inequality proof using contrapositon

[noblegas]

If x^2+2*x<0, then x<0.


Proof:


Assume that x is a real number and Using contraposition, supposed that x>=0(Assume ~Q). then x>=0 => x^2>=0. adding 2x on both sides of the inequality, the inequality becomes x^2+2*x>=2*x. Since x>=0 then 2*x>=0 and since x^2+2*x>=2*x, then x^2+2*x>=0(Therefore ~P). Since x>=0 => x^2+2*x>=0(~Q=> ~P), then x^2+2*x<0 => x<0.(P=>Q).

Is this how you would write this proof?

 

[stapel]

If x^2+2*x<0, then x<0.

Proof: Assume that x is a real number and Using contraposition, supposed that x>=0(Assume ~Q). then x>=0 => x^2>=0.

Strictly speaking, x² is not non-negative "because" x is non-negative. The square would be non-negative regardless of the value of x. A better statement (that is, one more likely to be graded completely) might be to say "In all cases, we have x² > 0. Then..." and so forth.

adding 2x on both sides of the inequality, the inequality becomes x^2+2*x>=2*x. Since x>=0 then 2*x>=0 and since x^2+2*x>=2*x, then x^2+2*x>=0(Therefore ~P).

Logically speaking, it might be better to state the "since" part before you actually use it: "By assumption, x > 0. Therefore, 2x > 0. Adding to either side of x² > 0, we have x² + 2x > 0 + 2x > 0 + 0 = 0."

Then conclude with your contrapositive conclusion.

In other words, your logic is pretty good; some added experience with the statement of your logic is all that's necessary. Nice work!

 

[Deleted member 4993]

If x^2+2*x<0, then x<0.


Proof:


Assume that x is a real number and Using contraposition, supposed that x>=0(Assume ~Q). then x>=0 => x^2>=0. adding 2x on both sides of the inequality, the inequality becomes x^2+2*x>=2*x. Since x>=0 then 2*x>=0 and since x^2+2*x>=2*x, then x^2+2*x>=0(Therefore ~P). Since x>=0 => x^2+2*x>=0(~Q=> ~P), then x^2+2*x<0 => x<0.(P=>Q).

Is this how you would write this proof?

Another way:

x^2+2*x<0 → x(x+2)<0 →

either x <0 and x+2 > 0 .....................(1) condition satisfied

or x > 0 and x + 2 < 0 ............... Impossible