Consider two natural numbers $n_1$ and $n_2$, out of which one is twice as large as the other. We are not told whether $n_1$ is larger or $n_2$, we can state following two propositions:

**PROPOSITION 1: **The difference $n_1-n_2$, if $n_1 >n_2$, is different from the difference $n_2-n_1$, if $n_2 >n_1$.

**PROPOSITION 2: **The difference $n_1-n_2$, if $n_1 >n_2$, is the same as the difference $n_2-n_1$, if $n_2 >n_1$.

Moving on the proofs:

**PROOF OF PROPOSITION 1: **Let $n_1 > n_2$, then $n_1=2n_2$. Therefore, $n_1-n_2=2n_2-n_2=n_2$. That’s, the difference is $n_2$.

Let $n_2 >n_1$, then $n_1=\dfrac{1}{2} n_2$. Therefore, $n_2-n_1=n_2-\dfrac{1}{2}n_2=\dfrac{1}{2}n_2$. That’s, the difference is $\dfrac{1}{2}n_2$.

$\Rightarrow$ The proposition is true. $\Box$

**PROOF OF PROPOSITION 2: **Let $n_1 >n_2$, then $n_1-n_2 = n$, where $n$ is a fixed natural number.

If $n_2 >n_1$, then $n_2-n_1$ again equals to $n$.

Therefore, the proposition is true. $\Box$.

If you re-read the propositions, you’ll find that these are actually contradicting each others. But as both can’t be true, which one should be treated as the correct one?

If you think that proposition 2 is true, then you need to reconsider and read again the propositions. Proposition 2 seems correct at first instance, but it isn’t. *Take an example, *let $n_1=40$ and $n_2=20$, then $n_1-n_2=20$. Again, let $n_1=10$ and $n_2=20$, then $n_2-n_1=10$. Hence, the proposition 1 is true. $\Box$

*Reference: *“A curious paradox”, R. Smullyan Download As PDF*Related*

Excellent Reader, Good Orator and Bad Writer. Talk about WordPress, PHP and Mathematics.

## Single Comment