# Answer to Question #3408 in Real Analysis for junel

Question #3408

Show that there does not exist a rational number t such t

^{2}= 3.Expert's answer

__Proof(by contradiction)__

Suppose that t∈Q satisfies t

^{2}= 3. We can write t = p/q for some p,q ∈N so that p and q have no common factors. Therefore

3 = t

^{2}= (p/q)

^{2}= p

^{2}/q

^{2},

p

^{2}= 3q

^{2}

Then 3 divides by p

^{2}.

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 => p

^{2}= 9n

^{2}-6n + 1 = 3(3n -2)n + 1

=> p

^{2}-1 = 9n

^{2}- 6n + 1 = 3(3n -2)n

and hence

p

^{2}-1 is divisible by 3 and therefore p

^{2}is not divisible by 3

a CONTRADICTION. Similarly, if p + 2 = 3n for some n∈N, then

p = 3n -2 , p

^{2}= 9n

^{2}- 12n + 4 = 3((3n- 4)n + 1) + 1 ,

p

^{2}- 1 = 9n

^{2}- 6n + 1 = 3((3n -2)n + 1)

and hence p

^{2}-1 is divisible by 3 and therefore p

^{2}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 3q

^{2}= p

^{2}= (3n)

^{2}= 9n

^{2}, q

^{2}= 3n

^{2}

Then, q

^{2}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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