1.if a and b are any two odd positive integers such that a> b then prove that one of the 2 numbers a+b/2 and a-b/2 is odd other is even. 2.prove that if a positive integers of the form 6q+5 then it is of the form 3q+2 for some integer q ,but not conversely,
Asked by Preethika .S.Nair | 5th May, 2013, 05:57: PM
1. Let the numbers be 2x + 1 and 2y +1
Add them gives 2x + 2y + 2 and if you divide by 2 gives x + y + 1
Subtract them give 2x - 2y and if you divide by 2 gives x - y
Add the two answers gives 2x + 1 which is an odd number. Since adding two even numbers gives an even number and adding two odd numbers gives an even number then the two numbers cannot both be even or both odd.
therefore one must be odd and one even.
2.
I CASE
let x be any no. divided by 6
x=6q,6q+1,6q+2,6q+3,6q+4,6q+5
considering x=6q+5
x=6q+5
=6q+3+2
=3(2q+1)+2
let 2q+1 be m
x=3m+2
thus 6q+5 can be expressed in form of 3q+2.(3m+2)
II CASE
let x be any no. divided by 3
x=3q,3q+1,3q+2
when x=3q+2
=3q+2
even if we multiply it by 2 or 3 we get x=6q+4,or x=9q+6
in noway can we get x=6q+5.
hence proved that if any positive integer is of form 6q+5,then it is of form 3q+2 for some positive integer q,but not conversely.
Add them gives 2x + 2y + 2 and if you divide by 2 gives x + y + 1
Subtract them give 2x - 2y and if you divide by 2 gives x - y
Add the two answers gives 2x + 1 which is an odd number. Since adding two even numbers gives an even number and adding two odd numbers gives an even number then the two numbers cannot both be even or both odd.
therefore one must be odd and one even.
I CASE
let x be any no. divided by 6
x=6q,6q+1,6q+2,6q+3,6q+4,6q+5
considering x=6q+5
x=6q+5
=6q+3+2
=3(2q+1)+2
let 2q+1 be m
x=3m+2
thus 6q+5 can be expressed in form of 3q+2.(3m+2)
II CASE
let x be any no. divided by 3
x=3q,3q+1,3q+2
when x=3q+2
=3q+2
even if we multiply it by 2 or 3 we get x=6q+4,or x=9q+6
in noway can we get x=6q+5.
hence proved that if any positive integer is of form 6q+5,then it is of form 3q+2 for some positive integer q,but not conversely.
Answered by | 6th May, 2013, 09:51: AM
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