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A Prime number is a number that has only two factors which are itself and one.
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Prime number is 180 what is the prime number of a and b?
180 is not a prime number.
Which is a prime numberWhich is a prime number A 21 B 31 C 57 D 91?
It is: B 31 is a prime number because it has only two factors which are itself and one
What number of Prime Minister was Lester B Pearson?
14th Prime Minister
Is 6 a prime?
6 is not a prime number b\c, it is divisible to 2 and 3
Which of these numbers is not a prime number A 21 B 31 C 41?
21 is not a prime.
Is two the first prime number?
yes b/c 1 is neither prime or composite
Which number has exactly 10 factors?
Any number of the form a*b^4 where a and b are different prime numbers, or c^9 where c is a prime, will have exactly 10 factors.
How do you prove by contradiction that square root of 7 is an irrational number?
If √7 is rational, then it can be expressed by some number a/b (in lowest terms). This would mean: (a/b)² = 7. Squaring, a² / b² = 7. Multiplying by b², a² = 7b². If a and b are in lowest terms (as supposed), their squares would each have an even number of prime factors. 7b² has one more prime factor than b², meaning it would have an odd number of prime factors. Every composite has a unique prime factorization and can't have both an even and odd number of prime factors. This contradiction forces the supposition wrong, so √7 cannot be rational. It is therefore irrational.
What is the prime factor of 293?
293 is a prime number, therefore the prime factorization of it would be 293, you cant do 293x1 because 1 is NOT a prime number therefore you cant do it because it wouldn't b a prime factorization that is your answer
What number Prime Minister was Richard B Bennett?
He was the 11th Prime Minister, since 1930 - 1935.
Which is not a prime number 31 47 49 53?
The only number which is not a prime number in that list is 49 - it is equal to seven squared.
Give a proof that the square root of 7 is an irrational number?
Proof by contradiction: suppose that root 7 (I'll write sqrt(7)) is a rational number, then we can write sqrt(7)=a/b where a and b are integers in their lowest form (ie they are fully cancelled). Then square both sides, you get 7=(a^2)/(b^2) rearranging gives (a^2)=7(b^2). Now consider the prime factors of a and b. Their squares have an even number of prime factors (eg. every prime factor of a is there twice in a squared). So a^2 and b^2 have an even number of prime factors. But 7(b^2) then has an odd number of prime factors. But a^2 can't have an odd and an even number of prime factors by unique factorisation. Contradiction X So root 7 is irrational.
