1

$a, b, c, d, e, f$ are natural numbers. $a$ and $b$ are co-prime(relatively prime). If this equation is true:

$$ a ^ c \times b ^ d = a ^ e \times b ^ f$$ Then does this mean that both these equations are always true: $$c = e ,d = f$$ ? Is it always true? What would the proof for it be? Would this be true if there were more terms?

Delta-u
  • 6,038
BorticFan
  • 13

2 Answers2

1

For natural numbers, yes it is always true. And the simple word 'proof' is that every natural number has exactly one factorization. So if you have the same primes, you will have to have same exponents to those primes, otherwise you would have more than one factorization. So there is only one solution to that equation, and therefore c=e and d=f.

It is not completely true though, if the exponents are allowed to be complex numbers, but I think it's out of scope of your question :)

KKZiomek
  • 4,012
  • Do you mean as in fundamental theorem of arithmetic? – BorticFan Apr 27 '18 at 17:41
  • The fundamental thing that you should know is that every natural positive number has exactly one prime factorization. If c was not equal to e and/or d wasnt equal to f, then it would mean that there are two different factorizations for some unknown number. But we know this ksnt true, so c must equal e and d must equal f, if we assume that both sides of the equation can be interpreted as a prime factorization of some unknown number – KKZiomek Apr 27 '18 at 17:44
  • I probably didnt explain it clearly at first. If a and b are prime, then $a^cb^d$ looks like a prime factorization of some number. Thats where i got this idea from – KKZiomek Apr 27 '18 at 17:46
0

We can simplify this by focussing on $a$:

$$k\cdot a ^ c = m\cdot a ^ e $$

with $k$ and $m$ coprime to $a$. Then the prime powers that contribute to $a^c$ on the left hand side can only be part of $a^e$ on the right hand side, so we must have $a^c=a^e$ (and of course $k=m$) and the rest of the analysis is similar, including if we have more pairwise coprime numbers to consider.

Joffan
  • 39,627