Solved: Irrationality of \sqrt{3}

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In summary, the conversation discusses the impossibility of the equation \sqrt{3}=a+b\sqrt{2}, where a and b are rational numbers and b is not 0. The conversation uses group, ring, or field theory to explain that this equation is impossible because the rationals form a field which is closed under addition, multiplication, and division. The conversation also discusses the irrationality of \sqrt{3} and provides a proof for it.
  • #1
ehrenfest
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[SOLVED] easy algebra problem

Homework Statement


Why is the following equation impossible:

[tex]\sqrt{3}=a+b\sqrt{2}[/tex]

where a and b are rational numbers and b is not 0. It seems so obvious... Feel free to use group, ring, or field theory in your answer.

Homework Equations


The Attempt at a Solution


EDIT: I can prove that [tex]\sqrt{3}[/tex] is irrational. Square both sides and rearrange to get
[tex]\frac{3-a^2-2b^2}{2ab}=\sqrt{2} [/tex]
which is impossible because the rationals form a field (which is closed under addition, multiplication, and division).
 
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  • #2
If 3= a+ b[itex]\sqrt{2}[/itex] then 3- a= b[/itex]\sqrt{2}[/itex] so [itex]\sqrt{2}[/itex]= (3- a)/b. a and b are rational numbers so ...
 
  • #3
well, i can prove that any rational(-non-zero) number b times sqrt(2) is irrational. Then again the sum of a rational number and an irrational one is irrational. So here we have the left side of the eq a rational number while the left side is an irrational one, which is not possible.
Am i even close?
 
  • #4
HallsofIvy said:
If 3= a+ b[itex]\sqrt{2}[/itex] then 3- a= b[/itex]\sqrt{2}[/itex] so [itex]\sqrt{2}[/itex]= (3- a)/b. a and b are rational numbers so ...

Is he asking for an answer, or he is just testing some of us? i got the feeling that the op already knows the answer, but rather wants to see who knows the reason, or am i wrong?
 
  • #5
ehrenfest said:

Homework Statement


Why is the following equation impossible:

[tex]\sqrt{3}=a+b\sqrt{2}[/tex]

where a and b are rational numbers and b is not 0. It seems so obvious... Feel free to use group, ring, or field theory in your answer.


Homework Equations





The Attempt at a Solution


I can prove that [tex]\sqrt{3}[/tex] is irrational. Square both sides and rearrange to get
[tex]\frac{3-a^2-2b^2}{2ab}=\sqrt{2} [/tex]
which is impossible because the rationals form a field (which is closed under addition, multiplication, and division).


Well, you edited it now, right, because before ther was a plain 3, while now ther is the square root of 3.
 
  • #6
Yes, I edited it, sorry. Thanks.
 
  • #7
ehrenfest said:
1.



The Attempt at a Solution


I can prove that [tex]\sqrt{3}[/tex] is irrational. Square both sides and rearrange to get
[tex]\frac{3-a^2-2b^2}{2ab}=\sqrt{2} [/tex]
which is impossible because the rationals form a field (which is closed under addition, multiplication, and division).


Well, yeah this loos fine to me, since a,b are rationals then they will also be rationals when squared, also -,+,* in the field of rationals are closed operations, so on the left side you have a rational nr. while on the right an irrational. this seems to me like a good contradiction, let's see what halls have to say.!
 

Related to Solved: Irrationality of \sqrt{3}

What is the irrationality of √3?

The irrationality of √3 refers to the fact that the number √3 cannot be expressed as a ratio of two integers. In other words, it cannot be written as a fraction with a finite number of digits after the decimal point.

How was the irrationality of √3 proven?

The irrationality of √3 was first proven by the ancient Greek mathematician, Pythagoras. He used a geometric proof to show that the length of the hypotenuse of a right triangle with sides of length 1 is an irrational number.

Why is the irrationality of √3 important?

The irrationality of √3 is important because it is one of the first numbers to be proven irrational and it has many practical applications in mathematics and science. It also helped pave the way for the discovery of other irrational numbers.

Can the irrationality of √3 be generalized to other square roots?

Yes, the proof for the irrationality of √3 can be generalized to other square roots. In fact, it can be extended to show that the square root of any non-perfect square is irrational.

Is there a pattern to the digits of √3?

No, there is no discernible pattern to the digits of √3. This is because it is an irrational number, meaning it has an infinite number of non-repeating digits after the decimal point.

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