Proving conditional statements -- Proof Writing 10

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[RandyKing]

interesting points: unpack P/repackQ is a nice mantra, and going against the simplicity gradient in proving an implication

[sillymel]

(7:11) For this definition of greatest common divisor to work, you need to restrict a and b to not both being zero or allow d to be nonnegative, not just positive, since gcd(0, 0) is generally considered to be 0. (Note that 0 divides itself (and only itself) as there definitely exists an integer k such that 0=0k. Now, this works for *any* integer k, but the definition of “divides” doesn’t specify that k has to be a *unique* integer, so we’re good. Notice that this means gcd(0, 0)=0 works since 0 divides both 0 and 0, and the only integer that 0 divides is 0 itself, so the greatest common divisor of 0 and 0 must be 0. This also doesn’t introduce any problems since if a and b are not both zero, then at least one of them is not divisible by zero since zero only divides itself, so zero can’t be the gcd in that case.)

(7:44) For this definition of least common multiple, you need to restrict l to being a nonnegative integer, since least common multiples are never negative. Note that if a or b is zero, then l must also be zero, since the only integer that zero divides is itself. This forces *every* common multiple to be zero as well, and this works since 0 divides 0, giving us lcm(0, a)=lcm(a, 0)=0. Also, note that if a and b are both nonzero, then a and b have a nonzero common multiple so lcm(a, b) can’t be zero as zero doesn’t divide anything other than itself. (Suppose n∈ℤ and 0|n. Then by the definition of “divides, ” there is an integer k such that n=0k. Since 0k=0, this means that n=0, which proves that 0 only divides itself, as we’ve shown earlier that 0|0.) Just to cover my bases, you could also restrict a and b to both be nonzero integers and l to be a natural number.

(13:19) Narration error: “two k squared plus two” should be “two k squared plus two k.”

(15:07) I was fine with you just asserting that every integer was either even or odd before since we didn’t have precise definitions of “even” and “odd” yet, but now that we do, you should prove that fact, or at least give a reason why it’s true given our definitions of “even” and “odd, ” before using it in a proof. I don’t know if you’ll be assigning that as homework for your students at some point, though, so for now I’ll just say that the proof involves the Division Algorithm. I have an actual proof in one of my comments on another video on this channel, so you can look for that if you want to see a proof.

(17:05) Using 2k-1 where k is a natural number as an alternate form for odd natural numbers requires you to prove that that form is equivalent to the standard form from the definition of “odd.” This a fairly easy proof, so for the same reasons as above, I won’t include it here. However, you don’t actually need this alternate form for an odd natural number here. Just because n is a natural number doesn’t mean *k* has to be (for that matter, nothing in the logic of the proof prevents you from lifting the restriction in the proposition that n be a natural number and letting it be any integer), and the logic used here works just as well with the standard definition of odd, resulting in a final value of -4k=4(-k) instead.

(25:22) “Both not zero” and “not both zero” are not the same thing. The one you want here is “not both zero, ” not “both not zero” (same thing at 25:30).

(27:39) You haven’t covered that notation for “does not divide” previously in this course.

(36:18) Narration error: “x is even and y is odd” should be “x is even *or* y is odd.”