Respuesta :
Given:
Expression is
[tex]3r^2-2r+4[/tex]
To prove:
If r is any rational number, then [tex]3r^2-2r+4[/tex] is rational.
Step-by-step explanation:
Property 1: Every integer is a rational number. It is Theorem 4.3.1.
Property 2: The sum of any two rational numbers is rational. It is Theorem 4.3.2.
Property 3: The product of any two rational numbers is rational. It is Exercise 15 in Section 4.3.
Let r be any rational number.
We have,
[tex]3r^2-2r+4[/tex]
It can be written as
[tex]3(r\times r)-2r+4[/tex]
Now,
3, -2 and 4 are rational numbers by property 1.
[tex]r^2=r\times r[/tex] is rational by Property 3.
[tex]3r^2\text{ and }-2r[/tex] are rational by Property 3.
[tex]3r^2+(-2r)+4[/tex] is rational by property 2.
So, [tex]3r^2-2r+4[/tex] is rational.
Hence proved.
We want to prove that if r is rational, then:
[tex]3*r^2 - 2r + 4[/tex] Is also rational.
The proof is below:
Remember that a rational number is any number that can be written as a quotient of two integer numbers.
Then we have:
[tex]r = \frac{a}{b}[/tex]
Where a and b are integers.
If we replace that in our equation, we get:
[tex]3*(\frac{a}{b}) ^2 - 2*(\frac{a}{b}) + 4\\\\\frac{3a^2}{b^2} - \frac{2a}{b} + \frac{4}{1}[/tex]
Note that if a is an integer, then 3*a^2 is also an integer.
If b is an integer, then b^2 is also an integer.
Now we can perform the sum of these fractions to get:
[tex]\frac{3a^2}{b^2} - \frac{2a}{b} + \frac{4}{1}\\\\\frac{3a^2 - 2ab + 4b^2}{b^2}[/tex]
And because integers are closed under the addition, then the numerator of that fraction is also an integer, then we have that the given expression can be written as a quotient of two integers, thus, it is a rational number.
If you want to learn more, you can read:
https://brainly.com/question/15815501
