Prove that 11x - 7 is even if and only if x is odd

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@martinhawrylkiewicz2025 2 жыл бұрын

Nice proof. You can also prove it directly the 2nd part which to me is much simpler I think. If 11x - 7 is even then 11x - 7 = 2n for some n in Z, but then 11x = 2n + 7 = 2(n + 3) + 1 for some integer n + 3 which means 11x is odd together with 11 being odd, means x is odd.

@fazex4185 2 жыл бұрын

Nice introduction to proofs, thanks!

@CoolAid988 2 жыл бұрын

Math sorcerer you are my idol. I want to be a mathematician like you. I want to be a math professor

@verypanda1801 2 жыл бұрын

Another great video! :)

@TheMarik766 2 жыл бұрын

Problem is 11x-7 but title is x-7

@TheMathSorcerer 2 жыл бұрын

Fixed thank you!!

@marriamhaji7440 Жыл бұрын

This helped alot❤

@lolzhunter 2 жыл бұрын

this one is actually quite easy relatively speaking, even*odd = even and odd*odd is odd, even-odd is odd and odd-odd is even, using these 4 rules you can quickly show that the only way for the end result to be even is if x is odd for example if x was even then that would be 11*even, and 11 is odd so odd*even, and that by definition gets you an even number, you then minus 7 which is odd which means even-odd which is odd, which shows that when x is even, you always get odd when x is odd you have 11 (which is odd) times an odd number, so odd*odd which gets you odd, then you minus 7 which we know is odd and you get odd-odd which is always even, therefore when x is odd you get even EDIT: since this proof is so general this also inadvertently proves that this rule holds for any equation in the format of odd*x±odd, where ± is plus or minus since the rules for odd even minus are the same for plus, so (2n-1)*x±(2m-1), n and m being any number you want since the result will be odd, getting the original equation the values would be n=6 and m=4

@andrewzhang5345 2 жыл бұрын

It's easier just to do 11x-7 = 0 (mod 2) if and only if x=1 mod 2

@lukasjuhrich503 2 жыл бұрын

Yes, that would work. However, note that the complexity of the proof doesn't completely vanish, it just moves to the statement that reduction mod 2 is a Ring homomorphism Z->Z/2Z, which is a prerequisite to your proof.

@lukasjuhrich503 2 жыл бұрын

However, I do agree that this proof is much more elegant as we don't have to prove both implications separately.

@andrewzhang5345 2 жыл бұрын

@@lukasjuhrich503 You don't need ring theoretic language to parse this. One way is just to recall that mod n induces an equivalence relation, and that the obvious operations on them are well defined. Or more ad hoc, you can just note that 11x - 7 = (10x - 6)+ (x-1), which is even if and only if x-1, i.e. x odd; which is in essence, what fancy modular arithmetic does.

@lukasjuhrich503 2 жыл бұрын

@@andrewzhang5345 „One way is just to recall that mod n induces an equivalence relation, and that the obvious operations on them are well defined.“ This is the same as stating that reduction mod n is a ring homomorphism, you just spelled out the definition. You're not really “omitting ring theory” here, you're just omitting the ring theoretic vocabulary that's attached.