# Prove that r and s must be irrational.?

sqrt(3)=r+s*sqrt(2)

### 1 Answer

- bimeateaterLv 71 decade agoFavorite Answer
r = ± 3 - √3, ± 1 - √3 so r IS irrational therefore it must be irrational

s = ±√[r^2 + (√3)r + 3/2], ± {√[r^2 - (2√3)r + 3]}/(√2) and substituting the values of r, one always gets an irrational result therefore it must be irrational

For the work? Square each side, move all terms without radicals to the left side, and square again. That rids you of radicals. Solve the resulting quartic equation:

r^4 - 4r^2s^2 - 6r^2 + 4s^4 - 12s^2 + 9 = 0

for either "r" or "s" to get a single variable. Then substitute to solve for a value for the other. That shows that one to be irrational. Then substitute the new results into the earlier solution to find the various possibilities for the other variable. That shows it to always end up irrational.

So both end up irrational.

(If one had other values in the original equation, for example 9 where 3 occurs, it looks like it would turn out that "r" could be rational (not gonna consider past that for "s" because it's not necessary), and so a generalized version of the equation might not always result in "r" and "s" being irrational.)

But they definitely are for this equation.

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