Consider two people \(A\) and \(B\) who do not have the same birthday, where \(A\) is younger than \(B\) . Show: There are exactly two age constellations \(a,b \in \mathbb{N}\) for which the following applies: \(2\cdot a = b\) . 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 an arbitrary 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.
In concrete terms, this means, for example: If your mother gave birth to you at the age of \(20\) , she is exactly twice as old as you at \(40\) and \(42\) . The case of whether and when your mother is \(n\) times as old is also interesting: Here you set \(n \in \mathbb{N}\) arbitrarily and get \(x_0 = \frac{d_0}{n-1} \in \mathbb{N} \Leftrightarrow d_0 = k (n-1)\) . This works if and only if the integer age difference \( \lfloor d \rfloor = d_0 \) is a multiple of \(n-1\) . For example, in the above case, your mother is \(24\) years old \(6\) times your age.