Solved: Irrationality of \sqrt{3}

[ehrenfest]

[SOLVED] easy algebra problem

Homework Statement

Why is the following equation impossible:

$$\sqrt{3}=a+b\sqrt{2}$$

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

 

[HallsofIvy]

If 3= a+ b$\sqrt{2}$ then 3- a= b[/itex]\sqrt{2}[/itex] so $\sqrt{2}$= (3- a)/b. a and b are rational numbers so ...

 

[sutupidmath]

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?

 

[sutupidmath]

If 3= a+ b$\sqrt{2}$ then 3- a= b[/itex]\sqrt{2}[/itex] so $\sqrt{2}$= (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?

 

[sutupidmath]

Homework Statement

Why is the following equation impossible:

$$\sqrt{3}=a+b\sqrt{2}$$

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 $$\sqrt{3}$$ is irrational. Square both sides and rearrange to get
$$\frac{3-a^2-2b^2}{2ab}=\sqrt{2}$$
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.

 

[ehrenfest]

Yes, I edited it, sorry. Thanks.

 

[sutupidmath]

1.

The Attempt at a Solution

I can prove that $$\sqrt{3}$$ is irrational. Square both sides and rearrange to get
$$\frac{3-a^2-2b^2}{2ab}=\sqrt{2}$$
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.!