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.
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 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.
@@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.
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
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.
Nice introduction to proofs, thanks!
Problem is 11x-7 but title is x-7
Fixed thank you!!
Math sorcerer you are my idol. I want to be a mathematician like you. I want to be a math professor
It's easier just to do 11x-7 = 0 (mod 2) if and only if x=1 mod 2
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.
However, I do agree that this proof is much more elegant as we don't have to prove both implications separately.
@@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.
@@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.
This helped alot❤
Another great video! :)
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
Why did yt recommend me this at 5am???
Lol so weird
Ahh I miss these types of proofs.
Yeah the fun easy ones hehehe:)
Show!
Why did I get recommended this? I strictly watch kawaii vtubers and leftist political content. Good shit though dude
Something is seriously wrong with you dude.
Same here
Lol!!!!
Time for u to study linear algebra