Let a and be any positive integer where a > b. Show that either of (a+b)/2 and (a-b)/2 is an even number.
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Given:a and b are two positive integers where a>b
To prove: a+b/2 and a-b/2 is odd and the other is even
proof:
We know that any positive integer is of the form q,2q+1
let a=2q+1 and b=2m+1 where q and m are some whole nos.
---> a+b/2=(2q+1)+(2m+1)/2
= 2q+2m+2/2
= 2(q+m+1)/2
=q+m+1 which is a +ve integr
now,on substituting the values of a and b in a-b/2
we get a-b/2=q-m
GIVEN, a>b
therefore, 2q+1>2m+1
2q>2m
q>m
therefore,a-b/2=(q-m)>0
Thus,a-b/2 is a +ve integer
Now we've to prove that a+b/2 and a-b/2 is either odd or even
assume that a+b/2 - a-b/2
=a+b-a+b/2=2b/2=b which is an odd +ve integer
Also we proved that a+b/2 and a-b/2 are +ve integers
We know that the diffrence of 2 +ve intg. is an odd no. if one of them is odd and the other is even(also,diff. between two odd and two even integ. is even)
Hence it is proved
Hope it helps!!!!
To prove: a+b/2 and a-b/2 is odd and the other is even
proof:
We know that any positive integer is of the form q,2q+1
let a=2q+1 and b=2m+1 where q and m are some whole nos.
---> a+b/2=(2q+1)+(2m+1)/2
= 2q+2m+2/2
= 2(q+m+1)/2
=q+m+1 which is a +ve integr
now,on substituting the values of a and b in a-b/2
we get a-b/2=q-m
GIVEN, a>b
therefore, 2q+1>2m+1
2q>2m
q>m
therefore,a-b/2=(q-m)>0
Thus,a-b/2 is a +ve integer
Now we've to prove that a+b/2 and a-b/2 is either odd or even
assume that a+b/2 - a-b/2
=a+b-a+b/2=2b/2=b which is an odd +ve integer
Also we proved that a+b/2 and a-b/2 are +ve integers
We know that the diffrence of 2 +ve intg. is an odd no. if one of them is odd and the other is even(also,diff. between two odd and two even integ. is even)
Hence it is proved
Hope it helps!!!!
sandossh:
ok
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