Consider two persons \(A\) and \(B\) who do not have the same birthday on the same day and who are \(A\) younger than \(B\). Show: There are exactly two age constellations \(a,b \in \mathbb{N}\) for which holds: \(2\cdot a = b\). We first set \(d \in \mathbb{R}^+\) as the age difference between \(A\) and \(B\) at the birth of \(A\) with \( d = d_0 + d_1 \), \( d_0 \in \mathbb{N}_0, d_1 \in \mathbb{R}, d_1 \in [0;1[\). We now consider any time \(x \in \mathbb{R}^+\) after the birth of \(A\) with \(x = x_0 + x_1\), \(x_0 \in \mathbb{N}_0, x_1 \in \mathbb{R}, x_1 \in [0;1[\).
At this point in time, according to the definition, \(a = \lfloor x \rfloor \) and \(b = \lfloor x+d \rfloor\). We now determine all \(x\) to which:
$$2 \lfloor x \rfloor = \lfloor x+d \rfloor \Leftrightarrow 2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor$$
case number one: \(0 \leq x_1 + d_1 < 1\):
Then $$2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor \Leftrightarrow 2 x_0 = x_0 + d_0 \Leftrightarrow x_0 = d_0.$$
This means that $$a = \lfloor x \rfloor = \lfloor x_0 + x_1 \rfloor = \lfloor d_0 + x_1 \rfloor = d_0$$ and $$b = \lfloor x + d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor = 2 d_0$$ the first age $$b = \lfloor x + d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + \underbrace{x_1 + d_1}_{< 1} \rfloor = 2 d_0$$ for.
2nd case: \( 1 \leq x_1 + d_1 < 2 \):
Then $$2 \lfloor x_0 + x_1 \rfloor = \lfloor x_0 + d_0 + \underbrace{x_1 + d_1}_{\geq 1} \rfloor \Leftrightarrow 2 x_0 = x_0 + d_0 + 1 \Leftrightarrow x_0 = d_0 + 1.$$
This means that $$a = \lfloor x \rfloor = \lfloor x_0 + x_1 \rfloor = \lfloor d_0 + 1 + x_1 \rfloor = d_0 + 1$$ and $$b = \lfloor x+d \rfloor = \lfloor x_0 + x_1 + d_0 + d_1 \rfloor = \lfloor 2 d_0 + 1 + \underbrace{x_1 + d_1}_{\geq 1} \rfloor = 2 d_0 + 2$$ the second desired age constellation.
Specifically, this means, for example: If your mother gave birth to you at the age of \(20\) years, she is exactly twice your age at \(40\) and \(42\) years. It is also interesting if and when she is \(n\) times that old: Here \(n \in \mathbb{N}\) arbitrarily and we get \(x_0 = \frac{d_0}{n-1} \in \mathbb{N} \Leftrightarrow d_0 = k (n-1)\) . This works exactly when the integer age difference \( \lfloor d \rfloor = d_0 \) a multiple of \(n-1\) , e.g. in the above case your mother is \(24\) years \(6\) times your age.