Show that any two consecutive integers are relatively prime

We $ $ have $\,n\mid \color{#c00}{\overbrace{n\!+\!1}^{\large m}}-1.\ $ Your (correct) proof easily generalizes widely as follows.

Table of Contents Show

What are the two consecutive numbers which are prime?
How do you prove that N 1 are relatively prime?
What does it mean if two integers are relatively prime?
What is the probability that two numbers are relatively prime?

Generally $\,n\mid \,k\:\color{#c00}m-1\,\Rightarrow\, k\,m-j\,n = 1,\ $ so $\,\ d\mid m,n\,\Rightarrow\, d\mid 1,\ $ so $\ \gcd(m,n) = 1$

$ $ i.e. $\ {\rm mod}\,\ n\!:\,\ m^{-1}\,$ exists $\,\Rightarrow\, \gcd(m,n) = 1\,$ (and conversely by Bezout, see here and here)

Remark $\ $ We can also use (a single step of) the Euclidean algorithm

$$ \gcd(km,n) = \gcd(jn\!+\!1,n) = \gcd(1,n) = 1\,\Rightarrow\, \gcd(m,n) = 1$$

Two consecutive numbers are relatively prime.

$\displaystyle (n, n+1)=1\rightarrow \alpha n+\beta (n+1)=1$

I am not sure what to do next to prove this.

Two consecutive numbers are relatively prime.

$\displaystyle (n, n+1)=1\rightarrow \alpha n+\beta (n+1)=1$

I am not sure what to do next to prove this.

$\displaystyle \alpha=-1,\; \beta=1 $

I see how that will justify the equality but is that ok to do for the proof and then just call it day and it is done?

I see how that will justify the equality but is that ok to do for the proof and then just call it day and it is done?

Let $\displaystyle g=(n,n+1) $

$\displaystyle g\mid n $ and $\displaystyle g\mid n+1 \implies g\mid (n+1)-n = 1 \implies g=\pm1 $. But since $\displaystyle g>0 $, we get $\displaystyle g=1 $.

I see how that will justify the equality but is that ok to do for the proof and then just call it day and it is done?

The answer is yes. In general: Let $\displaystyle a$ and $\displaystyle b$ be integers. There exists two integers $\displaystyle x$ and $\displaystyle y$ such that $\displaystyle ax+by=1$ if and only if $\displaystyle (a,b)=1$. In your case, $\displaystyle a=n$, $\displaystyle x=-1$ and $\displaystyle b=n +1$, $\displaystyle y=1$.

However, the reply you got from "chiph588" using the greatest common divisor ($\displaystyle g=(n,n+1)$) seems to me more elegant.

The answer is yes. In general: Let $\displaystyle a$ and $\displaystyle b$ be integers. There exists two integers $\displaystyle x$ and $\displaystyle y$ such that $\displaystyle ax+by=1$ if and only if $\displaystyle (a,b)=1$. In your case, $\displaystyle a=n$, $\displaystyle x=-1$ and $\displaystyle b=n +1$, $\displaystyle y=1$.

This is a special case of Bézout's identity.

What are the two consecutive numbers which are prime?

Example 1: The first 27 consecutive prime numbers (n = 1, ..., 27). They are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103.

How do you prove that N 1 are relatively prime?

If a prime p divides n then dividing n+1 by p leaves a remainder of 1, so p does not divide n+1. That means the factorizations of n and n+1 have no prime in common, so n and n+1 are relatively prime.

What does it mean if two integers are relatively prime?

Two integers are relatively prime if they share no common positive factors (divisors) except 1.

What is the probability that two numbers are relatively prime?

While the probability of a random number being prime decreases as the range of possible random numbers increases (Prime Number Theorem), the probability of two random numbers being relatively prime is 60.8% Is this something that is either well known by or trivially obvious to prime number gurus?