Show that root 2 is irrational
Answers
Answer:
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.
Answer:
Answer:Proof by contradiction
Answer:Proof by contradictionProof by contradiction (also known as reducto ad absurdum or indirect proof) is an indirect type of proof that assumes the proposition (that which is to be proven) is false and shows that this assumption leads to an error, logically or mathematically. Thus, the proposition is true. Famous results which utilized proof by contradiction include the irrationality of √2 and the infinitude of primes. This technique usually works well on problems where not a lot of information is known, and thus we can create some using proof by contradiction.