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How do you prove that n must divide m if n is square-free and n divides m squared?
Consider m, and its set of prime factors. e.g. m = 130 = 2 x 5 x 13 then m2 = (2 x 5 x 13)(2 x 5 x 13) n is a factor of m2. It must therefore be formed by the multiplication of a subset of the prime factors of m2. If this subset is contained in just one pair of brackets (see above) then it is contained in the prime factorisation of m, and so n is definitely a factor of m. [For instance, 2 x 5 = 10 is a factor of 1302 and 130. (2 x 5 x 13)(2 x 5 x 13)] If this subset spans both pairs of brackets, then n is a factor of m2 but not of m. If this is the case, however, n contains a repeated factor, and is not 'square-free'. [For instance, 2 x 2 x 5 = 20 is a factor of 1302 but not 130. (2 x 5 x 13)(2 x 5 x 13).But being divisible by 4, it is not square free.] To summarise: for a number to be divisible by m2 but not by m, it is necessarythat it has a repeated prime factor. For n to divide m, it is therefore sufficient (but not necessary) that n divides m2 and is square free.
What else can I help you with?
How do you prove the pythagorean theorem?
For any right angle triangle its hypotenuse when squared is equal to the sum of its squared sides.
Why would you divide 61 using prime numbers only?
To prove that 61 is a prime number.
Prove if a number is divisible by 3 then the sum of its digits must also be divisible by 3?
Suppose an (n+1)-digit number, X, is divisible by 3. Let X = a0*100 + a1*101 + a2*102 + ... + an*10n = a0 + a1*(1+9) + a2*(1+99) + ... + an*(1+999..9) = [a0 + a1 + a2 + ... + an] + [a1*9 + a2*99 + ... + an*999..9] 3 divides 9, 99, 999, etc so 3 divides each term in the second brackets (parentheses). Therefore 3 must divide the sum in the first brackets. That is, 3 must divide the sum of digits.
What is the decimal for one ninth?
The correct answer is .1 repeating because if you take a # like 4.86 square yards and find the feet for it you would have to divide by one ninth. If you need to cover a carpet with only that measurements and the area you need to cover is 110 feet squared you would multiply 110 by 4.86 then divide by 9 (feet in square yard) you would get 59.4 Then to prove it is .1 repeating multiply .1111 by 4.86 then multiply again by 110 you would get 59.39046. Round it and you get 59.4 there for the correct answer is .1 repeating.
How do you prove that 0.72 recurring as a fraction is 8 over 11?
Long division 8/11 = 8 divided by 11 .7272727272recurring
How do you prove cot squared theta plus cos squared theta plus sin squared theta scs squared theta?
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
How do you prove the pythagorean theorem?
For any right angle triangle its hypotenuse when squared is equal to the sum of its squared sides.
If n is an odd integer prove 8 divides n of n-1?
9
What did newton discover in 1684?
He discovered that the exponential range divide by the quatum physics of a bioshpere would equal that of a ginkosphere divide by the square root of pi. But that was only true if the radius of the two sides of a triangle was squared to equal 1. Which I have been conducting experiments to prove this and it has been adequet so far.
How do you prove that if M isn't prime or under 4 M divides into M-1 factorial?
5
Prove that the square root of 5 is irrational?
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.
Prove the lemma for each integer n 3 divides n if and only if 3 divides n squared?
This is an immediate consequence of the Fundemental Theorem of Arithmetic. The "only if" part of the assertion is trivial because if n = 3k then n squared = 3(3k^2). If we know that every number can be written as the product of primes in exactly one way, then it follows that p is in the prime factorization of AB only if p is in the prime factorization of A or p is in the prime factorization of B. Take p = 3 and A = B = n, and that proves the theorem. The Fundemental Theorem is really not necessary in proving this assertion; an alternate proof using "enumeration of cases" is possible. If 3 does not divide n then n is of the form 3k+1 or 3k+2 for some integer k. It is easy to check that (3k+1)^2 and (3k+2)^2 are both not divisible by 3.
How do you prove the formula for the volume of a square pyramid?
You really can't "prove" the formula. You use it. You first square the base 'b'. Then, you multiply that number by the height 'h'. Then, you divide the product of the base squared and height by 3. Boom! You get your answer. In my school, we get a formula sheet with all the formulas we will need to use. If you didn't understand the description above, here is the formula for a square pyramid:1/3b2 h.Hope this helped!
How do you prove a number is divisible by 48?
You divide that number by 48, and if it comes out equal, it is divisable by 48.
Why would you divide 61 using prime numbers only?
To prove that 61 is a prime number.
How can you prove that 6.25 is a quarter of 25?
A quarter is got when you divide something by 4. So if you divide 25 by 4, you will get 6.25 and if you multiply 6.25 by 4, you will get 25.
Prove identity sin4x equals 4sinxcosx times 1 minus 2sin squared x?
sin4x=(4sinxcosx)(1-2sin^2x)
