Square root of 3 is irrational

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SUMMARY

The discussion centers on proving that the square root of 3 is irrational, paralleling the proof for the square root of 2. The user initially assumes sqrt(3) = p/q and derives the equation 3q^2 = p^2. Unlike the case for sqrt(2), the user notes that they cannot conclude that p^2 is even. The key insight is that if p^2 = 3q^2, then both p and q must be multiples of 3, leading to a contradiction, thereby establishing the irrationality of sqrt(3).

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  • Understanding of rational and irrational numbers
  • Familiarity with proof by contradiction
  • Basic algebraic manipulation
  • Knowledge of properties of even and odd integers
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  • Study the proof of the irrationality of sqrt(2) in detail
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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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