MHB Is 2√(7)+4 an Irrational Number?

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The discussion centers on proving that the expression 2√(7) + 4 is irrational. It explains that if this expression were rational, then √(7) would also be rational, leading to a contradiction. The argument uses the properties of rational numbers and prime factorization to demonstrate that both integers a and b would have a common factor of 7, which contradicts their definition as having no common factors. The initial confusion about the term "continued decimal number" is clarified, emphasizing that the nature of the decimal representation is not the reason for the irrationality of the expression. Ultimately, the conclusion is that 2√(7) + 4 is indeed an irrational number.
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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.
 
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HallsofIvy said:
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.

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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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