Question

Logic: write the converse and contrapositive of the following statement:
for all real numbers x, if -10 my answers: converse: for all real numbers x, if x+1>0 then -1x>=0

Answer 1

@triciaal @Nnesha

Answer 2

@zepdrix

Answer 3

Converse is correct :)
We have make some clever adjustments for the Contrapositive though.

Answer 4

The problem we is that \(\large\rm x+1<0\) is not the negation of \(\large\rm x+1>0\)

Answer 5

x+1>=0 maybe

Answer 6

You need to also include the 0 value (since you didn't at the start.).
\(\large\rm \neg\left[x+1>0\right]\equiv \left[x+1\le0\right]\)

Yes good good :)

Answer 7

Oh maybe I misread your answer too quickly hehe

Answer 8

We flip the sign, but we also include 0.

Answer 9

ooh so x+1>=0 would make it correct right

Answer 10

No.
You didn't flip your inequality sign.

x+1 > 0
Should change to
x+1 <= 0

Answer 11

But we also need to fix the other inequality.
That one is a bit more difficult.

Answer 12

\[\large\rm -1<x\le0\]Drawing a number line will really help on this one.

Answer 13

ok then .. if i undertand correctly for all real numbers x, if x+1 <=0 then -1>=x>0

Answer 14

Do you understand how I graphed the inequality?
It's this shaded region in the middle for \(\large\rm -1<x\le0\) ya?