Prove that √2 not a rational number
Assume that √2 is a rational number, that is, it can be expressed as a ratio of two integers a and b in the lowest terms:
√2 = a/b
Squaring both sides, we get:
2 = a^2 / b^2
Multiplying both sides by b^2, we get:
2b^2 = a^2
This means that a^2 is even. Therefore, a must be even, because the square of an odd number is always odd. Let a = 2c, where c is an integer:
2b^2 = (2c)^2
Simplifying, we get:
2b^2 = 4c^2
Dividing both sides by 2, we get:
b^2 = 2c^2
This means that b^2 is even, and therefore b must also be even. However, this contradicts our initial assumption that a and b are in the lowest terms, because they both have a common factor of 2. Therefore, our assumption that √2 is a rational number must be false.
Hence, we can conclude that √2 is an irrational number.
Property of real numbers
There are many properties of real numbers, but some of the most basic ones are:
1. Commutative property of addition: a + b = b + a
This means that we can add two real numbers in any order and the result will be the same.
2. Associative property of addition: (a + b) + c = a + (b + c)
This means that we can group real numbers in any way we want when adding them and the result will be the same.
3. Identity property of addition: a + 0 = a
This means that if we add zero to any real number, the result is the same real number.
4. Inverse property of addition: a + (-a) = 0
This means that the sum of any real number and its opposite is zero.
5. Commutative property of multiplication: ab = ba
This means that we can multiply two real numbers in any order and the result will be the same.
6. Associative property of multiplication: (ab)c = a(bc)
This means that we can group real numbers in any way we want when multiplying them and the result will be the same.
7. Identity property of multiplication: a x 1 = a
This means that if we multiply any real number by one, the result is the same real number.
8. Inverse property of multiplication: a x 1/a = 1
This means that the product of any real number and its reciprocal is one, as long as the real number is not zero.
9. Distributive property: a(b + c) = ab + ac
This means that we can distribute the multiplication of a real number to the sum of two or more real numbers and the result will be the same.
Prove that |a+b|≤|a|+|b| where a,b are real numbers
To prove that √2 is not a rational number, we will use a proof by contradiction.
Assume that √2 is a rational number. This means that it can be expressed as a fraction a/b, where a and b are positive integers that have no common factors other than 1, and b is not equal to zero.
So, we have √2= a/b.
Squaring both sides of the equation, we get 2= (a^2)/ b^2.
Rearranging the equation, we have 2b^2 = a^2.
This implies that a^2 is even, since 2 is a factor of 2b^2.
By the fundamental property of integers, if a^2 is even, then a must also be even.
So, we can write a = 2k, where k is an integer.
Substituting back into the equation, we get 2b^2 = (2k)^2 = 4k^2.
Dividing both sides by 2, we have b^2 = 2k^2.
This implies that b^2 is even, which means that b is also even.
However, if both a and b are even, they have a common factor of 2, which contradicts our initial assumption that a and b have no common factors other than 1.
Thus, our assumption that √2 is rational is false.
Therefore, √2 is not a rational number.
To prove that √2 is not a rational number, we need to use proof by contradiction.
Assume that √2 is a rational number, which means it can be expressed as a fraction a/b, where a and b are positive integers with no common factors (other than 1) and b is not equal to zero.
If √2 = a/b, then we can square both sides of the equation to obtain:
2 = (a^2) / (b^2)
Cross-multiplying gives us:
2 * (b^2) = (a^2)
This implies that a^2 is an even number (since it is divisible by 2). Therefore, a must also be even, because the square of an odd number is odd.
Let's express a as 2k, where k is an integer. Substituting it back into the equation, we get:
2 * (b^2) = (2k)^2
2 * (b^2) = 4k^2
Dividing both sides by 2 gives us:
b^2 = 2k^2
This means that b^2 is also divisible by 2, and therefore, b must also be even.
However, we initially assumed that a and b have no common factors (other than 1). But now, we find that both a and b are divisible by 2, contradicting our assumption.
Hence, our assumption that √2 is a rational number is false. Therefore, √2 is irrational.