Question

how to prove that the product of two rational numbers is rational?

Answer 1

via direct proving

Answer 2

i suppose the start would be

p = two numbers are rational
q = their product is rational

Answer 3

for direct proving i assume both are true yes?

Answer 4

so if i assume p is true, i let a/b be the first number and c/d be the second number where a/b and c/d are in simplest terms

Answer 5

then i would assume q is true

Answer 6

so \[\frac ab \times \frac cd = \frac ef\]
where e/f is in simplest form

Answer 7

so if i simplify this... i get \[\frac {ac}{bd} = \frac ef\]

Answer 8

I think it is simpler than that, assuming you are just working from the definition of rational.

First is a/b, second is c/d, product is ac/bd which is rational (by definition)

Answer 9

how is ac/bd rational by definition?

Answer 10

Well, if you want to state that an integer multiplied by an integer is an integer, you can, I suppose.

Answer 11

i'll have to prove ac/bd is in simplest terms for it to be considered rational right?

Answer 12

No, as long as it can be expressed as integer over integer it's rational.

Answer 13

but the proving of sqrt 2 as irrational arrives with an integer/integer but it is still considered irrational because it wasn't in simplest form

Answer 14

That's a different question altogether because you can only assume that sqrt 2 is rational, you don't know that it is.

Answer 15

And the point of that proof is to arrive at a contradiction.

Answer 16

how is it any different?

Answer 17

Definition of rational is integer over integer and you are given that your starting point is rational.

Answer 18

so tell me why sqrt 2 is irrational if its proof arrived with an integer/integer solution?

Answer 19

You assume rationality and then show that that leads to a contradiction, therefore the assumption is wrong and so sqrt 2 is irrational.

Answer 20

why is it contradiction?

Answer 21

the last step looks like a/b wherea and b are even right?

Answer 22

they're still both integers....but just because it's even/even it's not rational...

Answer 23

http://www.homeschoolmath.net/teaching/proof_square_root_2_irrational.php

Let's look at a proof so that we know we are talking about the same thing...

Answer 24

In the box....

Answer 25

ah yes. it said specifically....forgot about that part

Answer 26

the proof i know of just said a/b is in simplest form...didn't say anything about even/even

so it's just contradiction because a/b is even/even and yet simplest form?

Answer 27

Yes, that is impossible because you could just divvide by 2

Answer 28

Here is a different proof of irrationality of sqrt 2

http://www.math.hmc.edu/funfacts/ffiles/30005.5.shtml