Qwalasela abantu ababini \(A\) kunye \(B\) abangenawo umhla wokuzalwa ofanayo, apho \(A\) emncinci kuno \(B\) . Bonisa: Kukho amakroza amabini kanye ngobudala \(a,b \in \mathbb{N}\) apho oku kulandelayo kuyasebenza: \(2\cdot a = b\) . Saqala sabeka \(d \in \mathbb{R}^+\) njengomahluko weminyaka phakathi \(A\) kunye \(B\) ekuzalweni kuka \(A\) kunye \( d = d_0 + d_1 \) , \( d_0 \in \mathbb{N}_0, d_1 \in \mathbb{R}, d_1 \in [0;1[\) Ngoku siqwalasela ixesha elingenasizathu \(x \in \mathbb{R}^+\) emva kokuzalwa kuka \(A\) kunye \(x = x_0 + x_1\) , \(x_0 \in \mathbb{N}_0, x_1 \in \mathbb{R}, x_1 \in [0;1[\) .
Okwangoku ngeli xesha, ngokwenkcazo, \(a = \lfloor x \rfloor \) kunye \(b = \lfloor x+d \rfloor\) . Ngoku simisela zonke \(x\) ezizigcinileyo:
$$2 \lfloor x \rfloor = \lfloor x+d \rfloor \Leftrightarrow 2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor$$
Ityala lokuqala: \(0 \leq x_1 + d_1 < 1\):
Emva koko $$2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor \Leftrightarrow 2 x_0 = x_0 + d_0 \Leftrightarrow x_0 = d_0.$$
Oku kuthetha ukuba $$a = \lfloor x \rfloor = \lfloor x_0 + x_1 \rfloor = \lfloor d_0 + x_1 \rfloor = d_0$$ kwaye $$b = \lfloor x + d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor = 2 d_0$$ iqela lokuqala lobudala $$b = \lfloor x + d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor = 2 d_0$$ .
Ityala lesibini: \( 1 \leq x_1 + d_1 < 2 \):
Emva koko $$2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + d_0 + \underbrace{x_1 + d_1}_{\geq 1} \rfloor \Leftrightarrow 2 x_0 = x_0 + d_0 + 1 \Leftrightarrow x_0 = d_0 + 1.$$
Oku kuthetha ukuba $$a = \lfloor x \rfloor = \lfloor x_0 + x_1 \rfloor = \lfloor d_0 + 1 + x_1 \rfloor = d_0 + 1$$ yaye $$b = \lfloor x+d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + 1 + \underbrace{x_1 + d_1}_{\geq 1} \rfloor = 2 d_0 + 2$$ iqela lesibini leminyaka elifunayo.
Ngokwemigaqo ebambekayo, umzekelo, oku kuthetha: Ukuba umama wakho wakuzala xa uneminyaka \(20\) ubudala, umdala ngokuphindwe kabini kunawe kwi \(40\) kunye \(42\) yeminyaka ubudala. Imeko yokuba kwaye nini umama wakho \(n\) amaxesha amadala nawo anomdla: Apha ubeka \(n \in \mathbb{N}\) ngokungenasizathu kwaye ufumane \(x_0 = \frac{d_0}{n-1} \in \mathbb{N} \Leftrightarrow d_0 = k (n-1)\) . Oku kusebenza ukuba kwaye kuphela ukuba umahluko wexesha elipheleleyo \( \lfloor d \rfloor = d_0 \) luphinda-phindo lwe \(n-1\) . Umzekelo, kule meko ingasentla, umama wakho \(24\) iminyaka \(6\) ngokuphindwe ubudala bakho.