Tixgeli laba qof \(A\) iyo \(B\) oo aan dhalan isla maalin iyo \(A\) kayar \(B\) . Muuji: Dhab ahaan waxa jira laba xiddigood oo da 'ah \(a,b \in \mathbb{N}\) , oo khuseeya: \(2\cdot a = b\) Waxaan marka hore dejinay \(d \in \mathbb{R}^+\) sida farqiga da'da ee u dhexeeya \(A\) iyo \(B\) dhalashada \(A\) leh \( d = d_0 + d_1 \) , \( d_0 \in \mathbb{N}_0, d_1 \in \mathbb{R}, d_1 \in [0;1[\) . Waxaan hadda tixgelinaynaa qodob aan macquul ahayn waqtiga \(x \in \mathbb{R}^+\) dhalashada \(A\) kadib \(x = x_0 + x_1\) , \(x_0 \in \mathbb{N}_0, x_1 \in \mathbb{R}, x_1 \in [0;1[\) .
Waqtigan xaadirka ah, qeexitaan ahaan, \(a = \lfloor x \rfloor \) iyo \(b = \lfloor x+d \rfloor\) . Waxaan hadda go'aamineynaa dhammaan \(x\) wixii xajinaya:
$$2 \lfloor x \rfloor = \lfloor x+d \rfloor \Leftrightarrow 2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor$$
Kiiska 1aad: \(0 \leq x_1 + d_1 < 1\):
Kadib $$2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor \Leftrightarrow 2 x_0 = x_0 + d_0 \Leftrightarrow x_0 = d_0.$$
Tani waxay ka dhigan tahay in $$a = \lfloor x \rfloor = \lfloor x_0 + x_1 \rfloor = \lfloor d_0 + x_1 \rfloor = d_0$$ iyo $$b = \lfloor x + d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor = 2 d_0$$ da'da koowaad $$b = \lfloor x + d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor = 2 d_0$$ .
Kiiska 2aad: \( 1 \leq x_1 + d_1 < 2 \):
Kadib $$2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + d_0 + \underbrace{x_1 + d_1}_{\geq 1} \rfloor \Leftrightarrow 2 x_0 = x_0 + d_0 + 1 \Leftrightarrow x_0 = d_0 + 1.$$
Tani waxay ka dhigan tahay in $$a = \lfloor x \rfloor = \lfloor x_0 + x_1 \rfloor = \lfloor d_0 + 1 + x_1 \rfloor = d_0 + 1$$ iyo $$b = \lfloor x+d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + 1 + \underbrace{x_1 + d_1}_{\geq 1} \rfloor = 2 d_0 + 2$$ kooxda labaad ee da 'da.
Gaar ahaan, tan macnaheedu waa, tusaale ahaan: Haddii hooyadaa kugu dhasho da'da \(20\) sano, waxay dhab ahaan labanlaab tahay da'daada \(40\) iyo \(42\) sano. Sidoo kale waa wax xiiso leh haddii iyo goorta ay \(n\) jeer ay duug tahay: Halkan \(n \in \mathbb{N}\) si aan macquul ahayn oo waxaan helnaa \(x_0 = \frac{d_0}{n-1} \in \mathbb{N} \Leftrightarrow d_0 = k (n-1)\) . Tani waxay si sax ah u shaqeysaa marka farqiga da'da isku dhafan \( \lfloor d \rfloor = d_0 \) tiro badan tahay \(n-1\) , tusaale ahaan kiiska kore hooyadaadu waa \(24\) sano \(6\) jeer da'daada.