Proofs using contrapositive or contradiction

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ash25
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Please Help! proofs using contrapositive or contradiction

Homework Statement



Prove using contrapositive or contradiction:
For all r,s∈R,if r and s are positive,then √r+ √s≠ √(r+s)
 
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I am so confused at this point. I have been working on this assignment for hours and I think that I am just over thinking it.
 


I haven't been working on this particular problem for hours, I have been working on the entire assignment for hours and I think that I am starting to blend everything together.
 


ash25 said:
I haven't been working on this particular problem for hours, I have been working on the entire assignment for hours and I think that I am starting to blend everything together.

Here at the PF, you must show some effort in order to receive tutorial help. What other similar questions have you been working on, and how did you solve them?
 


This is one of the first problems that I completed from this long assignment.


The equations y=-x^2 and y=-x+5 have no real solutions in common.
Proof:Suppose the 2 equations y=-x^2+2x+2 and y=-x+5 have at least one real solution,c.
→ -c^2+2c+2=-c+5
→c^2-3c+3=0
→c=(3±√(9-12))/2 not real
this is a contradiction therefore the 2 equations y=-x^2+2x+2 and y=-x=5 have no real solutions in common
 


Well, the method i would use to solve this problem is;

Proof it by contradiction. Just assume that it is correct, and then try to prove that the statement is true. If you arrive at a contradiction, you know that your original statement is true - that they're different from each other. You can do it like this;

sq(r)+sq(s)=sq(r+s)
Raise both sides of the equation to power 2;
(sq(r)+sq(s))^2=sq(r+s)^2
Which gives you;
s+r+2sq(s)sq(r)=s+r
And since you have that both r and s are positive, obviously it follows that;
s+r+2sq(s)sq(r)>s+r --> sq(r)+sq(s)≠sq(r+s)

cheers :P
 


Thank you very much! My head can now stop hurting! :) I knew I was over thinking this problem I really appreciate your time! Thanks again!