On December 20, 2023, Brent will stop offering office hours.
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Comment on What Must Be True?
Regardless of p, assuming q
q-1 or q-2: one of the two has to be even.
Thus, the result must be even.
Is this correct?
That is correct, devroy.
That is correct, devroy. Great work.
Hi, quick question; the
I'm just plugging in 2 for an
I'm just plugging in 2 for an even value and 1 for an odd value, and then performing the necessary calculations.
HOWEVER, once I see that one of the pieces in the product equals zero, I immediately know that the final result will equal zero.
For example, when we plug in 2 for p and q in the first expression, we get 2(2-2)(2-1)
Once I recognize that 2-2 = 0, then we have 2(2-2)(2-1) = 2(ZERO)(2-1), which means the final product will be zero.
understood. thanks :)
Hi Brent,
Thanks for the great videos. I was wondering if we should look to choose numbers that make the equation equal zero or should we try to avoid them? Can this be a trap for some question is what I was concerned about?
I'm glad you like the videos,
I'm glad you like the videos, Harshbir!
What equation are you referring to?
Sorry, I was referring to the
Thanks!
Thanks!
For me, it's much easier to test (plug in) nice even and odd numbers (like 0's and 1's) than it is to use E's and O's and then have to apply the different rules for even and odd numbers.