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Something interesting concerning parity...

by sEsposito
Tags: interesting, parity
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sEsposito
#1
Sep18-10, 03:02 PM
P: 154
Is there a proof or way of proving that all even numbers (taking into account the definition of an even number as [tex]n=2k[/tex]) end in 0,2,4,6, or 8?
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Gokul43201
#2
Sep18-10, 07:49 PM
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What do you think? Care to try showing that numbers ending in 1,3,5,7, or 9 can not be even?
TwilightTulip
#3
Sep19-10, 05:05 AM
P: 24
Rewrite n as 10a+b, where 0<=b<10.

Then n is even means 2 divides 10a+b, i.e. 2 divides b (let 10a+b=2k and solve for b). Hence b=0,2,4,6,8

sEsposito
#4
Sep19-10, 10:09 AM
P: 154
Something interesting concerning parity...

Quote Quote by TwilightTulip View Post
Rewrite n as 10a+b, where 0<=b<10.

Then n is even means 2 divides 10a+b, i.e. 2 divides b (let 10a+b=2k and solve for b). Hence b=0,2,4,6,8
An explanation such as this is alright for proving that an integer [tex]n[/tex] is even if [tex]n=2k[/tex], for some integer [tex] k.[/tex] Thus, we have to use our knowledge of even numbers to say that they end in an integer that is divisible by 2. A satisfactory explanation, no doubt, just not what I'm looking for.

Is there a way to show that the fact that all even numbers end in 0,2,4,6, or 8 implies the definition of an even number ([tex]n=2k[/tex], where [tex]k[/tex] is an integer)?
Prologue
#5
Sep19-10, 06:50 PM
P: 185
Any number that is a multiple of ten is even and divisible by 2. Then any number (base ten) denoted by ...dcba (where a is the ones place, b in the tens, etc.) will be divisible by two if a is divisible by 2 because ...+d*10^3+c*10^2+b*10^1+a represents the number and division is linear. All terms except a are multiples of ten always and therefore divisible by 2, then all that is left is a.


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