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
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?
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:
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??