Prove that the square root of 5 is irrational?
just a fun exercise in math
- Anonymous1 decade agoFavorite Answer
If √5 is rational, then it can be expressed by some number a/b (in lowest terms). This would mean:
(a/b)² = 5. Squaring,
a² / b² = 5. Multiplying by b²,
a² = 5b².
If a and b are in lowest terms (as supposed), their squares would each have an even number of prime factors. 5b² 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 √5 cannot be rational. It is, therefore, irrational.
- Christine PLv 51 decade ago
Here's a link to "The square root of any prime number is irrational". Since 5 is a prime number, then its square root is irrational
- Anonymous4 years ago
Suppose sqrt(2)+sqrt(5) is rational, say sqrt(2) + sqrt(5) = q where q is a rational number. Then, sqrt(5) = q - sqrt(2) Now square both sides 5 = q^2 - 2q*sqrt(2) + 2 i.e. 2q*sqrt(2) = q^2 - 3 Now q is clearly positive so we can divide both sides by 2q to get sqrt(2) = (q^2-3)/(2q) But if q is rational then so is (q^2-3)/(2q), so this says that sqrt(2) is rational, contradiction.
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- Anonymous1 decade ago
"The first sixty significant digits of its decimal expansion are:
2.23606 79774 99789 69640 91736 68731 27623 54406 18359 61152 57242 7089... which can be rounded down to 2.236 to within 99.99% accuracy. As of April 1994, its numerical value in decimal had been computed to at least one million digits."
There is no definite way to determine that the square root of 5 is irrational, you can only disprove this fact. Since the number has been computed to 1 million digits, and nobody has been able to prove that it is rational, the square root of 5 is generally accepted as an irrational number.
- ironduke8159Lv 71 decade ago
To prove: The square root of 5 is irrational. In other words, there is no rational number whose square is 2.
Proof by contradiction: Begin by assuming that the thesis is false; that is, that there does exist a rational number whose square is 2.
By definition of a rational number, that number can be expressed in the form c/d, where c and d are integers, and d is not zero. Moreover, those integers, c and d, have a greatest common divisor, and by dividing each by that GCD, we obtain an equivalent fraction a/b that is in lowest terms: a and b are integers, b is not zero, and a and b are relatively prime (their GCD is 1).
Now we have (a/b)^2 = 5, so a^2 = 5b^2
If b is even then b^2 is even and so 5b^2 is even. If b is odd then b^2 is odd, but 5b^2 is even. So since 5b^2 is even, a^2 is even. Thus a is even.
So we can say a = 2r and thus (2r)^2 = 5b^2
4r^2 = 5b^2
.8r^2 = b^2
Since .8r^2 is even, b^2 must be even, and thus b is even.
But since both a and b are even, a/b is not in its lowest terms which contradicts the assumption that a and b are
relatively prime. Therefore the assumption is incorrect, and there must NOT be a rational number whose square is 5.Source(s): Dr. Math
- 7 years ago
Yeah, whatever Louise said.
- Anonymous6 years ago
"just a fun exercise in math" <=> "it's from my homework assignment"
- 4 years ago
math is just for fun
- 5 years ago
i want a best answer frm u any of u guyzz