Final answer:
To prove the statements by contradiction, we assume the given statement is false and show that it leads to a contradiction. For statement a), we assume a+b<2√(ab) and prove it leads to a contradiction. For statement b), we assume 4 divides n²+2 and prove it leads to a contradiction. For statement c), we assume a²+4b+5=0 and show it leads to a contradiction.
Step-by-step explanation:
To prove each of the following statements by contradiction:
a) If a,b∈R >0 , then a+b≥2√(ab)
Assume a+b<2√(ab) for some positive a and b. Squaring both sides of the inequality, we get (a+b)²<4(ab).
Expanding the square, we have a²+2ab+b²<4ab, which simplifies to a²-2ab+b²<0.
The left-hand side can be factored as (a-b)²<0, but this is not possible since the square of a real number is always non-negative. Therefore, our assumption was incorrect and a+b≥2√(ab).
b) For all n∈Z,4∤(n²+2)
To prove this statement by contradiction, assume that there exists an integer n for which 4 divides n²+2.
So, we have n²+2=4k for some integer k. Rearranging the equation, we get n²=4k-2=2(2k-1).
This means that n² is even, which implies that n is also even. So, let's write n=2m, where m is an integer. Substituting this into the equation, we have (2m)²=2(2k-1), which simplifies to 4m²=4k-2.
Dividing both sides by 2, we get 2m²=2k-1. This means that 2k-1 is even, but this is not possible since an odd number subtracted by 1 is always even. Therefore, our assumption was incorrect and 4 does not divide n²+2.
c) For all a,b∈Z, we have a²+4b+5≠0
To prove this statement by contradiction, assume that there exist integers a and b for which a²+4b+5=0.
Subtracting 5 from both sides, we have a²+4b=-5. Rearranging the equation, we get a²=-4b-5.
Since the left-hand side is a square, a² is always non-negative. But the right-hand side is negative, which means that our assumption leads to a contradiction. Therefore, a²+4b+5≠0.