Is 2√(7)+4 an Irrational Number? Exploring the Proof of Its Irrationality

[nycfunction]

See picture for question and answer.

 

[HallsofIvy]

What, exactly, do you mean by "continued decimal number"? If you mean simply that it is not a terminating decimal, a rational number such as 1/3 has that property. That's not the reason $2\sqrt{7}+ 4$ is irrational.
First, the rational numbers are "closed under subtraction and division" so if $x= 2\sqrt{7}+ 4$ were rational so would be $\sqrt{7}= (x- 4)/2$. And if $\sqrt{7}$ were rational then there would exist integers, a and b, with no common factors, such that $\sqrt{7}= \frac{a}{b}$. From that $7= \frac{a^2}{b^2}$ and $a^2= 7b^2$. That is, $a^2$ has a factor of 7 and, since 7 is a prime number, a has a factor of 7. a= 7n for some integer n so $a^2= 49n^2= 7b^2$. Then $b^2= 7n^2$ so, as before, b has a factor of 7, contradicting the fact that a and b has no factors in common.

 

[nycfunction]

What, exactly, do you mean by "continued decimal number"? If you mean simply that it is not a terminating decimal, a rational number such as 1/3 has that property. That's not the reason [tex]2\sqrt{7}+ 4[/tex] is irrational.

First, the rational numbers are "closed under subtraction and division" so if [tex]x= 2\sqrt{7}+ 4[/tex] were rational so would be [tex]\sqrt{7}= (x- 4)/2[/tex]. And if [tex]\sqrt{7}[/tex] were rational then there would exist integers, a and b, with no common factors, such that [tex]\sqrt{7}= \frac{a}{b}[/tex]. From that [tex]7= \frac{a^2}{b^2}[/tex] and [tex]a^2= 7b^2[/tex]. That is, [tex]a^2[/tex] has a factor of 7 and, since 7 is a prime number, a has a factor of 7. a= 7n for some integer n so [tex]a^2= 49n^2= 7b^2[/tex]. Then [tex]b^2= 7n^2[/tex] so, as before, b has a factor of 7, contradicting the fact that a and b has no factors in common.

See attachment.