- #1

imDooiNGMATH

- 5

- 1

- Homework Statement
- Prove ##wt(x+y) \leq wt(x) + wt(y)##. where "wt(x)" is referring to the weight of a specific code word

- Relevant Equations
- ##0 \leq wt(x)##, ##0 \leq wt(y)##

I'm trying to prove the following:

##wt(x+y) \leq wt(x) + wt(y)##, where "wt(x)" is referring to the weight of a specific code word.

Proof:

For two code words ##x, y \in F^{n}_2##, we have the inequalities ##0 \leq wt(x)## and ##0 \leq wt(y)##. Adding these together, we have ##0 \leq wt(x) + wt(y)##. From here, adding ##wt(x+y)## to both sides, we have ##wt(x+y) \leq wt(x) + wt(y) + wt(x+y)##. From this, it is clear that ##wt(x) + wt(y) \leq wt(x) + wt(y) + wt(x+y)##. Thus, ##wt(x+y) \leq wt(x) + wt(y) \leq wt(x) + wt(y) + wt(x+y)##.

At first, I was thinking I should do cases where ##x = y## and ##x \neq y##, but reading over it, this seems sufficient.

##wt(x+y) \leq wt(x) + wt(y)##, where "wt(x)" is referring to the weight of a specific code word.

Proof:

For two code words ##x, y \in F^{n}_2##, we have the inequalities ##0 \leq wt(x)## and ##0 \leq wt(y)##. Adding these together, we have ##0 \leq wt(x) + wt(y)##. From here, adding ##wt(x+y)## to both sides, we have ##wt(x+y) \leq wt(x) + wt(y) + wt(x+y)##. From this, it is clear that ##wt(x) + wt(y) \leq wt(x) + wt(y) + wt(x+y)##. Thus, ##wt(x+y) \leq wt(x) + wt(y) \leq wt(x) + wt(y) + wt(x+y)##.

At first, I was thinking I should do cases where ##x = y## and ##x \neq y##, but reading over it, this seems sufficient.