Question

prove that root 3 is irrational number​

Answer 1

Answer:

it cannot be expressed in the form of p/q where p,q are rational, hence it is irrational

Step-by-step explanation:

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Answer 2

I'm trying to do this proof by contradiction. I know I have to use a lemma to establish that if x is divisible by 3, then x2 is divisible by 3. The lemma is the easy part. Any thoughts? How should I extend the proof for this to the square root of 6?

Say √3 is rational. Then √3 can be represented as a/b, where a and b have no common factors.

So 3=a²/b²

and 3b²=a².

Now a² must be divisible by 3, but then so must a (fundamental theorem of arithmetic).

So we have 3b²=(3k)² and 3b²=9k² or even b²=3k² and now we have a contradiction.

What is the contradiction?

suppose √3 is rational,

then √3 = ab for some (a,b) suppose we have a/b in simplest form.

√3=a/b

= a²=3b²

if b is even, then a is also even in which case a/b is not in simplest form.

if b is odd then a is also odd.

Therefore:

a = (2n+1)

or, b= (2m+1)

or,. (2n+1)² = 3(2m+1)²

or, 4n² + 4n + 1 = 12m² + 12m + 3

or, 2n² + 2n = 6m² + 6m + 1

or, 2( n²+ n) = 2( 3m² + 3m) + 1

Since (n^2+n) is an integer, the left hand side is even. Since (3m^2+3m) is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.