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Proofs
Proof means sufficient evidence to establish the truth of something. It is obtained from deductive reasoning, rather than from empirical arguments. Proof must show that a statement is true in all cases, without a single exception.
1,294 Questions
Proof that there are no three consecutive primes except 3 5 and 7?
Let p, p+2, and p+4 be three consecutive odd numbers. Then, if p ≡ 0 (mod 3), p is divisible by 3; if p ≡ 1 (mod 3), then p+2 is divisible by 3, and if p ≡ 2 (mod 3), then p+4 is divisible by 3.
That is, at least one of p, p+2, and p+4 must be divisible by 3. The only prime that can be divisible by 3 is 3 itself; for any other positive integer divisible by 3 must have another factor, making it composite. This gives the possibilities
-1, 1, 3
1, 3, 5
3, 5, 7
The first two are eliminated because -1 and 1 are not primes, leaving 3, 5, 7 as the only set of three consecutive integers that are all prime.
One way to answer this question is from first principles. Let
q = a + bi + cj + dk
and
v = e + fi + gj + hk
Then,
qv
= ae - bf - cg - dh
+ i(af + be + ch - dg)
+ j(ag - bh + ce + df)
+ k(ah + bg - cf + de)
= ae - bf - cg - dh + afi + bei + agj +cej + ahk + dek
+ i(ch - dg)
+ j(df - bh)
+ k(bg - cf)
and (keeping in mind that reals commute under multiplication)
vq
= ae - bf - cg - dh
+ i(af + be - ch + dg)
+ j(ag + bh + ce - df)
+ k(ah - bg + cf + de)
= ae - bf - cg - dh + afi + bei + agj + cej + ahk + dek
- i(ch-dg)
- j(df-bh)
- k(bg-cf)
Let m = ae - bf - cg - dh + afi + bei + agj +cej + ahk + dek and n = i(ch - dg) + j(df - bh) + k(bg - cf). Then,
qv = m + n
and
vq = m - n.
For qv to equal vq, we must have n = 0. For n to equal zero, each of its parts must equal zero also, that is,
ch = dg
df = bh
bg = cf
so that either
c/g = d/h
d/h = b/f
b/f = c/g
i.e. (bi + cj + dk) is a multiple (not necessarily integer) of (fi + gj + hk)
or
b = 0, c = 0, d = 0,
or
f = 0, g = 0, h = 0.
Therefore, the nonreal part of v must be a multiple (possibly zero) of the nonreal part of q. More concisely, quaternions q and v commute under multiplication if and only if
v = pq + r
where p and r are real.
A neater proof, requiring more background, can also be given.
Quaternions are often regarded as the combination of a scalar part (the real part of the quaternion) and the vector part (the part containing i, j, and k.) That is, the quaternion a + bi + cj + dk corresponds to the combination of the scalar a and the vector
qv = ac + ad + bc - b·d + b×d.
In the product qv, the order of the factors in each of these five products is reversed. That is, the product is now
vq = ca + da + cb - d·b + d×b.
The first four products commute, whereas the last (a cross product) does not. Therefore,
vq = ac + ad + bc - b·d + d×b.
so that qv = vq if and only if b×d = d×b. It is known that the cross product is anticommutative, meaning that b×d = -d×b. Therefore, this condition reduces to b×d = θ, the zero vector, meaning that one is a scalar multiple of the other (i.e. they lie along the same line.) This gives the same result as the purely algebraic derivation above.
How old would you be at a million minutes?
You'd be 1,000,000 minutes old.
However, in larger units some assumptions will have to be made:
As months do not all have the same number of days, when the 1,000,000 minutes started would affect the number of days, so an average number of mins per month will be used; then:
525,949 min ≈ 1 year
→ 1,000,000 mins ≈ 1,000,000 ÷ 525,949 years = 1 year 474,051 mins
525,949 mins ÷ 12 months ≈ 43,829.1 mins/month
→ 474,051 mins ≈ 474,051 ÷ 43,829.1 months = 10 months 35760 mins
1440 mins = 1 day
→ 35760 mins = 35760 ÷ 1440 days = 24 days 1200 mins
60 mins = 1 hour
→ 1200 mins = 1200 ÷ 60 hours = 20 hours
→ 1,000,000 mins ≈ 1 year, 10 months, 24 days, 20 hours.
Or approx 1 year, 10 months, 25 days old.
A counter example occurs when somebody makes a claim that all members of some category of things have a particular property, and then someone else proves that the claim is not true by showing an example of a thing in the category that does not have the property claimed.
For example, if someone claimed that "All presidents of the United States are dead white men", then Barack Obama would be a counter-example because he is a president of the United States but isn't a dead white man.
For another example, if someone claimed that all mammals bear live young, the echidna and the platypus would be counter-examples because they are mammals that lay eggs.
For another example, if someone claimed that the United States is the only country that has never defaulted on its debts, Australia and Tuvalu would be counter-examples. The claim is logically equivalent to saying that all countries in the category of being not the USA have defaulted on a debt at least once,....
For another example, if someone claimed that all prime numbers are odd, "2" would be a counter example. Or if someone claimed that all odd numbers are prime "9" would be a counter-example.
In short, a counter-example to a proposition or claim is an example that proves that the proposition or claim is not true.
How can you prove e-infinity equals 0?
I think you mean e to the (- infinity) power. The proof would be a limit proof.
The limit (as n-->infinity) of [( en) ] = 0
You should have some other limits in class that you have proven.
Show that your limit is less than one of those given for all values of n then you have your proof.
For instance, if you already know that lim (as n-->infinity) of [(1/n) ] = 0
then for n = 1, 1/e1 < 1/1 true
for n = 2, 1/e2 < 1/2 true
Then assume that it is true for n = k
so for n = K, 1/eK < 1/K assumed true
therefore: eK > K
multiply both by e
e(k+1) > ke
but we know ke > k+1 because e>1
SO: e(k+1) > ke > k+1 now take the reciprocal (reverses the inequalities)
1/e(k+1) < 1/ke < 1/ (k+1)
by transitive prop of inequalities eliminate the middle term
so that 1/e(k+1) < 1/ (k+1) this proves the case for n=K+1 and therefore will be true for all values of n since k was never a specified value.
And if: 1/e(k+1) < 1/ (k+1) by one of the properties of limits, since the lim of 1/n is zero, then the lim of 1/en is also zero when n --> infinity.
Proof for the midpoint theorem 7.5?
The midpoint theorem says the following:
In any triangle the segment joining the midpoints of the 2 sides of the triangle will be parallel to the third side and equal to half of it
In typing how are net words per minute calculated?
To determine net words a minute:
Total words (5 characters = 1 word), divided by number of minutes of the timing, minus 2 for each error = nwpm
So, if you typed 75 words in 3 minutes with 2 errors on the timing, you would calculate net words per minute as follows:
75 words, divided by 3 (number of minutes) = 25 gwpm (gross words per minute) minus 4 (2 errors) = 21 nwpm - (net words per minute)
Minkowski inequality for p equals 2?
For a general Lp space: In the notation of Lp norms:
Let f and g be Lp functions, then:
f+gp <= fp+gp
Specifically for p=2, using integrals, we have (where "S" means integral):
(S(f+g)2)1/2 <= (S(f)2)1/2+(S(g)2)1/2
and again, replacing p with 2 will yield the definition is a general Lp space.
What are congruence theorems and postulates?
If the sides AB, BC and CA of triangle ABC correspond to the sides DE, EF and FD of triangle DEF, then the two triangles are congruent if:
- AB = DE, BC = EF and CA = FD (SSS)
- AB = DE, BC = EF and angle ABC = angle DEF (SAS)
- AB = DE, angle ABC = angle DEF, angle BCA = angle EFD (ASA)
If the triangles are right angled at A and D so that BC and EF are hypotenuses, then the triangles are congruent if
- BC = EF and AB = DE (RHS)
- BC = EF and angle ABC = angle DEF (RHA).
Formally, the difference between two numbers, x and y, is abs(x - y) which is never negative. If the two numbers are the same then the difference is 0 else it is positive.
If you define the difference as x - y then it is
positive if x > y,
0 if x = y and
negative if x < y.
How many people would there be on each team if there was 24 boy and 30 girls?
Assuming the teams had the same number of people on each, each team would have 27 people.
(24+30) = 54/2 = 27
Diagonals bisect each other at what angels for a rhombus?
They bisect each other at an angle of 90 degrees
Does a parallelogram have diagonals that bisect each other?
They do in some parallelograms, not in others.
Can a square have rounded corners?
NO! Then it wouldn't be a square! It would be another shape, definetely not a square!
Which of the following best describes a bisector of an angle?
The set of all points in a plane that are equidistant from the two sides of a given angle
What symbol represents contradiction?
In Formal Logic proofs, the contradiction is represented with an inverted T (or upside-down T) as follows: ┴
The contradiction symbol can be introduced at any time a logical contradiction is encounterd, for example, all of the the following contradictory logical statements (using different symbols) can be replaced with the contradiction symbol:
The ball is completely blue and the ball not completely blue.
P ^ ¬P
P & ~P
P & !P
P AND NOT P
Surface area = 2*(L*B + B*H + H*L) cubic unitswhere
L = length
B = Breadth
H = Height.
What fractions are a perfect square?
None. Perfect squares, by definition, are the squares of counting numbers and these cannot be fractions.
Can you tell how to submit an attempted proof of Fermat's Last Theorem?
PIERRE DE FERMAT' S LAST THEOREM.
CASE SPECIAL N=3 AND.GENERAL CASE N>2. .
THE CONDITIONS.Z,X,Y,N ARE THE INTEGERS . Z*X*Y*N>0.N>2.
Z^3=/=X^3+Y^3 AND Z^N=/=X^N+Y^N.
SPECIAL CASE N=3.
WE HAVE
(X^2+Y^2)^2=X^4+Y^4+2X^2*Y^2.
BECAUSE
X*Y>0=>2X^2*Y^2>0.
SO
(X^2+Y^2)^2=/=X^4+Y^4.
CASE 1. IF
Z^2=X^2+Y^2
SO
(Z^2)^2=(X^2+Y^2)^2
BECAUSE
(X^+Y^2)^2=/=X^4+Y^4.
SO
(Z^2)^2=/=X^4+Y^4.
SO
Z^4=/=X^4+Y^4.
CASE 2. IF
Z^4=X^4+Y^4
BECAUSE
X^4+Y^4.=/= (X^2+Y^2.)^2
SO
Z^4=/=(X^2+Y^2.)^2
SO
(Z^2)^2=/=(X^2+Y^2.)^2
SO
Z^2=/=X^2+Y^2.
(1) AND (2)=> Z^4+Z^2=/=X^4+Y^4+X^2+Y^2.
SO
2Z^4+2Z^2=/=2X^4+2Y^4+2X^2+Y^2.
SO
(Z^4+Z^2+2Z^3+Z^4+Z^2-2Z^3)=/=(X^4+X^2+2X^3+X^4+X^2-2X^3)+)(Y^4+Y^2+2Y^3+Y^4+Y^2-2Y^3)
SO IF
(Z^4+Z^2+2Z^3)/4=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4
=> (Z^4+Z^2-2Z^3)/4=/=(Z^4+Z^2-2Z^3)/4+(Z^4+Z^2-2Z^3/4)
AND
SO IF
(Z^4+Z^2-2Z^3)/4=(Z^4+Z^2-2Z^3)/4+(Z^4+Z^2-2Z^3)./4
=> (Z^4+Z^2+2Z^3)/4=/=(Z^4+Z^2+2Z^3)/4+(Z^4+Z^2+2Z^3)/4
BECAUSE
(Z^4+Z^2+2Z^3)/4 - (Z^4+Z^2-2Z^3)/4 =Z^3.
SO
Z^3=/=X^3+Y^3.
GENERAL CASE N>2.
Z^N=/=X^N+Y^N.
WE HAVE
[X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=X^(N+1)/2+Y^(N+1)/2+ H.
BECAUSE X*Y>0=>H>0.
SO
[X^(N-1)/2+Y^(N-1)/2]^(N+1)/(N-1)=/= X^(N+1)/2+Y^(N+1)/2
CASE 1. IF
Z^(N-1)/2=X^(N-1)/2+Y^(N-1)/2
SO
[Z^(N-1)/2]^(N+1)/(N-1)=[X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1).
BECAUSE
[X^(N-1)/2+Y^(N-1)/2 ]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2.
SO
[Z^(N-1)/2]^(N+1)/(N-1)=/=X^(N+1)/2+Y(N+1)/2.
SO
Z^(N+1)/2=/=X^(N+1)/2+Y^(N+1)/2.
CASE 2. IF
Z^(N+1)/2=X^(N+1)/2+Y^(N+1)/2
SO
[Z^(N+1)/2]^(N-1)/(N+1)=[X^(N+1)/2+Y^(N+1)/2 ]^(N-1)/(N+1)
BECAUSE
[X^(N+1)/2+Y^(N+1)/2](N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2.
SO
[Z^(N+1)/2]^(N-1)/(N+1)=/=X(N-1)/2+Y^(N-1)/2.
SO
Z^(N-1)/2=/=X(N-1)/2+Y^(N-1)/2..
SO
(1) AND (2)=> Z^(N+1)/2+Z^(N-1)/2=/=X^(N+1)/2+Y^(N+1)/2+X^(N-1)/2+Y^(N-1)/2.
SO
2[Z^(N+1)/2+Z^(N-1)/2]=/=2[X^(N+1)/2+Y^(N+1)/2]+2[X^(N-1)/2+Y^(N-1)/2.]
SO
[Z^(N+1)/2+Z^(N-1)/2+2Z^N ]+[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]+[X^(N+1)/2+X^(N-1)/2-2X^N ]+[Y^(N+1)/2+Y^(N-1)/2+2Y^N ]+[Y^(N+1)/2+Y^(N-1)/2-2Y^N ]
SO IF
[Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2+2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4=>
[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4
AND
IF
[Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=[X^(N+1)/2+X^(N-1)/2-2X^N ] /4+ [Y^(N+1)/2+Y^(N-1)/2-2Y^N ]/4
=>
[Z^(N+1)/2+Z^(N-1)/2+2Z^N ]/4=/=[X^(N+1)/2+X^(N-1)/2+2X^N ]/4 + [Y^(N+1)/2+Y^(N-1)/2+2Y^N ]/4
BECAUSE
[Z^(N+1)/2+Z^(N-1)/2+2Z^N ] /4- [Z^(N+1)/2+Z^(N-1)/2-2Z^N ]/4=Z^N.
SO
Z^N=/=X^N+Y^N
HAPPY&PEACE.
Trantancuong.
