Direct Proof: Prove that if 5x - 7 is odd, then 9x + 2 is even

Prove that if 5x - 7 is odd, then 9x + 2 is even

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[wiggles]

I like the modular arithmetic approach, with everything mod 2. Assume 5x-7=1 => x+1=1 => x=0 => 9x+2 = 0+2 = 0, so 9x+2 is a multiple of 2.

[lucaterle]

How do I know when to use direct proof vs proof by contradiction or proof by cases, etc.? Is there a method for this, or will I just pick it up with experience?

[rageprod]

Way cleaner than mine (I have no practice with proofs)

5x-7 = 2k+1, for k in Z
5x = 2k+8
x = (2k+8)/5

y = 9x+2
= 9[(2k+8)/5] + 2
= (18k+82)/5 (which is an integer, because 9x+2 is an integer)

5y = 18k + 82
= 2(9+41) (which is even)

5y is even, but 5 is odd, so y must be even. QED.

[devondevon]

Since odd times even=even AND an even - odd = odd then for 5x-7
to be odd then 'x' must be even. And if the minuend '5x' is even then the value 5x-7 is odd since and even - odd = odd. Since "9x" is even given that 'x' is even, given that even times odd=even then the value 9x+2 must be even since an even + even=even.

[midknightcrisis]

Am I missing sth? 5x-7 is odd => 5x is even (bc only an even number minus seven is odd) => x is even (bc only an even number times 5 is even) => 9x is even (same reason) => 9x + 2 is even?
Idk there's a way to formalize this in a simple way, but that's basically what things boil down to, right?

[thiagovieira]

My solution :

5x-7 is odd

9x+2 =5x-7+4x+9
9x+1 =5x-7+4x+8
9x+1=odd+ 4(x+8)

4(x+8)... even number (because is divisible by 2)

9x+1=odd +even
9x+1=2k+1+2m (k and m are no negative integers)
9x+2=2k+2m+2
9x+2=2(k+m+1)=2p (p is a no negative integer)
9x+2 is even

[greenfinmusic]

if 5x-7 is odd,
then 5x must be even (since if 5x was odd, you'd have odd minus odd, which yields even; therefore contradiction, therefore 5x must be not odd).

if 5x is even, that means x must be even (since if x was odd, you'd have odd times odd, which yields odd; therefore contradiction, therefore x must be not odd)

if x is even, then 9k must be even, since odd times even equals even.
if 9k is even, then 9k+2 must be even, since even + 2 = even.

therefore, if 5x-7 is odd, then 9k+2 must be even.

QED

[aidanfogleman]

Here I was thinking about it graphically, like "both of those functions are neither odd nor even" lmao

[jennifersilves]

Why do you just get to add numbers to one side of an equation? I followed on first watch, then started thinking about it.

[wilhufftarkin]

Are these still from Spivak's Calculus book?

[zackariasstrindlund]

how is 9x +2 proven to be even? 5x-7 we know is even for a k in (k2+1), putting that equal to 9x+2 would just show that for a k 9x+2 is odd??

[labiribiri]

You lost me at the whole "this is the same as 2k+1". I thought the point of a proof is to make it clear to the audience?

Im sorry, but if you can tell me how this isnt a "obviously" moment in your proof im going to assume its not directed to the general audience.

[perfectionist]

I also thought the same way you did 😂 when i saw some steps

[yogesh]

Anyone pls suggest a book for
"Order relations""
Urgent⚠️

[omnomchomsky]

Proof before looking at video.

If 5x - 7 is odd, then 9x + 2 is even.

5x - 7 =
1: (2 * 2 + 1) * x - (2 * 3 + 1)

9x + 2 =

2: (2 * 4 + 1) * x - (2 * 1)

In order of 5x - 7 to be odd, x must be even:

( 2 * 2 + 1) * 2(k) - (2 * 3 + 1)
10(k) - (7) = 2(5k) - (2*3) + 1 = 2( 5k - 3) + 1

(2 * 4 + 1) * 2(k) - (2 * 1)
18(k) - 2 = 2(9k -1)

2(5k-3) + 1 is always odd by definition 2k + 1

2(9k -1) is always even by definition 2k

Therefor if 5x - 7 is odd, 9x + 2 is even.

[affirmajim]

5x–7 ≡ 1 mod 2
5x ≡ 8 ≡ 0 mod 2
(5+2•4)x ≡ 9x ≡ 0 mod 2
9x+2 ≡ 2 ≡ 0 mod 2
therefore, (9x+2) is even