Sol.
If a and b are two odd positive integers such that a > b, then prove
+
a
b
2
–
a
b
2
that one of the two numbers
and
is odd and the other is even.
∴
+
a
b
2
=
+
(2m + 3)
(2m + 1)
2
=
+
2m
2m
2
+
4
Q.
[ where m and n are + ve integers ]
=
4m
2
+
4
=
2
2
+
2)
=
2m
+
2
[ which is even ]
(2m
Let a
=
2m + 3
and
=
2m + 1
∴
+
a
b
2
If a and b are two odd positive integers such that a > b, then prove
+
a
b
2
–
a
b
2
that one of the two numbers
and
is odd and the other is even.
Now,
–
a
b
2
=
–
(2m + 3)
(2m + 1)
2
=
+
2m
3
2
–
2m
–
1
Q.
=
3
2
–
1
=
2
2
=
1
[ which is odd ]
∴
+
a
b
2
=
+
2m
2
and
1
–
a
b
2
=
∴
One of them is odd and the other is even
Hence Proved
1
[ where m and n are + ve integers ]
Let a
=
2m + 3
and
=
2m + 1
=
2m
+
2
[ which is even ]
∴
+
a
b
2
Sol.