritten as a mixed number. 1 + (7*N +1)/(14*N + 3), and the mixed number is still irreducible. Thoughts?

it is the key to an explanation

'cannot be reduced any further' does mean no further 'cancelling' is possible

From your algebraic division, the problem now becomes proving that (7*N + 1)/(14*N +3) is irreducible

Re write the denominator as (7*N +1) + (7*N + 1) + 1

so now it is (7*N + 1)/((7*N +1) + (7*N +1) + 1)

= (7*N +1)/(2*(7*N +1) +1)

let 7*N + 1 = X (as N is defined as an integer, it follows that X is a (bigger) integer)

= X/(2X+1)

so we are looking to see if X and 2X +1 can have any common factor other than 1

They cannot, as the 'right hand side' is a multiple of 'left hand side, PLUS JUST 1

So 1 is the only possible common factor

QED

Originally posted by st dominics preview
it is the key to an explanation

'cannot be reduced any further' does mean no further 'cancelling' is possible

From your algebraic division, the problem now becomes proving that (7*N + 1)/(14*N +3) is irreducible

Re write the denominator as (7*N +1) + (7*N + 1) + 1

so now it is (7*N + 1)/((7*N +1) + (7*N +1) ...[text shortened]... is a multiple of 'left hand side, PLUS JUST 1

So 1 is the only possible common factor

QED

So in the ending step we are comparing two algebraic expressions to see if they share any common factors. What is wrong with stating that the numerator and denominator of the original problem are not individually factorable, so there cannot exist a common factor, greater than 1 between them, which makes the expression irreducible...?

hi Joe

the argument is more than just share common factors.

The denominator is twice the numerator plus 1. There is no number that 'X' can take to make these 2 share a common factor. The 'plus 1' is the key.

To give an example, 3N + 2 and 5N share no common factors

but if N=6, one is 20, the other 30, and they clearly 'cancel'

Hope thats ok!

Originally posted by st dominics preview
hi Joe

the argument is more than just share common factors.

The denominator is twice the numerator plus 1. There is no number that 'X' can take to make these 2 share a common factor. The 'plus 1' is the key.

To give an example, 3N + 2 and 5N share no common factors

but if N=6, one is 20, the other 30, and they clearly 'cancel'

Hope thats ok!

Yep, that cleared it up. Thank you.

Originally posted by joe shmo
Yep, that cleared it up. Thank you.

no problem ~ I probably rushed the last bit of the proof, to show that N and 2N + 1 can have no common factors also depends on the single 'N'.

It is a separate proof altogether ~ if you try some integers for N you soon see the pattern (though a pattern is not a proof!)