Question

Let x and y be some integers. Consider the following statement: If xis odd and y is odd, then xy is odd. Proof. Assume that x is odd and y is odd. Then there exists some k integers such that x=2k+1 and m=2k+1. Then xy=(2k+1)(2k+1)=4k2+4k+1=2(2k2+2k)+1. Since 2k2+2k is an integer because k is an integer then xy is odd. The proof is correct but the statement is incorrect The statement is correct, but the proof is incorrect. The statement and proof are incorrect. The statement and the proof are correct.

Answer 1

hence proved If number is 2t + 1 where t belongs to integer, then it is odd integer.

Odd number integers = 2k + 1, where k is integer

Even number integer = 2k

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.

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

learn more about of odd number here

brainly.com/question/21330408

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