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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I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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