Question

Two numbers are said to be coprime if their highest common factor is 1. Prove that any two consecutive positive integers are always coprime.

Answer 1

give me a min

Answer 2

let n,p,q be integers , such that p>q>0 and n>1

then n*q and n*p are integers and are co-prime

Let's assume that it's possible they are consecutive positive integers - so:

n*q - n*p = 1
n(q-p)=1
q-p=1/n

but, 1/n is not an integer , so q and p cannot both be integers-
contradicting our assumption.

Answer 3

n*q and n*p are not coprime... they have n as a common factor and n>1, so their highest common factor cannot be 1.

Answer 4

Doesn't this sort of automatically follow from the division algorithm?

Answer 5

OOOps.. I misread what co-prime means, and just proved something else then ...

Answer 6

Please elaborate estudier.

Answer 7

how about something along these lines :)

Let p,n be integers and p be a factor of both n and (n+1)

so:
n/p is an integer
(n+1)/p is an integer

(n+1)/p=n/p+1/p = integer + 1/p

integer + 1/p --> p=1

Answer 8

p=1 is the only case where:
integer + 1/p will be an integer as well.

Answer 9

Good!
A simpler proof is using the Euclidean algorithm to find the highest common factor of two numbers.
gcd(n+1,n) = gcd(n,1) = 1
follows immediately like this! Thanks though!

Answer 10

Thanks, that Euclidean algorithm rocks !