If a^2 is divisible by 7 then a is divisble by 7 to? how???

Answer 1

a can be written as a product of primes.
and
a^2 is the same product squared.
if 7 (which is a prime) divides a^2, then 7 has to be one of the primes in the list, so 7 divides a also. .

(for example: 21= 3x7, 21^2 = 3^2 x 7^2 ).

Answer 2

For cases where a is a whole integer,

a^2 is divisible by 7 that means a^2/7 is also a whole number

since a^2 is a whole number, it can be broken down to be a product of prime numbers one of them being 7

a^2 = 7 x p1 x p2 x p3 ... x pn - where p1 - pn are prime numbers

we can take all the prime numbers and substitute P

a^2 = 7 x P
a^2/7 = P
or 1/7 x a^2 = P

since P is a whole number (product of prime numbers) and a^2 is also a whole number, there has to be another 7 in a^2 for the term a^2/7 to be a whole number. If there isn't then we would end up with a fraction and not a whole number.

so now we know that a^2 = 7 x 7 x P

sqrt(a^2) = sqrt(7^2 x P)
sqrt(a^2) = sqrt(7^2) x sqrt(P)
a=7 x sqrt(P)

since a is a whole number then sqrt(P) is also a whole number and we can just substitute n = sqrt(P) where n is a whole number

a = 7 x n

therefore a is divisible by 7

Answer 3

"a is divisible by 7" means we are dealing with integers. Any number is divisible by 7 if we are in the domain of all reals, so it becomes a meaningless statement.

So, if a is an integer such that a²/7 is also an integer, then we know that 7 is a factor of a². Because sqrt(7) is not an integer, we see that 7² must be a factor of a². If 7² is a factor of a², then 7 must be a factor of a.

Answer 4

Do the division of a by 7, say a =7q + r You raise it to the square and you get a^2 = 49 q^2 + 14 q*r + r^2. If a^2 is a multiple of 7, so is r^2 because 49 q^2 + 14 q*r is a multiple of 7. Now r^2 is 0,1,4 9, 16,25 or 36 because r<7. Therefore r has to be 0, which means a=7q.

Answer 5

Claim: For an integer a, let a^2 be divisible by 7. Then "a" is also divisible by 7.

Proof: (by contradiction) Let's supposed a is NOT divisible by 7. Then "a" does not contain 7 as a prime factor. a would contain prime factorization of

a = c[1]p[1]*c[2]p[2]*...c[n]p[n], for p[1]...p[n] prime numbers, and c[1]...c[n] whole numbers. (I'm denoting everything in square brackets as a subscript). None of p[1] .... p[n] can be equal to 7.

If we square both sides, then

a^2 = {c[1]}^2{p[1]}^2 * .... * {c[n]}^2{p[n]}^2

And we essentialy double the factors and whole numbers.
Since none of p[1] .... p[n] is equal to 7, it follows that a^2 is not divisible by 7. This is a contradiction.

Therefore, a is divisible by 7.

Answer 6

With remainder zero?
a=7, then
a*a=7*7 that is divisible by 7 , i.e. a=7 is divisible by 7 .
So, 7 is the minimum number that divides a^2 and divide "a" with remainder zero.
You can take 7,14,21,28,....you will get the result.
So, i think it is true.

Answer 7

a^2 = a * a
For an expression to be divisible by 7, one of its factors must be divisible by 7. there is only one factor in this example; so, both a^2 and a * a are divisible by 7.

Answer 8

Here is Puggy's proof in plain language: If a is not divisible by 7, then a is a product of a bunch of prime numbers, none of which is 7.

Then a² is a product of the squares of a bunch of prime numbers, none of which is 7. Because 7 is prime, no combination of any of the numbers within this bunch of, none of which is 7, will multiply to 7. Therefore, 7 is not a factor of a².

Answer 9

If a square is divisible by 7
this implies a square contains 7 square as it is a x a

Answer 10

Because, if a is an integer, it can only be a product of 7. (7, 14, 21, etc.) Other wise a^2 cannot be divisible by 7.

MDP