How to do these question ? Will be much appreciate if I can have a details expla

Question

How to do these question ?

Will be much appreciate if I can have a details explanation for each part...

Thank you.

For each of these statements, write down the hypothesis and the conclusion, and then either prove the statement is true, or find a counter-example to show it is false: (a) If the sum k + m of the integers k and m is even then both k and m are even. (b) If x and y are positive rational numbers then x/y is a rational number. (c) for every positive integer n. (d) For all integers x and y it is true that . (e) If n is any positive integer then n2 + n + 41 is prime.

Explanation / Answer

(a)- given sum k+m of integers k and m is even,
hypothesis: let's assume sum of integers k and m is 2P wher p is >=0

so we have k + m = 2P
so k = 2P - m

if P=1 and m =1

then k = 1,

so we have two odd integers 1 nd 1 whose sum is even which is 2

so above statement is wrong ANSWER

(b)- we know that In mathematics, a rational number is any number that can be expressed as the quotient or fraction p/q of two integers, with the denominator q not equal to zero.Since q may be equal to 1, every integer is a rational number.

so now if x is rational and y is rational then there might be the case when y = 0 since q is rational

so given statement is wrong, if x and y are rational then x/y might not be rational

c)- let's say y= n2 - 12 n +35

we can rewrite it as,

y =n2 - 12n + 36 - 1

y= (n -6)2 - 1

since (n -6)2 is always >=0

so if (n -6)2 = 0 for n=6

y = -1 which is less than 0

so given statement is wrong

d)- to prove |x+y| <= |x| + |y|

using the property of absolute value operator, we can have four cases as following

if x>0 and y <0
then |x| + |y| =x +y which is greater or equal to  |x+y| ,
if x <0 and y <0
|x| + |y| = x +y which is equal to  |x+y| ,
if x >0 and y >0
then  |x| + |y| = x +y which is equal to  |x+y| ,
if x<0 and y >0
|x| + |y| = x + y which is greater or equal to  |x+y| ,

hence this is true for all the four cases.

hence given statement is true.

|x+y| <= |x| + |y|

e)-prove by counter example

For f(n) = n^2 - n + 41:

f(41) = 41^2 - 41 + 41 = 41^2 = 41*41 which is obvoiusly not priem

also, f(42) = 42^2 - 42 + 41 = 42^2 - 1 = 42*43 not prime

hence given statement is wrong

proved

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