Proving equivalence

[Lazarus]

Well I have to prove the following statements are equivalent:
a. y is a rational number
b. y/3 is a rational number
c. 2y+5 is a rational number
So a -> b -> c -> a

I'm not quite sure how you are suppose to prove something is rational however.
I started like this:

* y = q/r Where q & r are integers, r is not 0, no common factors other than 1

* y/3
q/r/3 ... (q/r)(1/3)= q/3r ... Which is rational because everything is integer math?

* 2y+5 ... 2(q/r) + (5/1) ... same reason

This doesn't seem to work like the ones I've done before with proving something is even (2n) or odd (2n+1). Is there some step I'm not getting?

 

[Denis]

Well I have to prove the following statements are equivalent:
a. y is a rational number
b. y/3 is a rational number
* y = q/r Where q & r are integers, r is not 0, no common factors other than 1

* y/3
q/r/3 ... (q/r)(1/3)= q/3r ... Which is rational because everything is integer math?

If everything is integer as you state, then y divides 3,
so can be represented 3q where q is an integer ;
then y/3 is simply q.

 

[soroban]

Hello, Lazarus!

Prove the following statements are equivalent:

a. \(\displaystyle y\) is a rational number\(\displaystyle \;\;\;\)b. \(\displaystyle \frac{y}{3}\) is a rational number\(\displaystyle \;\;\;\)c. \(\displaystyle 2y\,+\,5\) is a rational number

You have the right idea . . . use the definition of a rational number.

(a) Given: \(\displaystyle y\) is a rational number.
Then \(\displaystyle y\) is the ratio of two integers: \(\displaystyle \frac{m}{n}\)

Dividing \(\displaystyle y\) by 3, we get: \(\displaystyle \,\frac{y}{3}\,=\,\frac{m}{3n}\)
\(\displaystyle \;\;\)Is this a rational number? \(\displaystyle \;\)Is it the ratio of two integers?
The numerator is an integer. \(\displaystyle \;\)(Given)
The denominator \(\displaystyle 3n\) is an integer. \(\displaystyle \;\)(The integers are closed under multiplication.)
\(\displaystyle \;\;\)Therefore, \(\displaystyle \frac{y}{3}\) is a rational number.


(b) Given: \(\displaystyle \frac{y}{3}\) is a rational number.
Then \(\displaystyle \frac{y}{3}\) is the ratio of two integers: \(\displaystyle \frac{p}{q}\)

Multiplying by 6, we get: \(\displaystyle \,6\,\cdot\frac{y}{3} \,=\,2y\,=\,\frac{6p}{q}\)
\(\displaystyle \;\;\)Is this a rational number?
The numerator is an integer. \(\displaystyle \;\)(Integers are closed under multiplication)
And the denominator is an integer.
\(\displaystyle \;\;\)Hence, \(\displaystyle 2y\) is a rational number.

Add 5: \(\displaystyle \,\frac{6p}{q}\,+\,5\:=\:\frac{6p\,+\,5q}{q}\)
\(\displaystyle \;\;\)Is this a rational number?
The numerator is an integer. \(\displaystyle \;\)(Integers are closed under add'n/mult'n)
And the denominator is an integer.

\(\displaystyle \;\;\)Therefore, \(\displaystyle 2y\,+\,5\) is a rational number.


Get the idea? . . . I'll let you do the third part.