Proofs

It is the same as finding the square root.

Trigonometric functions, exponential functions are two common examples.

yes it can, if you know how to use or have mathematica have a look at this demo http://demonstrations.wolfram.com/ApproximationOfDiscontinuousFunctionsByFourierSeries/

It is proved by contradiction (reductio ad absurdum), a powerful type of proof in mathematics.

Assume that the square root of 2 is rational and is equal to a/b where a and b are integers.

Squaring both sides gives:

2=a2/b2

a2=2b2

Since a2 is even, it implies that a is even

So, replacing a by 2k where k is an integer, we have:

(2k)2=2b2

b2=2k2

Since b2 is even, it implies that b is even, which is in contradiction to our first statement: a/b is in lowest terms.

Thus, the square root of 2 is irrational. (Q.E.D)

Note: It can be proved that if a2 is even, then a is also even.

Proof:

Odd numbers are of the form 2n+1, where n is an integer.

When an odd number is squared, we have:

(2n+1)2 = 4n2+4n+1 = 2(2n2+2n) +1 = 2y+1

where y = 2n2+2n

y is also an integer; so, 2y+1 is an odd number, which in turn means that the square of any odd number is also odd.

Therefore, if the square of a number is even, the number cannot be odd; it has to be even. (Q.E.D)

Neither true nor false. Some theorems can be proven using geometric arguments and methods, others cannot.

everything

one method is simply writing out all the factors of both 48 and 84:

48's factors: 1,48,2,24,4,12,16,3,8,6

84's factors:1,84,2,42,3,28,4,21,6,14,7,12

therefore by comparison greatest factor is 12.

The only other way i know is by reducing to prime factor form:

48 is 2^4 x 3 and 84 is 2^2x3x7

You can then see the highest common factor will be by using 2^2 x 3 which is common in the prime factors of 48 ad 84 (2^2 x 3 = 12)

to prove x|-|y≤|x-y| have to look at 4 cases:

a) x>0, y>0

b) x<0, y<0

c) x>0, y<0

d) x<0, y>0

(to save typing it out over and over again, i have shortened absolutes to abs)

For: a) the values dont matter both sides will always have the same value.

b) let x=-a, y=-b. Because of the abs around x and y, the left will be a-b and the right b-a so the left and right will have the same number but opposite signs, so after taking the abs on both sides, they will always be equal to one another.

c) let x=a, y=-b. The left will be a-b and the right a+b. if b<a then both sides are positive and the statement holds. if a<b then left is negative and right positive, but because you're starting from above 0 when you take away b (on the left side) once you take the abs left will be less than the right. when a=b the left is just 0 and right is positive so statement still holds.

d) let x=-a, y=b. left is a-b again, but the right is-a-b so always negative. if a<b then a-b is negative but not as low as -a-b so left is less than right after taking abs. if b<a then a-b is positive, but moving closer to 0 and -a-b moves further away from 0 so left is less than right again after taking abs. If a=b then then the left is always 0 and after taking abs the right is 2a therefore the statement holds for all 4 cases. Q.E.D.

Yes. Not only that, they are counting numbers.

5040

There is no upper bound to the sum of the numbers in the Fibonacci sequence; both the last number in the series and consequently the sum of all these numbers can be made as large as desired by continuing the series to sufficiently many numbers.

3

23 and 29

2222= 3 mod 7

5555 = 4 mod 7

2222^5555 + 5555^2222 = (3^5)^1111 + (4^2)^ 1111

as it is a^n + b^n and n is odd so it is divisible by (a+b)

3^5 + 4^ 2 = 259

hence by 7 as 7 is a factor of 259

Yes, here's the proof.

Let's start out with the basic inequality 25 < 30 < 36.

Now, we'll take the square root of this inequality:

5 < √30 < 6.

If you subtract all numbers by 5, you get:

0 < √30 - 5 < 1.

If √30 is rational, then it can be expressed as a fraction of two integers, m/n. This next part is the only remotely tricky part of this proof, so pay attention. We're going to assume that m/n is in its most reduced form; i.e., that the value for n is the smallest it can be and still be able to represent √30. Therefore, √30n must be an integer, and n must be the smallest multiple of √30 to make this true. If you don't understand this part, read it again, because this is the heart of the proof.

Now, we're going to multiply √30n by (√30 - 5). This gives 30n - 5√30n. Well, 30n is an integer, and, as we explained above, √30n is also an integer, so 5√30n is an integer too; therefore, 30n - 5√30n is an integer as well. We're going to rearrange this expression to (√30n - 5n)√30 and then set the term (√30n - 5n) equal to p, for simplicity. This gives us the expression √30p, which is equal to 30n - 5√30n, and is an integer.

Remember, from above, that 0 < √30 - 5 < 1.

If we multiply this inequality by n, we get 0 < √30n - 5n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √30p < √30n. We've already determined that both √30p and √30n are integers, but recall that we said n was the smallest multiple of √30 to yield an integer value. Thus, √30p < √30n is a contradiction; therefore √30 can't be rational and so must be irrational.

Q.E.D.

There are very many different mathematical definitions of distance: the Euclidean metric, the Minkovski metric are two common examples. The proof will be different.

Some are possible, others are not.

Tourism in Australia is a good thing because it generates a lot of income for Australian towns. The bad thing about it is that many people end up getting injured from snakes, scorpions, and other wildlife.

I will outline a way to prove it for you. I will also five a simple vector proof for those that have studied vectors. For the first proof, one can often cite some of these as known facts or refer to theorems in a text. 1. First show that a rhombus is a parallelogram 2. Next, using the above, show that diagonals of the rhombus divide it into 4 congruent triangles. 3. Last, use CPCTC and not that all 4 middle angles are congruent so that are 90 degrees. From this is is easy to say that the diagonals are perpendicular. Hints. to prove 1, use the fact that all 4 sides of the rhombus are congruent and then use SSS to find two congruent triangles. Then use CPCTC to show that the angles are the same and find a transversal. Look at same side interior angles cut by that transversal and say something about them being parallel. 2. Use SSS again and find 4 congruent triangles and look at the diagonals. I will help more by giving you another proof using vectors that is really much more straightforward. A rhombus is a quadrilateral with all sides having equal length. This means that if two vectors, a and b that form the corner of a rhombus, then the magnitude of a and b are equal The diagonals of the parallelogram are precisely a+b and a-b. Now look at the dot product of a+b and a-b and see that it is zero and remember that a dot product of zero means the vectors are perpendicular or orthogonal The first part is a pure synthetic geometry approach and if anyone need more help to finish that, just ask, The second part is a vector proof which is elegant because it is so simple.

Hi All,

I think the answer for this question is,

Testing method :Verification(review) and Validation

Testing types : Basically static and dynamic-> in dynamic testing further can be classified into structural(white box) and functional(black box) testing

Testing Technique : in white box testing we have 1) loop coverage 2)statement coverage 3) condition coverage 4) decision coverage

in black box testing we have 1) Equilance partioning, 2)boundary value analysis 3) error guessing

Testing Levels : Unit testing,

integration testing systemtesting ,

Acceptance testing

OMG more easy... (Without equations)

percent increase = [amount * (1 + x%)]

x% = x/100; 6% its (6/100) = .06

200000*1.06 = 212,000

If the angles are congruent, they will be less than 360 degrees.

A segment need not be a bisector.

No theorem can be used to prove something that may not be true!

Yes. A regular tessellation can be created from either an equilateral triangle, a square, or a hexagon.

A line can be tangent to a circle in which case it intersects it in one point, it can intersect it in two points, or no points at all.

So the choices are 0,1 or 2.