Proof by Contrapositive: If 5x - 7 is odd, then x is even
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- เผยแพร่เมื่อ 13 ก.ย. 2024
- 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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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
Nice quick proof upon which other types of odd / even problems can be modeled.
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.
Well done. Very helpful and clearly explained. Thanks!
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
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
Beautiful explanation ❤️
Really enjoying this kind of videos!
It can be done by contradiction as well
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.
no lol. 5x-7 = (4x-6) + (x-1), the sum of an even number and x-1. Hence x-1 must be odd, i.e., x even.
@@andrewzhang5345 If my method seems a bit difficult then you could go for the cubic which might be a bit easier. An odd number cubed is odd, so if (5x-7) is odd, then (5x-7)^3 is odd and then you can set (5x-7)^3=2k+1 and solve for the integer solutions of x.
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.
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
How does this constitute a proof? How do we know that the contrapositive being true makes the original statement true?
What are all the possible arrangements?
1. x is odd and 5x-7 is odd
2. x is odd and 5x-7 is even
3. x is even and 5x-7 is odd
4. x is even and 5x-7 is even
He showed 2. is true, so we can disregard 1.
And we don't want to know what happens if 4. is true, because the original question was „what happens to x if we assume 5x-7 to be odd?“
So we can disregard starement 4.
The only options left are 2. and 3.
Therefore they must both be true
Yes, he only showed 2. to be true, but since the value of these terms must be either even or odd, those are the only possible arrangements
They are logically equivalent
@@sehr.geheim So is contrapositive only possible when you have an iff relationship between properties like even and odd? With a more complicated expression it's not immediately obvious how arrangement 2 negates arrangement 1, unless it were worded like "x is always odd when 5x-7 is even"
@@tacticaltaco7481 The contrapositive of P->Q is ~Q->~P. They are logically equivalent. All you need is P->Q, and if you have that, then you know that ~Q->~P, or vice versa. So if you want to prove a statement of the form P->Q, you can assume ~Q and deduce ~P, and that constitutes a proof.
5x-7=(4x-7) + x. 4x-7 is odd so if 5x-7 is odd x must be even
Clever.
❤️♥️
This is proof by contradiction, not contrapositive
Nope, it's the contrapositive.