yep, that's all I've got to say, we know surely that $y + yf(y)$ goes over all real numbers once as well, and that's about all I got.
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Why do you think that $f(y)y$ would be constant? Also, does "$y+f(y)y+1$ goes over all real numbers once" mean exactly once or just at least once? – Varun Vejalla Jun 02 '23 at 19:37
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If $f(y)=\frac{k}{y}$, where $k$ is some constant, then $f$ cannot be bijective. – Anurag A Jun 02 '23 at 19:39
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Choose your favorite bijection function $g(y)$ and set $ y + f(y)y+1 = g(y)$. Is $f(y) y = g(y) - 1 - y $ always a constant? – Calvin Lin Jun 02 '23 at 19:56
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@CalvinLin my point was to give a hint that $f$ cannot be bijective. I wasn't suggesting a function. – Anurag A Jun 02 '23 at 20:04
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@AnuragA Yes, I realized that after posting and editing, hence deleted the comment. Sorry! – Calvin Lin Jun 02 '23 at 20:05
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@CalvinLin my favorite bijection function is $g(y) = y$, then $g(y) - 1 - y = y - 1 - y = -1$ and thus it is always a constant. – 11235 Jun 02 '23 at 20:05
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@11235 Great ,what about for any bijective function? Can you find one where it is not a constant? – Calvin Lin Jun 02 '23 at 20:07
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alright, I was pulling your leg a bit, I admit, yep $g(y) = ay$ for any $a \ne 1$ is not constant – 11235 Jun 02 '23 at 20:11