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
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.
Step 1: The square root of 2 is a decimal number that cannot be exactly represented as a fraction.
Step 2: To determine if it is rational or irrational, we need to check if it can be expressed as a fraction (rational) or not (irrational).
Step 3: Assume that the square root of 2 is rational and can be expressed as a fraction, p/q, where p and q are integers with no common factors other than 1.
Step 4: Squaring both sides of the equation, we get 2 = (p/q)^2.
Step 5: Cross-multiplying, we get 2q^2 = p^2.
Step 6: Since the right side is a perfect square, the left side must also be a perfect square.
Step 7: The prime factorization of 2 is 2^1. We can see that the exponent of 2 on the left side is 1, which means q^2 must be divisible by 2.
Step 8: This implies that q must also be divisible by 2.
Step 9: Let q = 2k, where k is an integer.
Step 10: Substituting this back into the equation, we get 2(2k)^2 = p^2.
Step 11: Simplifying, we get 8k^2 = p^2.
Step 12: Similar to step 7, the left side is divisible by 2, so p^2 must also be divisible by 2.
Step 13: This implies that p must also be divisible by 2.
Step 14: Let p = 2m, where m is an integer.
Step 15: Substituting this back into the equation, we get 8k^2 = (2m)^2.
Step 16: Simplifying, we get 8k^2 = 4m^2.
Step 17: Dividing both sides by 4, we get 2k^2 = m^2.
Step 18: By the same reasoning in steps 7 and 12, k and m must both be divisible by 2.
Step 19: However, we assumed that p and q have no common factors other than 1, so this is a contradiction.
Step 20: Therefore, our assumption that the square root of 2 is rational is incorrect.
Step 21: The square root of 2 is therefore irrational.
The square root of 2 is an irrational number. A rational number is a number that can be expressed as a fraction, where the numerator and denominator are both integers. An irrational number, on the other hand, cannot be expressed as a fraction, and its decimal representation goes on indefinitely without repeating.
To determine if the square root of 2 is rational or irrational, we can try to express it as a fraction. Suppose the square root of 2 is rational and can be written as a fraction a/b, where a and b are integers and b ≠ 0. We can square both sides of the equation to get 2 = (a/b)^2, which simplifies to 2b^2 = a^2. This implies that a^2 is an even number, since it is equal to 2 times another number (b^2).
Now, let's consider the possible cases for a^2 being even:
1. If a is even, then a^2 is also even. In this case, we can write a as 2c, where c is an integer. Substituting this into the equation, we get 2b^2 = (2c)^2, which simplifies to 2b^2 = 4c^2. Dividing both sides by 2, we obtain b^2 = 2c^2. This implies that b^2 is also even, and it means that both a and b have a common factor of 2. However, this contradicts our assumption that a/b is a fraction in lowest terms, meaning that the square root of 2 cannot be expressed as a fraction.
2. If a is odd, then a^2 is also odd. In this case, we can write a as 2c + 1, where c is an integer. Substituting this into the equation, we get 2b^2 = (2c + 1)^2, which simplifies to 2b^2 = 4c^2 + 4c + 1. Dividing both sides by 2, we obtain b^2 = 2c^2 + 2c + 1. This implies that b^2 is odd, which means that both a^2 and b^2 cannot have a common factor of 2.
In both cases, we arrive at a contradiction, which shows that our initial assumption that the square root of 2 is rational is false. Therefore, the square root of 2 is irrational.