Proof by Contrapositive: If 5x - 7 is odd, then x is even

In this short video I do a proof where I prove the contrapositive of the statement. The problem is to prove: If 5x - 7 is odd, then x is even. I hope this helps.

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[aioia]

I would have done this

5x - 7 = 2y + 1
5x - 2y = 8
let a = x - 2, b = y + 1
x = a + 2, y = b - 1
5(a + 2) - 2(b - 1) = 8
5a + 10 - 2b - 2 = 8
5a - 2b + 8 = 8
5a = 2b
then a = 2k, b = 5k for some integer k
so x = a + 2 = 2k + 2 = 2(k + 1) which is even

unnecessarily complicated but whatever

[acdude]

Nice quick proof upon which other types of odd / even problems can be modeled.

[michaeledwardharris]

Well done. Very helpful and clearly explained. Thanks!

[pepelefrog]

Excellent problem. The thing with this kind of problems for newcomers passionate by math modeling is that we are not as objective oriented as we should when trying to express the core of the problem: "given an expression, prove that the sum of its factors are odd". Thats the breaker. Using the corollaries for odd-even, the rest is history.

[isoid]

Simple(ish) explanation (or just how I would do it):

5x - 7 = odd
5x = odd + 7 (5x = odd + odd)
5x = even
5x/5 = EVEN/5
x = EVEN

(An even number *that can be divided by 5 with no remainder* will always divide into another even number
Since x is an integer, EVEN must be divisible by 5. numbers cleanly divisible by 5 can end in either a 0 or a 5, but since it is even it must end in a 0. Any number ending in a 0 divided by 5 will return an even number (including 0, which is technically even).
This means that 7 could be any odd number and this should still work.

Using the technical definitions of odd/even (k is an integer as in the video):
5x - 7 = 2k+1
5x = 2k+8
x = (2k+8)/5
2k+8 is even, so:
x = even/5
As explained above, even/5 = even, so:
x = even

Ok maybe not much more simple

[handsanitizer]

Maybe this is a dumb question, but how do I find out what area of math I'm weak at ?. That's what's plaguing me, is there a test online I can take to give me an idea of where I need to improve. I'm good at certain topics and bad at othera

[giovannicalafiore]

It can be done by contradiction as well

[wiggles]

An easy way is to use the fact that an odd number squared is odd. Assume 5x-7 is odd, so 25x^2 - 70x + 49 is odd, so set that equal to 2k+1, then rearrange to get 25x^2 - 70x + 49-2k-1=0. Then use the quadratic formula to get x = (1/5)*(7-sqrt(2k+1)) or x = (1/5)*(7+sqrt(2k+1)). Now we just need to figure out when 2k+1 is a perfect square so we get an integer for x. Note {2k+1} is the set of odd numbers, and just by guessing I can see the pattern that when k=2i^2+2i, for i in {0, 1, 2, 3, ...} 2k+1 is a perfect square, and this will get all possible perfect squares (i is an index, not the imaginary unit). So, setting k=2i^2+2i, we have

x = (1/5)*(7-sqrt(2(2i^2+2i)+1)) or x = (1/5)*(7+sqrt(2(2i^2+2i)+1))

=> x = (1/5)*(7-sqrt(4i^2+4i+1)) or x = (1/5)*(7+sqrt(4i^2+4i+1))

=> x = (1/5)*(7-(2i+1)) or x = (1/5)*(7+(2i+1))

=> x = (1/5)*(6-2i) or x = (1/5)*(8+2i)

=> x = (2/5)*(3-i) or x = (2/5)*(4+i).

For x to be an integer, we need 3-i to be a multiple of 5 in the case of the first root, and 4+i to be a multiple of 5 in the case of the second root, and in either case, when that happens, that factor of 2 still stays out in front, so in all cases, x is even. That completes this simple, direct way of proving the claim.

[alistairkentucky-david]

Can't we just say that if 5x - 7 is odd, then 5x - 7 + 7 is even (since odd + odd is even), thus 5x is even. Thus since 5x must have a factor of 2, and 5 =/= 2, x must have a factor of 2

[pedromooregaissler]

Why not do it this way: If 5x-7 is odd, then 5x-7+7 = 5x is even as the sum of two odd numbers. Since 5 does not have 2 in its prime factorisation, 5x is even only if x is even.

[phanphan]

5x-7=(4x-7) + x. 4x-7 is odd so if 5x-7 is odd x must be even

[tacticaltaco]

How does this constitute a proof? How do we know that the contrapositive being true makes the original statement true?

[darknova]

This is proof by contradiction, not contrapositive