# Twice the age of two people1017

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.

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