Signature of a even/odd transposition

[Matherer]

Homework Statement

(1 2 3 4 5 6 7)
(1 2 5 4 3 6 7)

is s, a permuation.
Find sgn(s).

Homework Equations

The sgn of a transposition is -1.

Sgn(s)=Product of [sgn(j)-sgn(i)]/(j-i) when 1<=i<j<=n

The Attempt at a Solution

I calculated the sgn(s) using the formula and got 1. But shouldn't it actually be -1, since s is a transposition? If not, why is it not a transposition when only two elements are swapped and the rest are stationary?

 

[ystael]

Sgn(s)=Product of [sgn(j)-sgn(i)]/(j-i) when 1<=i<j<=n

This formula is wrong; actually it's nonsense, since whenever $$1 \leq i < j \leq n$$, $$\mathop{\mathrm{sgn}} i = \mathop{\mathrm{sgn}} j = 1$$ so $$\frac{\mathop{\mathrm{sgn}} j - \mathop{\mathrm{sgn}} i}{j - i} = 0$$.

The formula you're thinking of is $$\mathop{\mathrm{sgn}} \sigma = \prod_{1 \leq i < j \leq n} \mathop{\mathrm{sgn}}(\sigma(j) - \sigma(i)) = \prod_{1 \leq i < j \leq n} \frac{\sigma(j) - \sigma(i)}{|\sigma(j) - \sigma(i)|}$$. This is the product of $$-1$$ once for every inversion of $$\sigma$$, where an inversion of $$\sigma$$ is a pair $$(i, j)$$ where $$i < j$$ but $$\sigma(i) > \sigma(j)$$.

If you use the correct formula, you will reach the correct result $$-1$$.