show that root p + root q is irrational
Answers
Answered by
4
First, we'll assume that √p + √q is rational, where p and q are distinct primes
√p + √q = x, where x is rational
Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.
(√p + √q)² = x²
p + 2√(pq) + q = x²
2√(pq) = x² - p - q
√(pq) = (x² - p - q) / 2
Now x, x², p, q and 2 are all rational, and rational numbers are closed under subtraction and division. So (x² - p - q) / 2 is rational.
But since p and q are both primes, then pq is not a perfect square and therefore √(pq) is not rational. But this is a contradiction. Original assumption must be wrong.
So √p + √q is irrational, where p and q are distinct primes
---------------------
We can also show that √p + √q is irrational, where p and q are non-distinct primes, i.e. p = q
We use same method: Assume √p + √q is rational.
√p + √q = x, where x is rational
√p + √p = x
2√p = x
√p = x/2
Since both x and 2 are rational, and rational numbers are closed under division, then x/2 is rational. But since p is not a perfect square, then √p is not rational. But this is a contradiction. Original assumption must be wrong.
So √p + √q is irrational, where p and q are non-distinct primes
---------------------
∴ √p + √q is irrational, where p and q are primes
√p + √q = x, where x is rational
Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.
(√p + √q)² = x²
p + 2√(pq) + q = x²
2√(pq) = x² - p - q
√(pq) = (x² - p - q) / 2
Now x, x², p, q and 2 are all rational, and rational numbers are closed under subtraction and division. So (x² - p - q) / 2 is rational.
But since p and q are both primes, then pq is not a perfect square and therefore √(pq) is not rational. But this is a contradiction. Original assumption must be wrong.
So √p + √q is irrational, where p and q are distinct primes
---------------------
We can also show that √p + √q is irrational, where p and q are non-distinct primes, i.e. p = q
We use same method: Assume √p + √q is rational.
√p + √q = x, where x is rational
√p + √p = x
2√p = x
√p = x/2
Since both x and 2 are rational, and rational numbers are closed under division, then x/2 is rational. But since p is not a perfect square, then √p is not rational. But this is a contradiction. Original assumption must be wrong.
So √p + √q is irrational, where p and q are non-distinct primes
---------------------
∴ √p + √q is irrational, where p and q are primes
Answered by
0
First, we'll assume that √p + √q is rational, where p and q are distinct primes
√p + √q = x, where x is rational
Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.
(√p + √q)² = x²
p + 2√(pq) + q = x²
2√(pq) = x² - p - q
√(pq) = (x² - p - q) / 2
Now x, x², p, q and 2 are all rational, and rational numbers are closed under subtraction and division. So (x² - p - q) / 2 is rational.
But since p and q are both primes, then pq is not a perfect square and therefore √(pq) is not rational. But this is a contradiction. Original assumption must be wrong.
So √p + √q is irrational, where p and q are distinct primes
---------------------
We can also show that √p + √q is irrational, where p and q are non-distinct primes, i.e. p = q
We use same method: Assume √p + √q is rational.
√p + √q = x, where x is rational
√p + √p = x
2√p = x
√p = x/2
Since both x and 2 are rational, and rational numbers are closed under division, then x/2 is rational. But since p is not a perfect square, then √p is not rational. But this is a contradiction. Original assumption must be wrong.
So √p + √q is irrational, where p and q are non-distinct primes
---------------------
∴ √p + √q is irrational, where p and q are primes
MARK ME AS BRAINLIEST
√p + √q = x, where x is rational
Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.
(√p + √q)² = x²
p + 2√(pq) + q = x²
2√(pq) = x² - p - q
√(pq) = (x² - p - q) / 2
Now x, x², p, q and 2 are all rational, and rational numbers are closed under subtraction and division. So (x² - p - q) / 2 is rational.
But since p and q are both primes, then pq is not a perfect square and therefore √(pq) is not rational. But this is a contradiction. Original assumption must be wrong.
So √p + √q is irrational, where p and q are distinct primes
---------------------
We can also show that √p + √q is irrational, where p and q are non-distinct primes, i.e. p = q
We use same method: Assume √p + √q is rational.
√p + √q = x, where x is rational
√p + √p = x
2√p = x
√p = x/2
Since both x and 2 are rational, and rational numbers are closed under division, then x/2 is rational. But since p is not a perfect square, then √p is not rational. But this is a contradiction. Original assumption must be wrong.
So √p + √q is irrational, where p and q are non-distinct primes
---------------------
∴ √p + √q is irrational, where p and q are primes
MARK ME AS BRAINLIEST
Similar questions