-4
$\begingroup$

The simple form of my question is given the equation:

$$\frac{a}{b} = \frac{c}{d}$$

Does this imply that: $ a = c$ and $b = d$ (I simply pair the numerators to themselves, and denominators to themselves).

If this is true, is there any basis? If not true, any thoughts or insights on why not?

The 'somewhat complex' form of my question stems from the Bayes Theorem. Given the Bayes Theorem defined as: $$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$ Then with shuffling of the elements, we have: $$\frac{P(A)}{P(B)} = \frac{P(A|B)}{P(B|A)}$$

Again, would this mean that: $P(A) = P(A|B)$ and $P(B) = P(B|A)$

I ask this question because I recently did an example math on conditional probability and noticed the above applied to my example math.

$\endgroup$
9
  • 5
    $\begingroup$ Have you ever looked at $\frac{1}{2}=\frac{2}{4} $ ? $\endgroup$
    – Kurt G.
    Commented May 27, 2022 at 8:39
  • 3
    $\begingroup$ @Aplateofmomos The question is perfectly valid in itself, but it is quite weird that it comes from someone who deals with Bayes theorem and conditional probabilities... Feels like the OP's "castle of maths" is built on sand rather than stone. $\endgroup$
    – TheSilverDoe
    Commented May 27, 2022 at 8:43
  • 1
    $\begingroup$ I would say in some sense mine is too, but even if that is the case, I don't think OP should be punished for knowing more about Baye's theorem when asking a question about arthimetic.. or is the idea to punish a person's study method? Doesn't make sense to me $\endgroup$
    – Cathartic Encephalopathy
    Commented May 27, 2022 at 8:46
  • 1
    $\begingroup$ That's a funny one haha @TheSilverDoe $\endgroup$
    – Joker
    Commented May 27, 2022 at 8:49
  • 2
    $\begingroup$ I am thinking that the problem here is not your (potentially) shaky background in elementary algebra but potential misunderstanding of the (probability) problem. Surely $P(A\mid B)$ is not necessarily the same as $P(A)$. However, there is an important case when it is. If $A$ and $B$ are independent events, then $P(A\mid B):=\frac{P(A\cap B)}{P(B)}=\frac{P(A)P(B)}{P(B)}=P(A)$ (if $P(B)\ne 0$). (In fact, the converse is also true.) So, could it be that, in the example you were studying, the events $A$ and $B$ were meant to be independent? $\endgroup$
    – user700480
    Commented May 27, 2022 at 9:34

1 Answer 1

Reset to default
5
$\begingroup$

No, it doesn't. Example:

$$\frac11 = \frac22$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .