How do you determine weather the square root of 3 is irrational
To determine whether the square root of 3 is irrational, we need to determine if it can be expressed as a fraction of two integers (a rational number) or not.
Assume, for contradiction, that the square root of 3 is rational. This means it can be written as the fraction a/b, where a and b are integers that have no common factors other than 1 and b is not equal to 0.
Taking the square of both sides, we get:
(sqrt(3))^2 = (a/b)^2
3 = (a^2)/(b^2)
Cross multiplying, we get:
3b^2 = a^2
This implies that a^2 is a multiple of 3. As a result, a must also be a multiple of 3. Let a = 3k, where k is an integer.
Substituting back into the equation, we get:
3b^2 = (3k)^2
3b^2 = 9k^2
b^2 = 3k^2
This implies that b^2 is also a multiple of 3, and therefore b must also be a multiple of 3.
Now, we have a contradiction, because both a and b are multiples of 3. However, we initially assumed that a and b have no common factors other than 1, which contradicts our assumption that the square root of 3 is rational.
Hence, we conclude that the square root of 3 is irrational.
To determine whether the square root of 3 is irrational, we can use a proof by contradiction.
Assume that the square root of 3 is rational, which means it can be expressed as a fraction in the form p/q, where p and q are integers and q is not equal to zero. We can also assume that p/q is simplified, meaning that p and q have no common factors other than 1.
Now, we can square both sides of the equation to get:
√3 = p/q
Squaring both sides gives us:
3 = (p/q)^2 = p^2/q^2
Multiplying both sides by q^2, we have:
3q^2 = p^2
This means that p^2 is divisible by 3, and therefore, p must also be divisible by 3. Let's say p = 3k, where k is an integer.
Substituting this into our equation, we get:
3q^2 = (3k)^2 = 9k^2
Dividing both sides by 3, we have:
q^2 = 3k^2
Now, we can see that q^2 is also divisible by 3, which implies that q is also divisible by 3.
However, this contradicts our assumption that p/q is a simplified fraction, as both p and q are divisible by 3.
Therefore, our initial assumption that √3 is rational is incorrect. Hence, we can conclude that the square root of 3 is irrational.