MHB Is 2√(7)+4 an Irrational Number?

  • Thread starter Thread starter nycfunction
  • Start date Start date
  • Tags Tags
    Rational
AI Thread Summary
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.
nycfunction
Messages
12
Reaction score
0
See picture for question and answer.
 

Attachments

  • MathMagic190527_1.png
    MathMagic190527_1.png
    11.5 KB · Views: 109
Mathematics news on Phys.org
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.
 
Last edited by a moderator:
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.

See attachment.
 

Attachments

  • MathMagic190527_2.png
    MathMagic190527_2.png
    12.1 KB · Views: 104
Last edited:
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Back
Top