Qwalasela abantu ababini \(A\) kunye \(B\) abangazalwanga kwangalemini inye kwaye \(A\) bancinci kuno \(B\) . Bonisa ukuba minye iminye iminyaka emibini yobudala \(a,b \in \mathbb{N}\) , esebenza kuyo: \(2\cdot a = b\) . Saqala ukuseta \(d \in \mathbb{R}^+\) njengomahluko wobudala phakathi kwe \(A\) kunye \(B\) ekuzalweni kwe \(A\) nge \( d = d_0 + d_1 \) , \( d_0 \in \mathbb{N}_0, d_1 \in \mathbb{R}, d_1 \in [0;1[\) . Ngoku sijonga inqaku elingenakuphikiswa ngexesha \(x \in \mathbb{R}^+\) emva kokuzalwa kwe \(A\) nge \(x = x_0 + x_1\) , \(x_0 \in \mathbb{N}_0, x_1 \in \mathbb{R}, x_1 \in [0;1[\) .
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.
Ngokukodwa, oku kuthetha, umzekelo: Ukuba umama wakho wakuzala eneminyaka eyi- \(20\) iminyaka, uyiphindaphindeke kabini iminyaka yakho \(40\) kunye \(42\) iminyaka. Kunika umdla ukuba kwaye nini na \(n\) amaxesha amdala: Nantsi \(n \in \mathbb{N}\) ngokungafanelekanga kwaye sifumana \(x_0 = \frac{d_0}{n-1} \in \mathbb{N} \Leftrightarrow d_0 = k (n-1)\) . Oku kusebenza kanye xa umahluko wobudala obupheleleyo \( \lfloor d \rfloor = d_0 \) nge \(n-1\) , o.k.t. kwimeko engentla umama wakho \(24\) iminyaka \(6\) amaxesha akho eminyaka.