let n be an element of Z. proove that n is an even integer if and only if n=2m with m being an element of Z.?

Answer 1

look Z is a set of integers. And when you have to prove " IF AND ONLY IF" then you have to prove by both sides i.e. once by considering N is even integer and another by choosing M as any integer. Read on you will understand. In simple words you have to prove that the converse is also true.


what are integers .... First thing integers are 1,2,3 .... upto infinity.
Next thing is zero a integer ???.... YES IT IS. And negative numbers like -1,-2,-3 ... upto minus infinity are integers.


Therefore,

Z = { -infinity, ..... , -3,-2,-1,0,1,2,3,4. ...... plus infinity }

if we take a number n from the set which is real and even. and let another number M (doesn't matter even or odd) from the set.

WE have to prove there is n=2m.

For any M, there is a number that is double that of M in the set and as it is multiplied by 2 it will be even.

Now proving converse,

M= N/2

As given in the question, n is even. Taking any even number N from the set Z. By dividing it by 2, it will be exactly dividible and hence will yield some number, even or odd we dont know but that will be in Z. Z is a set of Whole numbers or integers. hence you will get some M from the set.


HENCE PROVED