Question

if f is a factor of n then n/f is also a factor of n is it true or false if true give reason pelase and if false then give counter example thanks.btw i am in 7 grade please give answer on that behalf....please tellllll

Answer 1

Let,

$\longrightarrow n=fg$

where $n,\ f,\ g$ are positive integers.

This implies $f$ and $g$ are factors of $n.$

From the equation we get,

$\longrightarrow g=\dfrac{n}{f}$

Since $g$ is a factor of $n,$ so is $\dfrac{n}{f}.$

Thus the statement is true.

Let us check an example.

We know 4 is a factor of 20.

$\longrightarrow 20=4\times5$

We get 5 is also a factor of 20.

From this we get,

$\longrightarrow \dfrac{20}{4}=5$

Since 5 is a factor of 20, so is $\dfrac{20}{4}.$

Thus the statement is true.

Assume $\dfrac{n}{f}$ is not a factor of a positive integer $n.$ Let $\dfrac{n}{f}$ be a positive integer.

If a positive integer is not a factor of another positive integer, then their quotient (larger divided by smaller) won't be a positive integer.

E.g.: 3 is not a factor of 20. Therefore 20/3 is not a positive integer.

Thus we get that $\dfrac{n}{\left(\dfrac{n}{f}\right)}=f$ is not positive integer.

This implies $f$ can't be a factor of any positive integer.

This implies $f$ is not a factor of $n.$

Now we've proved by contradiction that the given statement is true.

Answer 2

False, because

${f}^{-1}f$

Here does not denote the f raise to the power -1, rather it is a inverse of a function which means when you put certain values in inverse function, you will get the pre-image of a function.

For instance,

Suppose

$f(x)={x}^{2}$

$Then \: it's {f}^{-1}(x)=\sqrt{x}$

Now, let feed the values.

$f(6)={6}^{2}=366$

${f}^{-1}(6)={\sqrt{6}}≠\frac{1}{f(6)}$

Because

$\frac{1}{f(6)}=\frac{1}{36}$

This is the best possible answer