We will discuss about the different properties of surds.
If a and b are both rationals and √x and √y are both surds and a + √x = b + √y then a = b and x = y
If a not equal to b, let us assume, b = a + m, where m (m ≠ 0) is a rational.
Now, by question, a + √x = b + √y
⇒ a + √x = a + m + √y
⇒ √x = m + √y, which is impossible (since a simple quadratic surd cannot be equal to the sum of a rational quantity and a simple quadratic surd).
Therefore, we must have, a = b.
When a = b then a + √x = b + √y ⇒ √x = √y ⇒ x = y.
Notes:
1. If a - √x = b - √y where a, b are both rationals and √x, √y are both surds, then proceeding as above we can show a = b and x = y.
2. If √x and √y are actually rationals (in the form of surds), then the relation a + √x = b + √y does not imply a = b and x = y.
For example, we have,
10 = 6 + 4 = 6 + √16 and 10 = 4 + 6 = 4 + √36
⇒ 6 + √16 = 4 + √36
Evidently we cannot have, 6 = 4 or 16 = 36.
This is due to the fact that √16 and √36 are not surds, they represent rational numbers.
3. If a + √x = b + √y where a, b are both rationals and √x, √y are both surds then, a = b i.e. rational parts of two sides are equal and x = y i.e., irrational parts of two sides are equal.
4. If a - √x = b - √y where a, b are both rationals and √x, √y are both surds then, a = b i.e. rational parts of two sides are equal and x = y i.e., irrational parts of two sides are equal.
5. If a + √x = 0, then a = 0 and x = 0.
6. If a - √x = 0, then a = 0 and x = 0.
7. If a + √x = b + √y then, a - √x = b - √y
8. If √(a + √x) = √b + √y then √(a - √x) = √b - √y
9. Identically, if √(a - √x) = √b - √y then √(a - √x) = √b - √y.
11 and 12 Grade Math
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