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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
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.
Matrix prove if Ax equals Bx then A equals B?
If x is a null matrix then Ax = Bx for any matrices A and B including when A not equal to B. So the proposition in the question is false and therefore cannot be proven.
What is the size of the interior angle of a regular pentagon?
108 degrees
The sum of all interior angles of a polygon is equal to the following:
180n-360, where n is the number of sides.
For n=5, the total of all of the angles is 180*5-360=900-360=540.
Since every angle in a regular polygon, each interior angle of a regular pentagon is 540/5=108 degrees.
Is a quadrilateral a parallelogram if both pairs of opposite sides are equal?
Yes.
Although the definition of a parallelogram is "a quadrilateral with both pairs of opposite sides parallel", the only way for a quadrilateral to include opposite sides of equal length is if the included angles are the same, and hence the sides are parallel.
(Hint : draw a diagonal to a parallelogram. You can show that one of the two triangles formed is the mirror image of the other, which immmediately proves that each pair of opposite sides is equal.)
Where were the first tessellations originally created?
in the roman times a man named tuffacuoo made it up he was the first man to make a tessellatios. sincerley,karmentyh
Given a three digit number n = "d1 d2 d3" (i.e. a number 0 <= n <= 999, where d1, d2 and d3 are it's digits), we can express n as 100d1 + 10d2 + 1d3. The number n' with the digits of nin reverse order would be: n' = "d3 d2 d1", which could be expressed as n' = 100d3 + 10d2 + 1d1. Subtracting n from n' (or vice versa) we get the following equation:
n - n' = (100d1 + 10d2 + 1d3) - (100d3 + 10d2 + 1d1)
resolving and rearranging we obtain:
n - n' = (100d1 - 1d1) + (10d2 - 10d2) + (100d3 - 1d3) = 99d1 - 99d3 = 99(d1 - d3)
Since d1 and d3 are integers (and d2 cancels out), we see that the result 99(d1 - d3) is always divisible by 99.
(The equation holds for all integers values for d1, d2 and d3, not just for the digits 0, 1, ... , 9, but we can no longer write it as a "three digit number". )
Example:
651 - 156 = (6*100 + 5*10 + 1*1) - (1*100 + 5*10 + 6*1) = 500 - 0 - 5 = 495 = 99*5
Do the numbers increase forever?
yes. Numbers never stop. They keep going and going. This is called google
