Jacoby Erickson

2022-10-11

For which natural numbers a fraction can be simplifed

I want to know how to check for which natural numbers fraction $\frac{7n+19}{7n+2}$ can be simplified. The correct answer is supposed to be $n=2k-1$

I do understand that the denominator and the numerator have to have a common divisor in order to simplify the fraction, but fail to see how this particular problem should be approached.

I want to know how to check for which natural numbers fraction $\frac{7n+19}{7n+2}$ can be simplified. The correct answer is supposed to be $n=2k-1$

I do understand that the denominator and the numerator have to have a common divisor in order to simplify the fraction, but fail to see how this particular problem should be approached.

Alannah Yang

Beginner2022-10-12Added 22 answers

$\frac{7n+19}{7n+2}=1+\frac{17}{7n+2}$

Thus, in order for $\frac{7n+19}{7n+2}$ to be simplified, we have that $\frac{17}{7n+2}$ can be simplified as well.

This implies that $17$ and $7n+2$ must have a common divisor that is not $1$. Since $17$ is prime, this implies that

$7n+2\equiv 0\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}17)$

This gives us that

$n\equiv 7\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}17)$

Thus, the answer is that $n=17k+7$ where $k\in \mathbb{Z}$, not $2k-1$ as stated, so your answer is incorrect.

Thus, in order for $\frac{7n+19}{7n+2}$ to be simplified, we have that $\frac{17}{7n+2}$ can be simplified as well.

This implies that $17$ and $7n+2$ must have a common divisor that is not $1$. Since $17$ is prime, this implies that

$7n+2\equiv 0\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}17)$

This gives us that

$n\equiv 7\phantom{\rule{1em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}17)$

Thus, the answer is that $n=17k+7$ where $k\in \mathbb{Z}$, not $2k-1$ as stated, so your answer is incorrect.

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