proof that root of three is irrational?

Answer 1

suppose it is not irrational. So it is rational.

Now, every rational number can be written as a quotient of two integers with no common divisor other than 1 between the numerator and denominator.

Thus, a/b = sqrt(3)
or a^2/b^2 = 3, by squaring both sides. Now,

a^2 = 3 b^2 which means that 3 divides a^2 and hence it also divides a. then a= 3k and a^2 = 9k^2

Thus, 9k^2 = 3 b^2 or b^2 = 3 k^2. Thus, 3 divides b^2 and b also.
Thus, 3 divides both a and b, which contradicts our assumption that there is no common factor dividing both a and b except for 1.

Thus, our assumption that sqrt(3) can be written as a quotient of two integers or that it is rational is wrong.

Therefore, it is irrational as was wished to be proved.

Answer 2

Assume that √(3) is rational then show that this leads to a contradiction.

√(3) is rational.
√(3) = a/b for some integers a and b. Also assume that a/b is in reduced form. i.e no common factors in a and b.
square both sides.
3 = a^2 / b^2
multiply both sides by b^2
3*b^2 = a^2
Since there's a 3 on the left, 3 must be a factor of a. so a^2 can be written as (3c)^2 = 3^2 * c^2 for some integer c.
3 * b^2 = 3^2 * c^2
Since there's a 3^2 on the right there needs to be another factor of 3 on the left. So b has 3 as a factor. But we already said that a and b don't have a common factor. Which gives us a contradiction.

Answer 3

This one is a classic. The most common proof is by contradiction.

Suppose sqrt(3) were rational. Then sqrt(3) = p/q, for some integers p and q, such that p/q is in lowest terms. In other words, p and q have no common factors except 1.

Squaring both sides of the equation, we get 3 = p^2/q^2. Multiplying by q^2, we get 3q^2 = p^2. Since q^2 is an integer, we can conclude that p^2 is divisible by 3. Since 3 is prime, we can also conclude that p is divisible by 3. Furthermore, since p is divisble by 3, p^2 must be divisble by 9.

So divide the equation by 9 to get (q^2)/3 = (p^2)/9. We know that (p^2)/9 is an integer since p^2 is divisible by 9. If the right side of the equation is an integer, so must be the left side. So (q^2)/3 is an integer. This means that q^2 is divisible by 3. Again, since 3 is prime, we can conclude that q is divisible by 3.

We have now shown that p and q are both divisible by 3. But we defined p/q to be in lowest terms, so that they share NO common factors. The common factor of 3 is a contradiction. Thus it cannot be true that sqrt(3) is a rational number. The only alternative is that sqrt(3) is irrational.

Answer 4

Let us assume that the √3 is rational.

Then there must be integers m, n such that they are relatively prime, (they do not share any common factors except 1), and m/n = √3. We also know that √3 is not an integer, so n =/= 1

If we square both sides, we get:
(m/n)^2 = (√3)^2
m^2 / n^2 = 3

Since m^2 and n^2 do not share any factors, then m^2 / n^2 cannot be reduced any further. Moreover, n =/= 1, so n^2 =/= 1, which implies that m^2 / n^2 is not an integer. But we know that 3 is indeed an integer, so our initial assumption has led us to a contradiction, so √3 cannot be rational.

Therefore, √3 is irrational.

Answer 5

Assume sqrt(3) is rational.
Then sqrt(3) = a/b where b <> 0, and a/b is in lowest form.
Then 3 = a^2/b^2
If a is even, then a^2 is even and so 3b^2 =a^2 is even.
If 3b^2 is even, then b can be either odd or even.
If both a and b even, then a/b is not inits lowest form as was assumed. Hence our assumption that 3 is rational is false.