the square root of 2 is an irrational number.
To prove this, we can use proof by contradiction.
Assume that the square root of 2 is a rational number. This means it can be expressed as a fraction a/b, where a and b are integers and b is not equal to 0. Since the square root of 2 is positive, both a and b must be positive.
Therefore, we can write:
sqrt(2) = a/b
Squaring both sides of the equation, we get:
2 = (a^2)/(b^2)
Cross-multiplying, we get:
2b^2 = a^2
This implies that a^2 is even, and therefore a must also be even (since the square of an odd number is odd).
Let a = 2k, where k is an integer.
Substituting this into the equation, we get:
2b^2 = (2k)^2
2b^2 = 4k^2
Dividing both sides of the equation by 2, we get:
b^2 = 2k^2
This implies that b^2 is even, and therefore b must also be even.
However, if both a and b are even, then they have a common factor of 2. This contradicts our assumption that a/b is a fraction in its simplest form.
Therefore, our assumption that the square root of 2 is rational is false. Hence, the square root of 2 is an irrational number.
Determine if square root of, 2, end square root
2
2
is rational or irrational and give a reason for your answer.
Answer
Attempt 1 out of 2
The number square root of, 2, end square root
2
2
is
because
1 answer
1 answer