Prove that the product of an arbitrary number of integers is odd, if and only if

Question

Prove that the product of an arbitrary number of integers is odd, if and only if all the integers used in the product are odd.

Hints: 1) A<=>B is same as both a=>B and B=>A 2) Even Integer can be expressed as 2n, where n is any integer. 3) Odd integer can be expressed as 2n+1, where n is any integer. 4) Prove A=> B is easy by contradiction, Assume all integers is not add, then atleast one even. 5) To prove B=>A first show it to be truefor product of 2 odd integers., now show induction to show it be true of an arbitrary number of integers.

Explanation / Answer

First we will prove that

A=>B

means we are given that product is odd

and we have to prove that all the integers used are odd.

We will prove it by contradiction

so suppose all the integers are not odd

and without loss of generality we can assume that there is one even no.

we can write even number as 2n

given n integers among them n-1 are of the form 2k+1 or odd integers

and one is of the form 2k1 or even integer

or

N1=2k1

N2=2k2+1

N3=2k3+1

.

.

.

.

.

.

Nn=2kn+1

so product is

N1.N2....................Nn = 2k.(2k1+1).........................(2kn+1)

= 2K

where K=k.(2k1+1)................(2kn+1)

product is even but we are given that product is odd.

here is contradiction .

so our assumption that not all the integers are odd is wrong

therefore

A=>B is true.

..................................................................................................................................................

Now we will prove B=>A

we know that we can write any odd integer as 2k+1 where k is an integer

now we are given n odd integers

and we have to prove that product is odd

we will prove it by induction

..........................................................................................................................................

BASE CASE:

n=2

suppose we have two odd integers

N1=2a+1

N2=2b+1

then

N1.N2=(2a+1).(2b+1)

=4ab+2(a+b)+1

   =2(ab+a+b)+1

=2(K)+1

where K = ab+a+b

so it is true for n=2

................................................................................................................................

INDUCTION HYPOTHESIS:

Now assume that it is true for n = k

means it is true for N odd intergers

so we can write

N1.N2......................Nk=2(m)+1 ......................(1)

where m is integer

............................................................

INDUCTION STEP:

Now we have to prove it for n=k+1

so we have to prove that

N1.N2...................Nk.Nk+1=2(m1)+1 where m1 is an integer

so in equation (1) multiply both side by Nk+1

N1.N2...................Nk.Nk+1=(2(m)+1)Nk+1

now we can write

Nk+1=2c+1 where c is an integer

therefore

N1.N2...................Nk.Nk+1=(2m+1)(2c+1)

=2(mc+m+c)+1

   =2(m1)+1

where m1=mc+m+c

so we are done

hence we can say that product of n odd integers is odd.

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