Respuesta :
The question is incomplete. The complete question is :
Each statement below involves odd and even integers. An odd integer is an integer that can be expressed as 2k+1, where k is an integer. An even integer is an integer that can be expressed as 2k, where k is an integer. Prove each of the following statements using a direct proof. (a) The sum of an odd and an even integer is odd. (b) The product of two odd integers is an odd integer.
Solution :
Odd number integers = 2k + 1, where k is integer
Even number integer = 2k
a). Odd integer + even integer
= 2k + 1 + 2k
= 4k + 1
= 2(2k) + 1
Let 2k = t, where t is integer
= 2t + 1
= Odd integer by definition
If number is 2t + 1 where t belongs to integer, then it is odd integer.
Hence proved.
b). Product of two odd integers : [tex]$2k_1 + 1,\ \ 2k_2 +1 $[/tex] where [tex]$k_1, k_2 $[/tex] belongs to integer.
[tex]$(2k_1+1)(2k_2+1)= 4k_1k_2+2k_1+2k_2+1$[/tex]
[tex]$2(2k_1k_2)+2(k_1+k_2)+1$[/tex]
Let [tex]$2k_1k_2 = a$[/tex] and [tex]$k_1+k_2 = b $[/tex]
here, a and b belongs to integers since [tex]$k_1$[/tex] and [tex]$k_2$[/tex] are integers.
We get:
2a+2b+1
= 2(a+b)+1
= 2l + 1, Let (a+b)=l and l belongs to integers.
It is odd integer by definition.
Hence proved.
