A blast from the past — a problem from the 1965 math olympiad

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[samsonblack]

"More equal than equal" – Dr. Robert Zombie

[SlipperyTeeth]

2:33 isn't a correct deduction. Consider 2/4 and 3/6. The simplest solution is to allow for k to be a rational number instead of an integer. And really, k could well be a real number.

[ArnaldoMandel]

Subtract 1 from each sids, you get fractions with 4z numerator. So, either z=0 or the denominators are equal. That immediately gets you x-y+z=0, without all these calculations.

[cHrtzbrg]

2:30 What about 2/4 = 3/6? (You can choose a rational k, though -- or just multiply both equations by their denominators, and see that they are equivalent -- assuming no denominator was zero)

[farfa]

0/0 could be 1, if you're brave enough.

[isuckatcodm]

I'm confused. What's the question asking for?

[abrahammekonnen]

Finally finished it! Interesting video, this is definitely a different flavor than your normal videos. I did like the comparison between the old and the new.

But I'm also not quite sure what to make of the final answer. So the fraction doesn't make any sense if z!=0, so z=0. And then what? I think I'm missing some subtle point here which I need to think about.

As usual thank you for the video.

[peterwaksman]

What am I missing? To say a/b and c/d are the same, means they are on a single line from the origin. The line's slope can be determined using (0, 0)

[sergeipetrov]

Thanks a lot for clear explication of the task! I got it. BTW, I have never thought about that point earlier.

[abrahammekonnen]

I do remember a video on equivalence relations, but i don't remember the ZxZ

[tcmxiyw]

You could just observe that the fraction is equal to 1 if the denominator is nonzero.

[terryendicott]

If a/b = c/d we get a-c = a - da/b = a(b-d)/b. Now divide a(b-d)/b by b-d. Done. True for all reals except bd equal to zero.

[pedroteran]

1965 must have been one of the very first Spanish MO.

[sawyersmith]

what is the problem I just see statements

[abrahammekonnen]

2:45 Note: As I'm watching this I realized that this reminds me of the discrete analouge to derivatives.

[wafikiri_]

Zero over zero is the only division by zero allowed. But it is undefined: any number can result.