Question

Show that there does not exist a rational number t such t[sup]2[/sup] = 3.

Answer 1

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).