Prove that √p is an irrational number , if p is not a perfect square.​

prove that √p is an irrational number , if p is not a perfect square.

Let us assume, to the contrary, that √p is rational. So, we can find coprime integers a and b(b ≠ 0) such that :-

=> √p = a/b

=> √p b = a

=> pb² = a² ….(i) [Squaring both the sides]

=> a² is divisible by p

=> a is divisible by p So, we can write a = pc for some integer c.

Therefore,

a² = p²c² ….[Squaring both the sides]

=> pb² = p²c² ….[From (i)]

=> b² = pc²

=> b² is divisible by p

=> b is divisible by p

=> p divides both a and b.

=> a and b have at least p as a common factor.

But this contradicts the fact that a and b are coprime. This contradiction arises because we have assumed that √p is rational. Therefore, √p is irrational.

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4th option is correct xy=-10

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the population of other specie of birds will decline or other specie of birds will find some alternate source of food to survive.

explanation:

according to the competitive exclusion principle, which is also known as gause's law, two specie living in the same niche and feeding on the same resource cannot co-exist stably.

as we have well studied the theory of natural selection which states that only those organisms survive which have better tendency to compete for food and adapt to the environmental conditions.

coming towards the question, what wi ll happen when one species outcompetes the other for the small fish , the other specie will not get enough food and will start to die or decline in population. eventually either other specie needs to find another alternative food source for itself or migrate at some other location where there is less competition otherwise it will just not exist anymore at that niche.

hope it !