Is 2√(7)+4 an Irrational Number?

[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 $$2\sqrt{7}+ 4$$ 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.

See attachment.