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How do you prove by contradiction that square root of 7 is an irrational number?
Wiki User
∙ 14y ago
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.
Wiki User
∙ 14y ago
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Prove that square root of 2 is an irrational number?
The square root of 2 is 1.141..... is an irrational number
How can you prove that the square root of three is an irrational number?
Because 3 is a prime number and as such its square root is irrational
Prove that square of 2 is irrational?
It is not possible to prove something that is not true. The square of 2 is rational, not irrational.
Why is 37 square root an irrational number?
The same reasoning you may have seen in high school to prove that the square root of 2 is not rational can be applied to the square root of any natural number that is not a perfect square.
How is the sum of a rational and irrational number irrational?
This can easily be proved by contradiction. Without loss of generality, I will take specific numbers as an example. The proof can easily be extended to any rational + irrational number. Assumption: 1 plus the square root of 2 is rational. (It is a well-known fact that the square root of 2 is irrational. No need to prove it here; you can use any other irrational number will do.) This rational sum can be written as p / q, where "p" and "q" are whole numbers (this is basically the definition of a "rational number"). Then, the square root of 2, which is equal to the sum minus 1, is: p / q - 1 = p / q - q / q = (p - q) / q Since the difference of two whole numbers is a whole number, this makes the square root of 2 rational, which doesn't make sense.
Prove that square root of 2 is an irrational number?
The square root of 2 is 1.141..... is an irrational number
How can you prove that the square root of three is an irrational number?
Because 3 is a prime number and as such its square root is irrational
Prove the square root of 2 not irrational?
This is impossible to prove, as the square root of 2 is irrational.
Prove that square of 2 is irrational?
It is not possible to prove something that is not true. The square of 2 is rational, not irrational.
How do you prove that root 2 root 3 is an irrational number?
It is known that the square root of an integer is either an integer or irrational. If we square root2 root3 we get 6. The square root of 6 is irrational. Therefore, root2 root3 is irrational.
Is the square root of 6 rational?
No; you can prove the square root of any positive number that's not a perfect square is irrational, using a similar method to showing the square root of 2 is irrational.
Is the square root of a 143 a irrational number?
Yes. The square root of a positive integer can ONLY be either:* An integer (in this case, it isn't), OR * An irrational number. The proof is basically the same as the proof used in high school algebra, to prove that the square root of 2 is irrational.
Prove that square root of 6 is irrational number?
If a/b=sqrt(6), then a2=6b2 On the other hand, given integers ''a'' and ''b'', because the valuation (i.e., highest power of 2 dividing a number) of 6b2 is odd, while the valuation of a2 is even, they must be distinct integers. Contradiction.
Why is 37 square root an irrational number?
The same reasoning you may have seen in high school to prove that the square root of 2 is not rational can be applied to the square root of any natural number that is not a perfect square.
How is the sum of a rational and irrational number irrational?
This can easily be proved by contradiction. Without loss of generality, I will take specific numbers as an example. The proof can easily be extended to any rational + irrational number. Assumption: 1 plus the square root of 2 is rational. (It is a well-known fact that the square root of 2 is irrational. No need to prove it here; you can use any other irrational number will do.) This rational sum can be written as p / q, where "p" and "q" are whole numbers (this is basically the definition of a "rational number"). Then, the square root of 2, which is equal to the sum minus 1, is: p / q - 1 = p / q - q / q = (p - q) / q Since the difference of two whole numbers is a whole number, this makes the square root of 2 rational, which doesn't make sense.
Is the square root of 5 an irrational or a rational number?
A rational number can be expressed as the ratio of two integers (the number one does not count). An irrational number cannot. The square root of 5 cannot be expressed as the quotient of two integers. This assertion is by experience that the square root of a prime number is irrational. There are several methods to prove the irrationality -- look for them elsewhere. For the sake of curiosity, my calculator gives the following result for it: 2.2360679774997896964091736687312762354406183596115257242708972454105209256378048994144144083787822749695081761507737835042532677244470738635863601215334527088667781731918791658112766453226398565805357613504175337850034233924140644420864325390972525926272288762995174024406816117759089094984923713907297288984820886415426898940991316935770197486788844250897541329561831769214999774248015304341150359576683325124988151781394080005624208552435422355561063063428202340933319829339597463522712013417496142026359047379... You can say that it is irrational by looking at the inexact answer.
Does there exist an irrational number such that its square root is rational?
No, and I can prove it: -- The product of two rational numbers is always a rational number. -- If the two numbers happen to be the same number, then it's the square root of their product. -- Remember ... the product of two rational numbers is always a rational number. -- So the square of a rational number is always a rational number. -- So the square root of an irrational number can't be a rational number (because its square would be rational etc.).
