How to prove???

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poiuy

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How can I prove that \(\displaystyle \sqrt{5}+\sqrt{6}+\sqrt{7}\) is irrational ?
 
poiuy said:
How can I prove that \(\displaystyle \sqrt{5}+\sqrt{6}+\sqrt{7}\) is irrational ?
Click to expand...
Do you know how to prove that \(\displaystyle \sqrt{5}+\sqrt{6}\) is irrational?
 
poiuy said:
How can I prove that \(\displaystyle \sqrt{5}+\sqrt{6}+\sqrt{7}\) is irrational ?
Click to expand...

Hint: Sum of two non-equal positive irrational numbers is irrational.

Let p and q be positive integers, and let
x = sqrt(p) + sqrt(q). Assume that x is a rational number and \(\displaystyle X \ne \ 0\).

Then x * (sqrt(p) - sqrt(q)) = p - q, hence
(p-q)/x = sqrt(p) - sqrt(q).

Adding these equations gives x + (p-q)/x = 2 sqrt(p).
But x + (p-q)/2 is rational, so 2 sqrt(p) is rational.
Therefore p is the square of an integer.

Similarly, if we subtract the equations then we obtain
x - (p-q)/x = 2 sqrt(q), which implies that q is the square
of an integer.
 
I think so pka :),
k is rational number
\(\displaystyle \sqrt{5}+\sqrt{6}=k\)
\(\displaystyle 11+2\sqrt{30}=k^{2}\)
\(\displaystyle \sqrt{30}=\frac{k^{2}-11}{2}\)
so i wrote \(\displaystyle \sqrt{30}\) asa rational numberwhich is a contradiction, so k must be irational, is it well?
 
Last edited:
poiuy said:
I think so pka :),
k is rational number
\(\displaystyle \sqrt{5}+\sqrt{6}=k\)
\(\displaystyle 11+2\sqrt{30}=k^{2}\)
\(\displaystyle \sqrt{30}=\frac{k^{2}-11}{2}\)
so i wrote \(\displaystyle \sqrt{30}\) asa rational numberwhich is a contradiction, so k must be irational, is it well?
Click to expand...
Well just expand that idea to three irrational numbers.
 
ok :)
\(\displaystyle \sqrt{5}+\sqrt{6}+\sqrt{7}\)
\(\displaystyle 18+2\sqrt{30}+2\sqrt{35}+2\sqrt{42}=k^{2}\)
\(\displaystyle \sqrt{30}+\sqrt{35}+\sqrt{42}=\frac{k^{2}-12}{2}\)
and I did as before, but I have a question:),is it enough?because on one side I have a sum of three elements
 
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