Prove that the number #(sqrt(2)+sqrt(17)-sqrt(19))# is irrational?
2 Answers
Here's a proof based on the rational roots theorem...
Explanation:
Consider the octic polynomial with zeros:
#sigma_1sqrt(2)+sigma_2sqrt(17)+sigma_3sqrt(19)#
where
We find:
#(x+sigma_1sqrt(2)+sigma_2sqrt(17)-sqrt(19))(x+sigma_1sqrt(2)+sigma_2sqrt(17)+sqrt(19))#
#=(x+sigma_1sqrt(2)+sigma_2sqrt(17))^2-19#
#=(x^2+2sigma_1sqrt(2)x)+sigma_2(2sqrt(17)x+2sigma_1 sqrt(34))#
Then:
#((x^2+2sigma_1sqrt(2)x)-(2sqrt(17)x+2sigma_1 sqrt(34)))((x^2+2sigma_1sqrt(2)x)+(2sqrt(17)x+2sigma_1 sqrt(34)))#
#=(x^2+2sigma_1sqrt(2)x)^2-(2sqrt(17)x+2sigma_1 sqrt(34))^2#
#=x^4+4sigma_1sqrt(2)x^3+8x^2-68x^2-136sigma_1sqrt(2)x-136#
#=(x^4-60x^2-136)+sigma_1(4sqrt(2)x^3-136sqrt(2)x)#
Then:
#((x^4-60x^2-136)-(4sqrt(2)x^3-136sqrt(2)x))((x^4-60x^2-136)+(4sqrt(2)x^3-136sqrt(2)x))#
#=(x^4-60x^2-136)^2-(4sqrt(2)x^3-136sqrt(2)x)^2#
#=x^8-152x^6+5504x^4-20672x^2+18496#
By the rational roots theorem, any rational zero of this octic is expressible in the form
That means that the only possible rational zeros are the integer factors of
Now
Since this is not an integer, it is not a rational zero of the octic.
Alternative proof using the irrationality of
Explanation:
Suppose:
#sqrt(2)+sqrt(17)-sqrt(19) = c#
Then:
#sqrt(2)+sqrt(17)-c = sqrt(19)#
Square both sides to get:
#19+2sqrt(34)-2csqrt(2)-2csqrt(17)+c^2 = 19#
Subtract
#2sqrt(34)-2csqrt(2)-2csqrt(17)+c^2 = 0#
That is:
#(2sqrt(2)-2c)sqrt(17) = 2csqrt(2)-c^2#
So:
#sqrt(17) = (2csqrt(2)-c^2)/(2sqrt(2)-2c)#
Square both side to get:
#17 = (8c^2-4sqrt(2)c^3+c^4)/(8-8sqrt(2)c+4c^2)#
Multiply both sides by
#136-136sqrt(2)c+68c^2 = 8c^2-4sqrt(2)c^3+c^4#
So:
#(4c^3-136c)sqrt(2) = c^4-60c^2-136#
So:
#sqrt(2) = (c^4-60c^2-136)/(4c^3-136c)#
Now we know that