Thread: Contrapositive proof

Hey New here on the forum. I have a small problem with some contrapositive proofs..

I have to make a contrapositive proof that:
∀x, y ∈ R : x · y < 1 ⇒ x < 1 ∨ y < 1

I can logically see what I'm supposed to do.. and that the statement is correct.. but how do i prove it?? I know that the normal way is to prove that .. but I'm a little stuck from there.. anyone here who can help?

So you must prove that For all and

Idk how rigororous you need this proof to be...

The general idea is Let with

So we can multiply both sides of by y to get

But , so

Originally Posted by Kasperloeye

Hey New here on the forum. I have a small problem with some contrapositive proofs..
I have to make a contrapositive proof that:
∀x, y ∈ R : x · y < 1 ⇒ x < 1 ∨ y < 1

Originally Posted by artvandalay11

So you must prove that For all and

Idk how rigororous you need this proof to be...

The general idea is Let with

So we can multiply both sides of by y to get

But , so

Thanks that really cleared things for me ^^ all hail the mighty proofmaster Hehe.. but then i have another question..

Is it possible to make a contrapositive proof to prove that n² <= n! ? and by that i mean for which nonnegative integers would the statement apply.

Originally Posted by Kasperloeye

Thanks that really cleared things for me ^^ all hail the mighty proofmaster Hehe.. but then i have another question..

Is it possible to make a contrapositive proof to prove that n² <= n! ? and by that i mean for which nonnegative integers would the statement apply.

Are you familiar with proof by induction?

You wouldn't use a contrapositive proof, you would use proof by induction. You're base case would be n=4 since this fails for n=2 as and , and , and

So above we showed for n=4

Now we assume for some n and show it is true for

So consider (we need to show this quantity is a natural number)



By our assumption so

So for some

And now since and and so

and so therefore and so therefore

so by induction we are done


Now thats just an outline, you would need to "know/prove" that the sum of two natural numbers is a natural number and that the product of two natural numbers is a natural number, and something should be said like if k=0, nothing changes and if k>1 then k is a natural number

You also need to be aware that is by definition equivalent to

So I'm not in a totally wrong direction when I did this:

Originally Posted by CompuChip

In your particular problem, you could consider a statement like "For all n >= 2, n² <= (n!)".

Hehe that's where it get's funny.. because n>= 2, n²<=n! is like 2²<=2! 4 <= 2 .. so it has to n /= 2


So Basis should be n=4 since

4²<=4!
16 <=24

SO: For n=k, k²<=k!

but if we take it with n= k+1 it's (k+1)² <=(k+1)!
(k+1)² <= (k+1)(k!)
(k+1) <= k!
but our hypothasis says k^2 <= k!, and because k+1 <= k^2 (for k>=4) then
k+1 <= k^2 <= k!.

right??

yes that's also valid but for a rigorous proof you would need to state that k+1>0 and then need to prove that statement k+1<=k^2

but yes your ideas and logic are sound