2006-09-13 11:25:37 · 6 answers · asked by thamal3 1 in Science & Mathematics ➔ Mathematics
If x is an even integer, then:
3x is even since odd*even = even.
So 3x+1 is odd because even + odd = odd.
5x is even since odd*even = even
5x+2 is even because even + even = even.
Now if x is an odd integer, then:
3x is odd since odd*odd = odd
So 3x+1 is even because odd + odd = even
5x is odd since odd*odd = odd
So 5x+2 is odd because odd + even = odd.
Therefore, 3x+1 and 5x+2 have opposite parity for any integer x.
2006-09-13 11:35:57 · answer #1 · answered by Anonymous · 0⤊ 1⤋
If x = 0 it is definitely true.
Consider the case that x is a positive integer.
5x+2 > 3x+1 obvious
(5x+2)-(3x+1) = 2x+1
2x + 1 is always odd
Therefore, if 5x+2 is even, 3x+1 will be odd since an even number minus an odd number is odd. Conversely, if 5x+2 is odd, 3x+1 will be even, since an odd number minus an odd number is even.
Therefore, for positive integers, 3x+1 and 5x+2 will be of opposite parity. An analogous argument would work for the negative integers.
2006-09-13 18:36:56 · answer #2 · answered by bruinfan 7 · 0⤊ 0⤋
3x and 5x will always be the same parity.
1 and 2 are of opposite parity
will leave you to fill the gaps
2006-09-13 18:41:13 · answer #3 · answered by smudgepuss 2 · 0⤊ 0⤋
Take the difference. Numbers of opposite parity have an odd difference.
2006-09-13 19:04:57 · answer #4 · answered by AnyMouse 3 · 0⤊ 0⤋
Put x=1,2,3,4,5,..................
3x +1=4,7,10,13,16,19,...............i
while
5x+2=7,12,17,22........................ii
The first number in 3x+1 is 4 i.e even while the first number in 5x+2 is odd. I f we See further second number is odd then second on the other is even and so on.It means if in the first number is odd then in second is even.
2006-09-13 18:41:16 · answer #5 · answered by Amar Soni 7 · 0⤊ 1⤋
stop getting people to do your hw for you. at least try it and tell us what you're having problems with.
2006-09-13 18:27:48 · answer #6 · answered by Anonymous · 1⤊ 1⤋