Square root of 3 is irrational

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The discussion focuses on proving that the square root of 3 is irrational, paralleling the proof for the square root of 2. The initial approach involves assuming sqrt(3) can be expressed as a fraction p/q and squaring both sides, leading to the equation 3q^2 = p^2. Unlike the case with sqrt(2), this does not directly show that p and q are even. Instead, it reveals that both p and q must be multiples of 3, suggesting a different approach is needed. The conclusion emphasizes that proving the irrationality of sqrt(3) may require considering divisibility by 3 rather than evenness.
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I am trying to prove sqrt(3) is irrational. I figured I would do it the same way that sqrt(2) is irrational is proved:
Assume sqrt(2)=p/q
You square both sides.
and you get p^2 is even, therefore p is even.
Also q^2 is shown to be even along with q.
This leads to a contradiction.
However doing this with sqrt(3), you get 3q^2=p^2. Now you can't show p^2 is even.Also now I am stuck. How do I continue from here?
 
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So maybe when going from 2 to 3, you don't need to use evenness but 'divisible by three' ;)
 
Thanks jacobrhcp. If p^2 =3q^2, then p must be a multiple of 3. The same goes for q^2 and q, both are multiples of 3.
 
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