OK: help on approaching:
n is integer and is a multiple of 4, then n is a multiple of 4.
Now I know when you plug in 6 - it disproves it - but I need to write a legitimate proof that disproves it and I'm unsure as to where to start.
OK: help on approaching:
n is integer and is a multiple of 4, then n is a multiple of 4.
Now I know when you plug in 6 - it disproves it - but I need to write a legitimate proof that disproves it and I'm unsure as to where to start.
You really only need one counterexample to disprove a statement, so showing that 6 makes the statement false is sufficient proof. You may observe that this statement would in general be false for any number of the form 4x + 2, where x is an integer, but 6 does the job. What you basically need to write is that 6 is an integer and 36 is a multiple of 4, but 6 is not a multiple of 4. It's that simple.
It would be appropriate to write:
Let n = 6.
Then the statement "If n is an integer and n^2 is a multiple of 4, then n is a multiple of 4" is false.
Hence, the statement is false in general, since for the statement to be true, it must be true for all acceptable values of n.
That is about as comprehensive a proof as I or anyone can give.
OK NEXT QUESTION:
Find a statement about whether or not the sum of 2 consecutive even integers is divisible by 4. Prove your statement.
SO: (my answer) For all integer x, if both x and x+2 are even and consecutive, then their sum is divisible by 4.
Proof: Let x and x+2 be two even consecutive integers.
So x=2k and x+2 = 2(k+2) = 4k+4 for some integer k.
Therefore, 2k+2k+4 = 4k+4 = 4(k+1) which is divisible by 4.
Does that make ANY sense???
see I was thinking that before - so what happens then?
Its false, therefore to write a statement that makes it false you get
There exists two integers (x?) such that both are not even and consecutive and their sum is not divisible by 4.
How do you prove that?
3 and 11: neither even nor consecutive and 3+11=14 witch is not divisible by 4 ... does THAT work?
You can write such a statement and prove it as you have shown, although I would claim that this statement is not about even consecutive integers. By using the generic algebra, you should arrive at the result that no two even consecutive integers have a sum that is divisible by four.
Yeah except I'm not really certain how to completely negate it all. I just know how to negate the specific one - ya know? Not "No integer x ..."
OK, a few little questions - sorry to bug you!
Determine whether sentence is a statement of open sentence or neither:
The plane is leaving in five minutes.
- technically it could be any plane - but the "the" in front suggests the specific one, therefore its a statement?
Get a note from your doctor.
-neither?
If n and m are even integers, then nm is even.
-statement - however it does have variables even though the truth doesn't depend on those variables because we've already specified that they have to be even
and:
for each statement, determine universal or existential quantifiers and rewrite as "for all..." or "there exits ... such that ..." Introduce variables when appropriate:
All positive real numbers have square roots.
-There exists a positive real number x such that x has a square root. -- yeah?
You obtain the negation of a statement by reversing all of its quantifiers. So, the negation of "For all pairs of two even consecutive integers, their sum is not divisible by four" is "There exists a pair of even consecutive integers whose sum is divisible by four." The two quantifiers "for all" and "is not" are replaced by "there exists" and "is". And the negation of "All positive numbers have a square root" is "There exists a positive number that does not have a square root."
No - ok: I switched problems on ya: not negation these are two separate problems
the very bottom one just wants you to write the statement and determine if its existential or universal. - I'm saying its existential - is that right?
and what about the ones above it - are my answers correct?