prove thatroot 2 is irrational
Answers
Answer:
Step-by-step explanation:
Let √2 be a rational number
Therefore, √2= p/q [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0
On squaring both sides, we get
p²= 2q² ...(1)
Clearly, 2 is a factor of 2q²
⇒ 2 is a factor of p² [since, 2q²=p²]
⇒ 2 is a factor of p
Let p =2 m for all m ( where m is a positive integer)
Squaring both sides, we get
p²= 4 m² ...(2)
From (1) and (2), we get
2q² = 4m² ⇒ q²= 2m²
Clearly, 2 is a factor of 2m²
⇒ 2 is a factor of q² [since, q² = 2m²]
⇒ 2 is a factor of q
Thus, we see that both p and q have common factor 2 which is a contradiction that H.C.F. of (p,q)= 1
Therefore, Our supposition is wrong
Hence √2 is not a rational number i.e., irrational number.
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 a itself is also an even number. Why? Because it can't be odd; if a 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 a 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.
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