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Consider two natural numbers $x^{(1)} \in \mathbb{N}$ and $y^{(1)} \in \mathbb{N}$ with the following relation: $x^{(1)}y^{(1)} > (x^{(1)} + 1)(y^{(1)} - 1)$.

I am wondering if exists a different pair of natural numbers $x^{(2)} \in \mathbb{N}$ and $y^{(2)} \in \mathbb{N}$ with $x^{(2)} > x^{(1)}$ and $y^{(2)} < y^{(1)}$ such that $x^{(1)}y^{(1)} > x^{(2)} y^{(2)} \geq (x^{(1)} + 1)(y^{(1)} - 1)$.

Thanks for any help.

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  • $\begingroup$ How about $x^{(2)}=x^{(1)}y^{(1)}$ and $y^{(2)}=1$ (assuming $y^{(1)}\gt 1$)? $\endgroup$
    – Henry
    May 28 '16 at 1:26
  • $\begingroup$ What does $\cdot^{(z)}$ mean? $\endgroup$ May 28 '16 at 2:00
  • $\begingroup$ @YoTengoUnLCD I think he means $ab > (a+1)(b-1)$ and $ab > cd \geq (a+1)(b-1)$ where $c > a -2$ and $d < b$ and $a, b, c, d \in \mathbb{N}^+$ $\endgroup$
    – coolcat007
    May 28 '16 at 2:19
  • $\begingroup$ Not always. Let $x_1 \geq y_1=1.$ $\endgroup$ May 28 '16 at 2:53
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If you mean in general for any $x^{(1)}\in\mathbb{N}$ and $y^{(1)}\in\mathbb{N}$, then no, because $y^{(1)}$ can be 0 and therefore there doesn't exist a $y^{(2)} < y^{(1)}$.

If you want just one example:

$x^{(1)} = 1$

$y^{(1)} = 1$

$x^{(2)} = 2$

$y^{(2)} = 0$

$x^{(1)}y^{(1)} = 1 >$

$x^{(2)}y^{(2)} = 0 \geq$

$(x^{(1)}+1)(y^{(1)}-1)=0$

For strictly larger: $x^{(1)}y^{(1)}>x^{(2)}y^{(2)}>(x^{(1)}+1)(y^{(1)}−1)$:

$x^{(1)} = 3$

$y^{(1)} = 2$

$x^{(2)} = 5$

$y^{(2)} = 1$

$x^{(1)}y^{(1)} = 6 >$

$x^{(2)}y^{(2)} = 5 >$

$(x^{(1)}+1)(y^{(1)}-1)=4$

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  • $\begingroup$ Thank you! Actually, I was looking for an example where $x^{(1)} < x^{(2)} - 2$. I edit the question. Sorry. $\endgroup$ May 28 '16 at 1:29
  • $\begingroup$ @BrunoFanzeres You're welcome. Also a nice sidenote: For $x^{(1)}, y^{(1)}, x^{(2)}, y^{(2)} \in [0,1000)$ there are $54261214$ with the strict property and $55987745$ with the property you described. $\endgroup$
    – coolcat007
    May 28 '16 at 1:29
  • $\begingroup$ @BrunoFanzeres for $x^{(1)} < x^{(2)} -2$, pick $x^{(1)} = 1, y^{(1)} = 1, x^{(2)} = 4, y^{(2)} = 0$, then $x^{(1)}y^{(1)} = 1 > x^{(2)}y^{(2)} = 0 \geq (x^{(1)} + 1)(y^{(1)} - 1) = 0$ $\endgroup$
    – coolcat007
    May 28 '16 at 1:36
  • $\begingroup$ forgot to mention. I am assuming $0 \notin \mathbb{N}$. Sorry, $\endgroup$ May 28 '16 at 1:56
  • $\begingroup$ @BrunoFanzeres So you mean $x^{(1)}, y^{(1)}, x^{(2)}, y^{(2)} \in \mathbb{N}^+$? Then the general case still doesn't hold because $y^{(1)}$ can be 1 and thus there doesn't exist a $y^{(2)} < y^{(1)}$. For the single example: $x^{(1)} = 4, y^{(1)} = 2, x^{(2)} = 7, y^{(2)} = 1$. $\endgroup$
    – coolcat007
    May 28 '16 at 2:12

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