Hi everyone, first post (woohoo)!

Anyways, on to the math.

I am reading a book on propositional logic and proof building to help me in my writing of proofs.

I ran into this problem and the author uses what he calls "the arithmetic rule for the sum of two quotients" and he is using it to rearrange a problem. I googled it, but I could not find anything!

Here is the problem:

.1n+.1=a/b+.1 and using this rule, he rearranged it to: a/b+1/10=(10a+b)/10b

Please explain how he got this, I would be most grateful!

PS: There is one more problem somewhere in the book that he used it..if this above is not substantial please tell me!

Originally Posted by Principia11
Here is the problem:
.1n+.1=a/b+.1 and using this rule, he rearranged it to: a/b+1/10=(10a+b)/10b
This one of the strangest post I have seen.
Surely anyone who would ask this question knows that $\displaystyle 0.1=\frac{1}{10}~?$

Originally Posted by Plato
This one of the strangest post I have seen.
Surely anyone who would ask this question knows that $\displaystyle 0.1=\frac{1}{10}~?$
I am asking how the person rearranged it. That first equation is not the entire problem, but is the first equation that the author goes on to rearrange. I want to know how he rearranged it the way he did.

I am confused about where the n went and how he arranged the a/b's.

"Use mathematical induction to show that all natural-number multiples of .1 are rational numbers"

The first element of this set, 0, is rational, because it can be expressed in the form a/b where "a" is an integer and "b" is a nonzero natural number. Simply let a = 0 and b = 1.

We know that there exists some integer "a" and some nonzero natural number "b" such that .1n = a/b

Therefore we can rewrite the above expression as .1n + .1 = a/b +.1

and then the author says: "Using the arithmetic rule for the sum of two quotients we can rearange the above as follows:

a/b +1/10 = (10a+b)/10b

$\displaystyle \displaystyle \frac{a}{b}+\frac{c}{d}= \frac{ad+cb}{bd}$
$\displaystyle \displaystyle \frac{a}{b}+\frac{c}{d}= \frac{ad+cb}{bd}$