E noʻonoʻo i nā poʻe ʻelua \(A\) a me \(B\) i hānau ʻole ʻia ma ka lā like a ʻo \(A\) kaikaina ma mua o \(B\) . Hōʻike aia aia ʻelua mau makahiki o nā hōkū \(a,b \in \mathbb{N}\) , no nā mea e pili ana: \(2\cdot a = b\) . Ua hoʻonohonoho mua mākou iā \(d \in \mathbb{R}^+\) ma ke ʻano he ʻokoʻa makahiki ma waena o \(A\) a me \(B\) i ka hānau ʻana o \(A\) me \( d = d_0 + d_1 \) , \( d_0 \in \mathbb{N}_0, d_1 \in \mathbb{R}, d_1 \in [0;1[\) . E noʻonoʻo mākou i kahi manawa kūpono i ka manawa \(x \in \mathbb{R}^+\) ma hope o ka hānau ʻana o \(A\) me \(x = x_0 + x_1\) , \(x_0 \in \mathbb{N}_0, x_1 \in \mathbb{R}, x_1 \in [0;1[\) .
I kēia manawa i ka manawa, e like me ka wehewehe ʻana, \(a = \lfloor x \rfloor \) a me \(b = \lfloor x+d \rfloor\) . Hoʻoholo mākou i kēia manawa \(x\) no nā mea paʻa:
$$2 \lfloor x \rfloor = \lfloor x+d \rfloor \Leftrightarrow 2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor$$
Hihia 1st: \(0 \leq x_1 + d_1 < 1\):
A laila $$2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor \Leftrightarrow 2 x_0 = x_0 + d_0 \Leftrightarrow x_0 = d_0.$$
ʻO kēia ka manaʻo ʻo $$a = \lfloor x \rfloor = \lfloor x_0 + x_1 \rfloor = \lfloor d_0 + x_1 \rfloor = d_0$$ a me $$b = \lfloor x + d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor = 2 d_0$$ ka $$b = \lfloor x + d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor = 2 d_0$$ makahiki mua a $$b = \lfloor x + d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor = 2 d_0$$ .
2 hihia: \( 1 \leq x_1 + d_1 < 2 \):
A laila $$2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + d_0 + \underbrace{x_1 + d_1}_{\geq 1} \rfloor \Leftrightarrow 2 x_0 = x_0 + d_0 + 1 \Leftrightarrow x_0 = d_0 + 1.$$
ʻO kēia ka manaʻo ʻo $$a = \lfloor x \rfloor = \lfloor x_0 + x_1 \rfloor = \lfloor d_0 + 1 + x_1 \rfloor = d_0 + 1$$ a me $$b = \lfloor x+d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + 1 + \underbrace{x_1 + d_1}_{\geq 1} \rfloor = 2 d_0 + 2$$ ka ʻelua mau makahiki i makemake ʻia.
ʻO ke kikoʻī, ke ʻano o kēia, no ka laʻana: Inā hānau kou makuahine iā ʻoe i ka makahiki o \(20\) mau makahiki, ʻelua mau manawa ʻo ia i kou makahiki i \(40\) a me \(42\) mau makahiki. Hoihoi nō hoʻi inā a ʻoiai ʻo ia ʻo \(n\) manawa i ʻelemakule ai: Eia \(n \in \mathbb{N}\) a loaʻa mākou \(x_0 = \frac{d_0}{n-1} \in \mathbb{N} \Leftrightarrow d_0 = k (n-1)\) . Hana pololei kēia i ka manawa o ka ʻokoʻa o ka makahiki integer \( \lfloor d \rfloor = d_0 \) maha o \(n-1\) , e laʻa me nā mea i luna aʻe nei ʻo kou makuahine he \(24\) mau makahiki \(6\) mau manawa i kou mau makahiki.