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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
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.
What type of figure is a parallelogram if the diagonals of a parallelogram are perpendicular?
A rhombus, square, kite or arrowhead.
Can a group contain its operation as an element?
Yes, as shown by this example:
S={*}
G=(S,*)
*:S X S-->S
*(*,*)=*
However, I could not find any nontrivial examples.
What is motion in 3 dimensions?
The 3 dimensions, normally, are:
- left-right,
- front-back, and
- up-down.
Any motion in all three of these dimensions - exemplified when riding a camel! - is motion in 3-D.
Total surface area of a cuboid?
With sides of length A, B and C units, the total surface area is
2*(AB + BC + CA) square units.
How can you prove that -1 equals 1?
We know -1=-1, make these two into different but equivalent fractions -1/1=1/-1, take the square root of both sides sqrt(-1/1)=sqrt(1/-1), the rule sqrt(x/y)=sqrt(x)/sqrt(y) is true so we can write... sqrt(-1)/sqrt(1)=sqrt(1)/sqrt(-1), simplify the fractions i/1=1/i, multiply both sides by i -1/1=i/i, i/i=1 and -1/1 is -1 so we can write... -1=1 There's your proof i=sqrt(-1) spoiler:i is imaginary
What is the name of the postulate that states through any 3 points a circle can be formed?
There cannot be such a postulate because it is not true.
Consider a line segment AB and let C be any point on the line between A and B. If the three points are A, B and C, there can be no circle that goes through them. It is so easy to show that the postulate is false that no mathematician would want his (they were mostly male) name associated with such nonsense.
Well, if one of the points approach the line that goes through the other two points, the radius of the circle diverges. The line is the limit of the ever-growing circles. In the ordinary plane, the limit itself does not exist as a circle, but mathematicians have supplemented the plane with infinity to "hold" the centres of such "straight" circles.
Spherical balls have no corners. The American football has 2 corners.
A ball (of any shape) may be sewn together of polygons each of which has several corners (vertexes).
Do the diagonals of a rhombus bisect each other?
Yes. Because the diagonals are perpendicular to each other and intersect at their midpoints, they bisect each other.
What is the AAA theorem and the SSS postulate?
There is no AAA theorem since it is not true.
SSS is, in fact a theorem, not a postulate. It states that if the three sides of one triangle are equal in magnitude to the corresponding three sides of another triangle, then the two triangles are congruent.
Where can i find Proof square root of 5 is irrational?
The proof that the square root of 5 is irrational is exactly the same as the well-known proof that the square root of 2 is irrational - except using 5 in place of 2. We can prove a more general result: the square root of any prime is irrational.
First of all, we require the lemma:
for any prime p, and integer x,
p|x2 ⇒ p|x
That is, if x2 is divisible by p, then so is x.
Proof:
The prime factorization of x2 necessarily contains p at least once, since it is divisible by p. But it also has to contain an even power of every prime, since it is the prime factorization of a square. Therefore, it contains p at least twice, and its square root, x, contains p at least once: that is, x is divisible by p.
Now, given a prime p, assume that its square root is rational. Then, it may be written in the form a/b, where a and b have no common factors (that is, the fraction a/b is in lowest terms). This is always possible for any nonzero rational number. Since this quantity is the square root of p, its square equals p, that is
(a/b)2 = p
a2/b2 = p
a2 = pb2
Now, pb2 is a multiple of p, so a2 must be too. And, using the result above, this means that a must be a multiple of p also. Thus, there exists an integer c such that
a = PC
Then,
(PC)2 = pb2
p2c2 = pb2.
Since p is not zero, we may divide both sides by p to obtain
PC2 = b2
That is, b2 is divisible by p also, and thus b is divisible by p.
Since a and b were both divisible by p, the fraction a/b could not have been in lowest terms, which contradicts our initial assumption. Therefore, the square root of p cannot possibly be a rational number. Since 5 is prime, the proof is complete.
