Question

Prove by Contraposition. Let x and y be real numbers. If [x+y >= 2], the [x>=1 or y>=1].

What I got so far is:

If -[x>=1 or y>=1], then -[x+y >= 2].

When x < 1 and y < 1, then x+y < 2.

................

Answer 1

I know how to express even and odd numbers in a general setting; but how to show less than?

Answer 2

Or did i do my negating wrong?

Answer 3

-(pvq) = -p ^ -q

Answer 4

Would I show that the largest possible values for x and y cant sum to greater then 2?

Answer 5

max(x,y) + max(x,y) < 2 ??

but then would i have to define the max function?

Answer 6

what is Contraposition,

and the [x>, notation, is that dirac notation or what?

Answer 7

p -> q; contras to -q -> -p

Answer 8

the brackets just help me to sort out the p and q

Answer 9

and -> is an arrow?

Answer 10

yes, its the conditional notation

Answer 11

ok

Answer 12

Let m be the limit of f(x)=x as x approaches 1 from the left; and let n be the limit of f(y)=y as y approaches 1 form the left.

...maybe?

Answer 13

- so what you have wrote in your question that you have got ALL are very trivially and
so is true

Answer 14

right, i understand they are trivial but i find reinventing the wheel to BE trivial :) What i ended up settling on was

Prove by Contraposition. Let x and y be real numbers. If [x+y >= 2], the [x>=1 or y>=1].

If -[x>=1 or y>=1], then -[x+y >= 2].

If x < 1 and y < 1, then x+y < 2.

Let m be the limit of f(x)=x as x approaches 1 from the left; and let n be the limit of f(y)=y as y approaches 1 from the left. Then the limit of (m+n), as x and y approach 1, approaches 2 from the left.

Answer 15

- sorry but i think that what we can prove in case of two functions like f(x)=x and f(y)=y realy trivially that is true so i think that those not will be true and trivially for and in case of their limit
- this is just my opinion