Prove that 5+√3 is an irrational number, given that √3 is irrational
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let us assum that 5-√3 is rational number so we can find two integers a , b. Where a and b are two co - primes number. So it arise contradiction due to our wrong assumption that 5 - √3 is rational number. Hence, 5 -√3 is irrational number
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Answer:
To prove that 5+√3 is irrational, we assume the opposite, i.e. that 5+√3 is rational. By definition, a rational number can be expressed as the quotient of two integers. We can therefore write:
5+√3 = p/q
Where p and q are integers with no common factors.
Rearranging the equation we get:
√3 = (p/q) - 5
By squaring the two sides we get:
3 = (p^2/q^2) - 10(p/q) + 25
Always multiplying by q^2 we get:
3q^2 = p^2 - 10pq + 25q^2
Since p^2 and 25q^2 are divisible by 5, p^2 - 10pq + 25q^2 is also divisible by 5. So 3q^2 must be divisible by 5.
Since 5 is prime and 3 is not divisible by 5, it follows that q^2 must be divisible by 5. So q must be divisible by 5.
Let q = 5k, where k is an integer. Substituting p^2 - 10pq + 25q^2 = 3q^2 into the equation, we get:
p^2 = 5(2q^2 - p^2/5)
Since p^2/5 is an integer, it follows that 5 divides p^2. Therefore 5 must also divide p.
This means that p and q have a common divisor of 5, which contradicts our assumption that p and q have no common divisor. From this we can conclude that 5+√3 is an irrational number.
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