Logarithm properties in functional equations Show that f(1/a) = -f(a)

[My Name is Earl]

Suppose that we have a function f(x) such that f(ab) = f(a)+f(b) for all rational numbers a and b.
(a) Show that f(1) = 0.
(b) Show that f(1/a) = -f(a).
(c) Show that f(a/b) = f(a) - f(b).
(d) Show that f(aⁿ) = nf(a) for every positive integer a.

For (a), if ab = 1 then a = 1/b and b = 1/a. Not sure how to proceed from here.

 

[MarkFL]

Suppose that we have a function f(x) such that f(ab) = f(a)+f(b) for all rational numbers a and b.
(a) Show that f(1) = 0.
(b) Show that f(a/b) = -f(a).
(c) Show that f(a/b) = f(a) - f(b).
(d) Show that f(aⁿ) = nf(a) for every positive integer a.

For (a), if ab = 1 then a = 1/b and b = 1/a. Not sure how to proceed from here.

For part (a), suppose we let \(b=1\)...what does this imply from the given definition of \(f\)?

 

[MarkFL]

I think part (b) should be:

Show that:

\(\displaystyle f\left(\frac{1}{a}\right)=-f(a)\)

For this, using what you stated for part (a), that is to let \(ab=1\), along with the result from part (a), will be a good strategy.

 

[My Name is Earl]

I think part (b) should be:

Show that:

\(\displaystyle f\left(\frac{1}{a}\right)=-f(a)\)

For this, using what you stated for part (a), that is to let \(ab=1\), along with the result from part (a), will be a good strategy.

You are correct - fixed it in original post. Thank you.

 

[My Name is Earl]

For part (a), suppose we let \(b=1\)...what does this imply from the given definition of \(f\)?

If b = 1 then f(a) = f(a) + f(1) therefore f(1) = 0. Thanks

 

[My Name is Earl]

I think part (b) should be:

Show that:

\(\displaystyle f\left(\frac{1}{a}\right)=-f(a)\)

For this, using what you stated for part (a), that is to let \(ab=1\), along with the result from part (a), will be a good strategy.

From part a: If ab = 1, then f(1) = 0 and b = 1/a
Substituting these values into f(ab) = f(a) + f(b)
0 = f(a) + f(1/a)
f(1/a) = -f(a)

 

[MarkFL]

From part a: If ab = 1, then f(1) = 0 and b = 1/a
Substituting these values into f(ab) = f(a) + f(b)
0 = f(a) + f(1/a)
f(1/a) = -f(a)

Excellent! Now, for part (c), I would begin by using the definition of \(f\) to write:

\(\displaystyle f\left(\frac{a}{b}\right)=f\left(a\cdot\frac{1}{b}\right)=f(a)+f\left(\frac{1}{b}\right)\)

Now, use the result from part (b)...what do you get?

 

[My Name is Earl]

Excellent! Now, for part (c), I would begin by using the definition of \(f\) to write:

\(\displaystyle f\left(\frac{a}{b}\right)=f\left(a\cdot\frac{1}{b}\right)=f(a)+f\left(\frac{1}{b}\right)\)

Now, use the result from part (b)...what do you get?

Let \(\displaystyle b = \frac{1}{b}\), then substituting into f(ab) = f(a) + f(b):
\(\displaystyle f(a\cdot\frac{1}{b}) = f(a) + f(\frac{1}{b})\)
From part (b) it follows that \(\displaystyle f(\frac{1}{b}) = -f(b)\), therefore
\(\displaystyle f(a\cdot\frac{1}{b}) = f(a) - f(b)\)

 

[MarkFL]

I would use an inductive prove for part (d). Consider the induction hypothesis \(P_n\):

\(\displaystyle f\left(a^n\right)=nf(a)\) where \(n\in\mathbb{N}\)

We see the base case \(P_1\) is true, because it leads to the identity:

\(\displaystyle f(a)=f(a)\)

So, applying the given rule for the function and \(P_n\), we can state:

\(\displaystyle f\left(a^{n+1}\right)=f\left(a\cdot a^{n}\right)=\,?\)

 

[My Name is Earl]

I would use an inductive prove for part (d). Consider the induction hypothesis \(P_n\):

\(\displaystyle f\left(a^n\right)=nf(a)\) where \(n\in\mathbb{N}\)

We see the base case \(P_1\) is true, because it leads to the identity:

\(\displaystyle f(a)=f(a)\)

So, applying the given rule for the function and \(P_n\), we can state:

\(\displaystyle f\left(a^{n+1}\right)=f\left(a\cdot a^{n}\right)=\,?\)

continuing from above...

\(\displaystyle = f(a) + f(a^{n})\)
\(\displaystyle = f(a) + n\cdot f(a)\)
\(\displaystyle = (n+1) \cdot f(a)\)

 

[MarkFL]

continuing from above...

\(\displaystyle = f(a) + f(a^{n})\)
\(\displaystyle = f(a) + n\cdot f(a)\)
\(\displaystyle = (n+1) \cdot f(a)\)

Yes good...and so you have derived \(P_{n+1}\) from \(P_n\), completing the proof by induction. (Yes)