proof 2.1

[jsk0004]

Can someone prove or disprove the following and explain the steps. (The number x is an even integer if and only if x^3+13 is odd)

[Ajinkya Kulkarni]

Hi, Thanks for your post. I am not into Discrete Structure but I have forwarded your message to rest of the GTAs. Some one will definitely reply to your post.

[shreyas]

Hey, I am not from CS14, but, the proof may go something like this:

Given, X^3 + 13 = ODD
=> X^3 is EVEN (as 13 is ODD, and to get a sum ODD, we should add an EVEN number to it, else the sum will be EVEN, like for ex 13 + 3 = 16[ODD + ODD = EVEN])
=> X is EVEN (as EVEN number multiplied with any more EVEN number(incl. itself) will be EVEN, if X was ODD then X^3 would be ODD)

Some one please correct or discuss if I am wrong.

[thejaswi.hr]

I do not know how to formally prove the statement by using the techniques taught in CS214, but logically I would do it this way:

An even number + An odd number = An odd number --------------(1)
An odd number + An odd number = An even number -------------(2)
An even number + An even number = An even number ---------(3)

An even number raised to any number is always even (4^2, 4^3 are both even numbers) ------ (4)
An odd number raised to an even number is even and an odd number raised to an odd number is always odd. ------ (5)

Now lets look at the statement: The number x is an even integer if and only if x^3+13 is odd
Let x^3 = y ---- (6)
So y + 13 is odd only if y is an even number (From (1))
But from (6) we know that y = x^3
From (4) and (5) we know that x^3 is even only iff x is even (3 is an odd number, so x should be an even number to have x^3 as even number. If x was to be an odd number then x^3 would be an odd number too!)

Hence, "The number x is an even integer if and only if x^3+13 is odd"

I am not sure if this is the way we prove such statements formally in CS214, so if we have a GTA for CS214 then it would be better if he/she takes a look at it.

- Thejaswi Raya

[Sui Gong]

Can someone prove or disprove the following and explain the steps. (The number x is an even integer if and only if x^3+13 is odd)

p:x is even
q: x^3+13 is odd
first we prove p->q:
if x is even, we can derive that x=2*m(m is an integer). then x^3=8*m^3, which is an even integer too. Then x^3+13=8*m^3+13=8*m^3+2*6+1=2*(4*m^3+6)+1, since 4*m^3+6 is an integer, and let k=4*m^3+6, then we know x^3+13=2k+1, then x^3+13 is odd.
second we prove q->p
if x^3+13 is odd, then x^3 should be even, then x should be an even integer to make this all correct.