सिद्ध कीजिए कि
√2
एक अपरिमेय संख्या है।
Answers
Answer:
उत्तर- यदि सम्भव हो, तो माना √2 एक परिमेय संख्या है।
तब मान √2 = m / n, H.C.F. (m, n) = 1, n≠ 0
⇒ m = √2n
⇒ m2 = 2n2 ….(1)
⇒ 2n2 एक समपूर्णाक है।
⇒ m2 एक समपूर्णांक है।
⇒ m एक समपूर्णांक है। ....(A)
= m = 2q, q∈ z ....(2)
(1) व (2) से
4q2 = 2n2
⇒ n2 = 2q2
⇒ n2 एक समपूर्णांक है।
⇒ n एक समपूर्णांक है। ....(B)
(A) तथा (B) ⇒ m तथा n दोनों ही समपूर्णांक है।
⇒ H.C.F. (m, n) # 1
अतः जो कि विरोधाभास है परिमेय होने का अतः √2 एक अपरिमेय संख्या है।
Answer:
Step-by-step explanation:
A proof that the square root of 2 is irrational
Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero.
We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.
From the equality √2 = a/b it follows that 2 = a2/b2, or a2 = 2 · b2. So the square of a is an even number since it is two times something.
From this we know that itself is also an even number. Why? Because it can't be odd; if itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don't believe me!
Okay, if itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.
If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:
2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2
This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!!
WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.