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Answer to Question #3408 in Real Analysis for junel

Question #3408
Show that there does not exist a rational number t such t2 = 3.
Expert's answer
Proof(by contradiction)
Suppose that t∈Q satisfies t2 = 3. We can write t = p/q for some p,q ∈N so that p and q have no common factors. Therefore
3 = t2= (p/q)2= p2/q2 ,
p2= 3q2
Then 3 divides by p2.
CLAIM: Show that 3 divides p. Prooceed by contradiction. It’s known that
either p, p + 1 or p + 2 is divisible by 3. Suppose that p + 1 = 3n for some n∈N.
Then
p = 3n -1 => p2= 9n2-6n + 1 = 3(3n -2)n + 1
=> p2-1 = 9n2- 6n + 1 = 3(3n -2)n
and hence
p2 -1 is divisible by 3 and therefore p2 is not divisible by 3
a CONTRADICTION. Similarly, if p + 2 = 3n for some n∈N, then
p = 3n -2 , p2= 9n2- 12n + 4 = 3((3n- 4)n + 1) + 1 ,
p2- 1 = 9n2- 6n + 1 = 3((3n -2)n + 1)
and hence p2 -1 is divisible by 3 and therefore p2 is not divisible by 3 ,a CONTRADICTION. Therefore, p must be divisible by 3. End of claim.
We can write (since 3 divides p) p = 3n; for some n∈N:
Therefore 3q2= p2= (3n)2 = 9n2 , q2= 3n2
Then, q2 is divisible by 3. By claim above, q must also be divisible by 3.
CONTRADICTION -
we assumed at the beginning that p and q have no common factors, but found here that both are divisible by 3.
Consequently, it must follow that t ∈Q (t is not a rational number).

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