Proofs

As tan(x)=sin(x)/cos(x)

and sin(pi/4) = cos(pi/4) (= sqrt(2)/2)

then tan(pi/4) = 1

You cannot find the first 15 prime numbers by dividing by 2 and 3. This method would not identify 25 as being a composite.

if divides both and , then it will also divide therefore

will divide thus

The last part comes from the fact that:

if gcd(x,y)=g, then .

As proof:

Since g|x and g|y, let x=kg, and y=jg, then we have so g|(mx+ny).

One way is to get the prime factorization of the number. If every prime occurs an even number of times, it is a square, otherwise, not.

Another is to estimate the square root of the number, and square it. If you get more than the number, try a lower estimate; if less, a higher one. Using interval bisection you very quickly zero in on the square root, if it is a whole number. If so, the number is a perfect square. Otherwise, you find 2 consecutive whole numbers between which is the square root, in which case it is not a perfect square.

Jagged lines have toothed/serated edges whereas diagonal lines can have any edge so long as they run diagonally,so it is possible to have a jagged diagonal line!

Its perimeter is half of its circumference plus its diameter

Global winds, insolation, large bodies of water and ocean currents.

One fathom is equal to six feet. Therefore, 620 fathoms is equal to 620 x 6 = 3720 feet.

No. The sq root of 84 is 9.16555515139. 81 is a perfect square, where the sq root is 9.

a false aurelia requires a moderate amount of light, and will thrive with more light. the amount of water it needs will be directly proportionate to how much light it receives. they prefer their soil to be moist at all times, if it starts to dry out the leaves will quickly begin losing their color. i have two in my bedroom, the smaller of which is right in front of a window, it is definitely the thirstiest of all my plants, neeeding to be watered 2-3 times per week. besides having many different plants, i worked as an interior landscaper for a few years

nope! that's a unique name for a person so it probably wouldn't be very popular which it isn't; also a little bit wierd.

The probability of getting 3 white balls in a draw of 5 balls with replacement from an urn containing white balls and black balls is always greater than the same test without replacement, because the number of white balls decreases when you draw a white ball and do not replace it.

The ratio of white to black with replacement is constant, and is always less than one, assuming there is at least one black ball. The ratio of white to black without replacement decreases each turn, and is still less than one, and is less than the previous ratio, unless the question asked about 2 white balls or less.

Solubility (in that solvent and at that temperature).

theorem always needs proof

First, by induction it is shown that 102n-1 ≡ -1 (mod 11) for all natural numbers n (that is, all odd powers of 10 are one less than a multiple of 11.)

i) When n = 1, then we have 102(1)-1 = 10, which is certainly congruent to -1 (mod 11).

ii) Assume 102n-1 ≡ -1 (mod 11). Then,

102(n+1)-1 = 102n+2-1 = 102×102n-1

102 ≡ 1 (mod 11) and 102n-1 ≡ -1 (mod 11), so their product, 102×102n-1, is congruent to 1×(-1) = -1 (mod 11).

Now that it is established that 10, 1000, 100000, ... are congruent to -1 (mod 11), it is clear that 11, 1001, 100001, ... are divisible by 11. Therefore, all integer multiples of these numbers are also divisible by 11.

A palindrome containing an even number of digits may always be written as a sum of multiples of such numbers. The general form of such a palindrome, where all the As are integers between 0 and 9 inclusive, is

A0 100 + A1 101 + ... + An 10n + An 10n+1 + ... + A1 102n + A0 102n+1

which may be rewritten as

A0 (100 + 102n+1) + A1 (101 + 102n) + ... + An (10n + 10n+1)

and again as

100 A0 (1 + 102n+1) + 101 A1 (1 + 102n-1) + ... + 10n An (1 + 101)

Each of the factors in parentheses is one more than an odd power of 10, and is hence divisible by 11. Therefore, each term, the product of one such factor with two integers, is divisible by 11. Finally, the sum of terms divisible by 11 is itself divisible by 11.

QED

The sign of the larger number will be the sign of the answer, if you're adding them together. If they're the same size, and of different signs, the answer will be zero.

This requires no skill in PROOF, go to a different category man!

What's the ratio of 20 to 120? 1/6 QED :P

Waht, wiyd u ask?

The proof is based on reduction ad absurdum.

Suppose sqrt(5) is rational.

That is sqrt(5) = p/q from some co-prime integers p and q. If they are not co-prime, simply divide both by their GCF.

Multiply by q and square both sides, 5q^2 = p^2.

5 divides the left hand side so 5 must divide the right hand side.

That is, 5 divides p^2 and since 5 is prime, 5 must divide p.

That is p = 5r for some integer r.

Substituting for p gives: 5q^2 = (5r)^2 = 25r^2

Dividing both sides by 5 gives q^2 = 5r^2

5 divides the right hand side so 5 must divide the left hand side.

That is, 5 divides q^2 and since 5 is prime, 5 must divide q.

But that means that p and q are not co-prime: contradiction!

Therefore sqrt(5) is irrational.

No. Axioms and postulates are statements that we accept as true without proof.

Leonardo Fibonacci of Pisa. Not fibinacci!

Yes, here's the proof.

Let's start out with the basic inequality 36 < 48 < 49.

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

6 < √48 < 7.

If you subtract all numbers by 6, you get:

0 < √48 - 6 < 1.

If √48 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 √48. Therefore, √48n must be an integer, and n must be the smallest multiple of √48 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 √48n by (√48 - 6). This gives 48n - 6√48n. Well, 48n is an integer, and, as we explained above, √48n is also an integer, so 6√48n is an integer too; therefore, 48n - 6√48n is an integer as well. We're going to rearrange this expression to (√48n - 6n)√48 and then set the term (√48n - 6n) equal to p, for simplicity. This gives us the expression √48p, which is equal to 48n - 6√48n, and is an integer.

Remember, from above, that 0 < √48 - 6 < 1.

If we multiply this inequality by n, we get 0 < √48n - 6n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √48p < √48n. We've already determined that both √48p and √48n are integers, but recall that we said n was the smallest multiple of √48 to yield an integer value. Thus, √48p < √48n is a contradiction; therefore √48 can't be rational and so must be irrational.

Q.E.D.