prove r is rational

[kory]

I'm trying to write a proof but I'm getting stuck.
I have to prove that if r is rational, then 2/7(3r+5)-9 is rational.

I let r = a/b by definition of rational.
then I substituted r for a/b
after doing some algebra I got down to 2/7(3a+5b/b)-9 but im not sure if my math is correct. I'm not sure where to go from here...

 

[JeffM]

First, what you wrote means [MATH]\dfrac{2}{7} * (3r + 5) - 9[/MATH],

but I suspect you meant [MATH]\dfrac{2}{7(3r + 5) - 9}.[/MATH]
I suggest that you repeat your problem paying careful attention to grouping symbols and order of operations.

[MATH]r \text { is a rational number} \implies \exists \text { integers } a \text { and } b \text { such that } b \ne 0 \text { and } r = \dfrac{a}{b}.[/MATH]
[MATH]\dfrac{2}{7(3r + 5) - 9} = \dfrac{2}{21r + 35 - 9} * 1 = \dfrac{2}{21r + 26} * \dfrac{b}{b} =[/MATH]
[MATH]\dfrac{2b}{21br + 26b} = \dfrac{2b}{21b * \dfrac{a}{b} + 26b} = \dfrac{2b}{21a + 26b}.[/MATH]
Now what?

 

[kory]

First, what you wrote means [MATH]\dfrac{2}{7} * (3r + 5) - 9[/MATH],

but I suspect you meant [MATH]\dfrac{2}{7(3r + 5) - 9}.[/MATH]
I suggest that you repeat your problem paying careful attention to grouping symbols and order of operations.

[MATH]r \text { is a rational number} \implies \exists \text { integers } a \text { and } b \text { such that } b \ne 0 \text { and } r = \dfrac{a}{b}.[/MATH]
[MATH]\dfrac{2}{7(3r + 5) - 9} = \dfrac{2}{21r + 35 - 9} * 1 = \dfrac{2}{21r + 26} * \dfrac{b}{b} =[/MATH]
[MATH]\dfrac{2b}{21br + 26b} = \dfrac{2b}{21b * \dfrac{a}{b} + 26b} = \dfrac{2b}{21a + 26b}.[/MATH]
Now what?

I meant 2/7∗(3r+5)−9...sorry about that

 

[pka]

I meant 2/7∗(3r+5)−9...sorry about that

After expanding we get: \(\dfrac{6r}{7} + \left(\dfrac{10}{7}-9\right)\). Now what does that suggest?

 

[kory]

2/7 * ( 3( a/b )+5 )-9
2/7 * ( 3a/b +5 )-9
2/7 * ( 3a +5b/b ) -9
(6a + 10b /7b) - 9
6a + 10b -63b / 7b
6a - 53b / 7b
is this correct so far?

 

[pka]

2/7 * ( 3( a/b )+5 )-9
2/7 * ( 3a/b +5 )-9
2/7 * ( 3a +5b/b ) -9
(6a + 10b /7b) - 9
6a + 10b -63b / 7b
6a - 53b / 7b
is this correct so far?

Why oh why oh, do students make things so complicated?
Do you know what combinations give rational numbers?

 

[lookagain]

. . .
is this correct so far?

Look at some of these amendments. Some grouping symbols (about nine pairs)
are needed, while the remaining ones are there for emphasis/clarification.
Increased spacing in certain places helps for better readability.

(2/7)[3(a/b) + 5)] - 9 =
(2/7)(3a/b + 5) - 9 =
(2/7)(3a + 5b)/b - 9 =
(6a + 10b)/(7b) - 9 =
(6a + 10b - 63b)/(7b) =
(6a - 53b)/(7b)

 

[lex]

2/7 * ( 3( a/b )+5 )-9
2/7 * ( 3a/b +5 )-9
2/7 * ( 3a +5b/b ) -9
(6a + 10b /7b) - 9
6a + 10b -63b / 7b
6a - 53b / 7b
is this correct so far?

Good. Just a final line of justification needed.
(Also, I suspect proper layout is easier for you when the answer is written out on paper, in normal mathematical format).

For a direct proof at this level, you're probably just looking to show that:

[MATH]\frac{2}{7}\left(3r+5\right)-9[/MATH] can be written as [MATH]\frac{c}{d}[/MATH] with [MATH]c[/MATH] and [MATH]d[/MATH] integers [MATH](d\neq0)[/MATH]
[MATH]r\ [/MATH] is rational so [MATH]r=\frac{a}{b}[/MATH] for some integers [MATH]a, b[/MATH] with [MATH]b\neq0[/MATH]
Then [MATH]\frac{2}{7}\left(3r+5\right)-9[/MATH]
[MATH]=\frac{2}{7}\left(\frac{3a}{b}+5\right)-9[/MATH]
[MATH]=\frac{6a}{7b}+\frac{10}{7}-9[/MATH]
[MATH]=\frac{6a}{7b}-\frac{53}{7}[/MATH]
[MATH]=\frac{6a}{7b}-\frac{53b}{7b}[/MATH]
[MATH]=\frac{6a-53b}{7b}[/MATH]
Since [MATH]a[/MATH] is an integer, then [MATH]6a-53b[/MATH] is also an integer.
Since [MATH]b[/MATH] is an integer and not 0, [MATH]7b[/MATH] is an integer and not 0.

Therefore [MATH]\frac{6a-53b}{7b}[/MATH] is a rational.
(i.e. [MATH]\frac{2}{7}\left(3r+5\right)-9[/MATH] is a rational. Q.E.D.)