Question

write the number p such that p square is irrational but p power 4 is rational


pllzzzzzzzzz help

Answer 1

Answer:

There is a number x such that x^2 is irrational but x^4 is rational, then what can x be?

Yes, in general there is at least one such number. More generally there should be an infinite number of such numbers, though none of them would be integers.

In general, if we take the fourth root of any prime number,

X=P−−√4 ,

we should find that,

X2=P−−√

is also irrational. Why? Because if X could be expressed as a rational number, say the ratio of integers m over n , then we could write,

P=X4=m4n4.

But we have already determined that P cannot have any integer factors other than itself and 1. Yet all prime numbers are greater than 1, so each factor of P , given as m/n , would need to be first greater than 1. Second, if we multiply both sides of the above equation by n4, we have,

m4=P×n4 .

But whatever number n is, it could only have an integer number of factors of P inside it, say Pk . And the same for m , say Pj . Now, looking only at the number of factors of P on the left and the right, we have,

P4j=P4k+1 .

And whatever integers j and k are, being integers they cannot produce an equality between the left and right sides because of the extra factor of P on the right side.

So any prime number will also be a solution to this problem. And any products of primes will also be solutions, as long as they are not to the fourth powers of P.

x4= rational number that is not a perfect square

x4 can be an integer:

x4=2,3,5,6,7,8,10,…

x2=2–√,3–√,5–√,6–√,7–√,8–√,10−−√,…

x=2–√4,3–√4,5–√4,6–√4,7–√4,8–√4,10−−√4,…

x4 can also be the ratio of 2 integers, where at least 1 integer is not a perfect square:

x4=23,29,43,163,…

x2=23−−√,2√3,23√,43√,…

x=23−−√4,2√43√,2√3√4,23√4,…

Answer 2

Answer:Answer

Open in antwrapp

Let p be a rational number and

b²

⇒ a² = pb²

..p divides a²

But when a prime number divides the product of two numbers, it must divide atleast one of them.

here a² = ax a

p divides a

Let a = pk

(pk)² = pb²

→ p²k² = pb²

⇒ b² = pk²

..p divides b²

But b² = bx b

..p divides b

Thus, a and b have atleast one common

multiple p

But it arises the contradiction to ourassumption that a and b are coprime. Thus, our assumption is wrong and pis irrational number.

Step-by-step explanation: