fraction $\frac{21n+4}{14n+3}$ , $n\in N$

Question

Prove tat for every natural number

$n$ the fraction

$\frac{21n+4}{14n+3}$ is irreducible

How can I beginning in this problems

I don't have any ideas to approach it

compute $\gcd(21n+4,14n+3)$ using Euclid. – user10354138 May 15 '19 at 14:19 — user10354138, May 15 '19 at 14:19

score 2 · Answer 1 · answered May 15 '19 at 14:20

2

$\gcd(21n+4,14n+3)=\gcd(7n+1,14n+3)=\gcd(7n+1,1)=1$

answered May 15 '19 at 14:20

CY Aries

Can u expline why !! $\gcd(21n+4,14n+3)=\gcd(7n+1,14n+3)$ – Ellen Ellen May 15 '19 at 14:24
Euclidean Algorithm. $\gcd(a,b)=\gcd(a-b,b)$ – CY Aries May 15 '19 at 14:25
If $d$ divides both $a$ and $b$, then it divides $a-b$ (actually, it divides $a-kb$ for any integer $k$). – CY Aries May 15 '19 at 14:26
Thank you very much sir – Ellen Ellen May 15 '19 at 14:34

score 1 · Answer 2 · answered May 15 '19 at 14:28

1

$$\frac{21n+4}{14n+3} = \frac{(14n+3)+(7n+1)}{14n+3} = 1+\frac{7n+1}{14n+3} = 1+\frac{7n+1}{2(7n+1)+1}$$

It can be seen that $7n+1$ and $2(7n+1)+1$ are coprime.

answered May 15 '19 at 14:28

minori minus

score 1 · Answer 3 · answered May 15 '19 at 14:34

1

$d\mid 14n\!+\!3,21n\!+\!4\,\Rightarrow\, d\mid 3(14n\!+\!3)-2(21n\!+\!4) = 1\ $ (by eliminating $n$)

answered May 15 '19 at 14:34

Bill Dubuque

Nice Sir , I like it thank you very much – Ellen Ellen May 15 '19 at 14:36
@user674299 This is essentially the Bezout identity that would result from performing the extended Euclidean algorithm to compute theur gcd. – Bill Dubuque May 15 '19 at 14:39

score 0 · Answer 4 · answered May 15 '19 at 14:39

Let's call d the largest number the numerator and denominator divides.

21n + 4 $\vdots$ d $\Rightarrow$ 2(21n + 4) = 42n + 8 $\vdots$ d

14n + 3 $\vdots$ d $\Rightarrow$ 3(14n + 3) = 42n + 9 $\vdots$ d

That means (42n + 9) - (42n + 8) = 1 $\vdots$ d

Therefore, d must be 1. And so, the fraction is irreducible.

score 0 · Answer 5 · answered May 15 '19 at 14:40

Suppose there is an integer $d$ which divides both the top and the bottom of $$\frac{21n+4}{14n+3}$$

Then $d$ divides the difference which is $7n+1$.

Therefore it divides $21n+3$ and it divides $21n+4$ hence it divides the difference which is $1$

Thus the only positive common divisor is $d=1$

5 Answers5