give a counterexample to each claim:
(a) if x, y are both irrational then xy is also irrational
(b) for all n ? n, 2n 2 + 5 is prime
(c) for all a, b, n ? n, if n | ab then n | a or n | b.

a)

b)  n = 5

c) n = 6, a = 4, b = 3

Step-by-step explanation:

Let . Here is a known irrational, and   where the number 2 is not only rational but integer. 

If you take , you will get 2(25) + 5 = 55 that is not prime, because 5 divides 55.

Here let n = 6, a = 4, b = 3. We can see that ab = 12, and of course 6 divides 12 (n | ab). But, also 6 does not divides 4 and  6 does not divides 3.

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.