Questions on Logic: Proofs answered by real tutors!

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Let and be two rational numbers, where a,b,c,d are integers. Then,



Since integers are closed under addition and multiplication, the numerator and denominator will both be integers. Hence, the sum of two rational numbers is a rational number. We can also say that rational numbers are closed under addition.

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Let p = a/b and q = c/d be two rational numbers, where a, b, c, d are integers.


Now add them:


p + q


a/b + c/d


(ad)/(bd)+(bc)/(bd)


(ad+bc)/(bd)


So p + q = (ad+bc)/(bd), which is a rational number (the numerator and denominator are both integers since integer addition and multiplication are both closed operations)


So this proves that the sum of any pair of rational numbers is a rational number.

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A person spins the pointer and is awarded the amount indicated by the pointer.
It costs $5 to play the game. The diagram shows a circle divided into 3 parts. The pointer points to the 1/2 section marked $2. The 1/4 section is marked $20, and the other 1/4 section is marked $5.
Determine:
The expectation of a person who plays the game.
E(x) = (1/2)2 + (1/4)20 + (1/4)5 = 1 + 5 + 5/4 = $7.25
------
The fair price to play the game.::$7.25
=============================================
Cheers,
Stan H.

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a) Quadrilateral can be divided into two triangles (each with vertices at the vertices of the quadrilateral) so the sum of the angles is 2*180 = 360

b) Same logic as a) but now we can divide into three triangles --> 3*180 = 540

c) 4*180 = 720

We can generalize to n sides and derive the sum of the angles of an n-gon to be 180(n-2).

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cents/apple
cents/apple x apples = cents
cents x dollar/cents = dollars
15 apples cost $2.25

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since schools will close per statement 4, then workers will lose their jobs per the statement 1.
the contra-positive to statement 2 is:
if workers lose their jobs then we did not save fuel.
since we did not save fuel, then there must be an energy crisis per statement 3.

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John

My calculator said it, I believe it, that settles it

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Presuming

where x is a positive integer. We know that the domain is {1,2,3,4,...} and the range is {...-2,-1,0,1,2,...}. Hence we can say that if x is even, the domain is {2,4,...} and the range is {1,2,3,...} (since the function is equivalent to f(x) = x/2). If x is odd, the domain is {1,3,...} and the range is {0,-1,-2,...}. There exists a one-to-one correlation between both disjoint domains and ranges, so the entire function is bijective.

However, we must be careful with one-to-one correlations with infinite sets such as {2,4,6,8,...} and {1,2,3,...} if x is even (e.g. look up "Hilbert's paradox of the Grand Hotel"). We can still define a function (e.g. f(x) = x/2) so we don't have to worry about this.

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Both must be irrational. Suppose that b-a is, on the contrary, a rational number r. Then a = b-r, contradiction because the sum/difference of two rational numbers is always rational (another way we can say this is that addition and subtraction are closed under rational numbers). The same logic applies to b+a = r.

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You have a slash / in there. That's not a logic operator.
Please repost and correct that error.

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i never did one like this before.
those instructors just get more inventive every year.
i think you need to use a truth table to determine the validity of this argument.
i read this as:
if (SvB) is true AND if (B->D) is true AND if (S->G) is true, THEN DvG is true.
this is equivalent to the statement:
(SvB) AND (B->D) AND (S->G) -> (DvG)
my truth table concerning this relationship is shown below:

in this table, T is true and . is false.
i used . rather than F because the difference between T and . shows up easier than the difference between T and F.
The first 4 columns are the truth tables for S, B,D,and G by themselves in relationship to each other.
Since there are 4 variable involved, there are 2^4 = 16 number of entries required.
the next column is SvB which translates to S or B.
S or B is true if S is true or if B is true.
it is only false if both S and B are false.
the next column is B->D which translates to B implies D, or: if B then D.
B implies D is true if B is true and D is true, or if B is false.
it is only false if B is true and D is false.
the next column is S->G which translates to S implies G, or: if S then G.
S implies G is true if S is true and G is true, or if S is false.
it is only false if S is true and G is false.
the next column is DvG which translates to D or G.
D or G is true if D is true or if G is true.
it is only false if both D and G are false.
the next column is (SvB)+(B->D)+S->G) which translates to SvB AND B->D AND S->G
it is only true if SvB and B->D and S->G are all true.
if any of them is false, then this category is also false.
the last column is ((SvB)+(B->D)+(S->G))->(DvG).
this says that is SvB is true and B->D is true and S->G is true, then DvG is true.
this last column is false only if SvB is true and B->D is true and S->G is true and DvG is false.
otherwise it is true.
since this last column is true in all circumstances, then it is reasonable to conclude that the fact that SvB and B->D and S->G are true implies that DvG is true.
There are no conditions where all 3 of them are true and DvG is not true.
i believe this is what you are looking for.
under the assumptions that i made about what the problem is addressing, the logic in the truth table proves that the original statements are true based on that interpretation.
good luck with this one.
hopefully this interpretation is correct.

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The error is here:
((x+y)(x-y)/(x-y))=(y(x-y)/(x-y))
Because we know that x=y
So x-y = 0
And we cannot divide by zero.

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m = y/x-b/x
-------
mx = y - b
y = mx + b
I'm convinced.

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Since we have the conclusion N, let's assume for the sake of argument that the opposite is true. In other words, let's assume that the conclusion is ~N. Our job is to show that a contradiction will arise, and if it does, then the opposite of ~N must be true (ie N is really the correct conclusion).

1.) (K > K) > R
2.) (R v M) > N                     / N
----------------
	3.)  ~N                     AIP
	4.)  ~(R v M)        2,3    Modus Tollens
	5.)  ~R & ~M         4      De Morgan's Law
	6.)  ~R              5      Simplification
	7.)  ~(K > K)        1,6    Modus Tollens
	8.)  ~(~K v K)       7      Material Implication
	9.)  ~~K & ~K        8      De Morgan's Law
	10.) K & ~K          9      Double Negation
11.) N                       3-10   IP

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Don't use x for multiplication. Use a dot, or *.

4*10^3*6*10^x all divided by 2.4*10-4 all equals 1.

Scientific notation of x times 10^y means you're multiplying x by that 10 raised to the y power. So deal with those 10 to some power as any base raised to an exponent.

24*10^(3+x) divided by 2.4 * 10^-4.

24/2.4 = 10 and (10^(3+x))/10^-4 = 10^(7+x) (when dividing bases with exponents, subtract the exponents. 3+x-(-4) = 7+x).

we now have 10*10^(7+x) = 1 (when multiplying bases with exponents, add the exponents. 10 is 10^1.)

10^(8+x) = 1 (any number to the 0 power is 1.)

10^(8+x) = 10^0 (If the same bases raised to different exponents are equal, then the exponents are equal.)

8+x = 0 (Subtract 8 from both sides.) x = -8

Check my profile and contact me for one on one online tutoring. I can help you understand this with one on one online tutoring. It's one thing to follow an example, but even better if you understand the concept.

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You'll have to come up with the reasons but it should go something like this:

~(J & K) -> ~J v ~K

~(L & M) -> ~L v ~M

Since J v L, ~(~J & ~L)

~(~J & ~L) -> ~K v ~M -> ~(K & M)

That last step, in English: Since J and L cannot both be false, one or the other or both of K and M must be false, so both K and M cannot be true.

John

My calculator said it, I believe it, that settles it

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You're correct. All you need to do now is factor it to 4(k^2 + 2k + 8). Since k^2 + 2k + 8 is an integer, the expression is divisible by 4.

Another way to prove it is to use modular arithmetic, or arithmetic dealing with remainders. If m is an odd integer, then m ≡ 1 mod 2 (i.e. m divided by 2 leaves a remainder 1. The ≡ sign means "equivalent"). Then m^2 ≡ 1 (mod) 4, 2m ≡ 2 (mod 4), and 5 ≡ 1 (mod 4). We can add modulos or residues the same way we add numbers; in this case, m^2 + 2m + 5 ≡ 1 + 2 + 1 ≡ 4 ≡ 0 (mod 4). Note that if a number is m mod n, we can subtract any multiple of n and still obtain an equivalent result. Since the expression is 0 mod 4, it is divisible by 4 (since it leaves remainder 0).

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Counterexample: Let a = 2, b = 7, c = 3, d = 13. Here, the conditions are met, but a+c divides b+d (5 divides 20).

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Make a conjecture about the sum of the angles of a convex polygon with 'n' sides and use mathematical induction to prove your conjecture.
----
Prove that an n sided quadrilateral
has sum of interior angles = (n-2)180 degrees
------------------------------------------------------
Step #1:
For n = 3, sum of angles = (3-2)180 = 180::::that is true
----
Step # 2:
Assume true for n = k sides, i.e,
sum of interio angles = (k-2)180
-----
Step # 3:
Prove true for n = k+1 sides
----
sum for k+1 = sum for k sides + 180
sum for k+1 = (k-2)180 + 180
Factor out the 180 to get:
sum for k+1 = (k-2+1)(180)
Therefore, sum for k+1 sides = [(k+1)-2)]180
Q.E.D.

Two positive integers that differ by 2 can always be written as k+1 and k-1,
k>=2

The difference of their squares is

(k+1)²-(k-1)² = 4k  

4k will be a perfect square if and only if k is a perfect square.

So let k = m²

Then (k+1)²-(k-1)² = 4k = 4m² = (2m)² = n².  Thus n = 2m

I.e., n can be and can only be any even positive integer.

Edwin

To prove:
P. R->~Q. P->Q therefore ~R

 P         given premise
 P->Q      given premise
----
∴Q         modus ponens
∴~~Q       double negation

  R->~Q    given premise
~~Q
-------
∴~R        modus tollens 

Edwin

 P -> Q Therefore ~Q -> ~P

P  Q  (P -> Q)  ~Q  ~P  (~Q->~P) 
T  T     T       F   F      T
T  F     F       T   F      F
F  T     T       F   T      T
F  F     T       T   T      T

They have the same truth table TFTT



Edwin

1.   P * P
2.   Q -> ~P    / :. ~Q
--------------------------
3.   P                   1      Simplification
4.   ~~P                 3      Double Negation
5.   ~Q                  2,4    Modus Tollens