prove that √2 is an irrational number
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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/b22=4k2/b22*b2=4k2b2=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/bwas 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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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/b22=4k2/b22*b2=4k2b2=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/bwas 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.
hope so it will help u
please mark my answer as brainlist
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Answer :
To prove,
√2 is an irrational no.
Proof :
Let √2 be a rational number in the form of p / q where q is not equal to zero at p and q are co-prime integers.
√2 = p/q
Whole sqauring both sides of this equation :-
2 = p^2/q^2
p^2 = 2q^2 (I)
From (I),
2 divided p^2
So, p divides p. (a)
Now , let p= 2k where k is any integer.
Substituting the values , we get :-
(2k)^2 = 2q^2
4k^2 = 2q^2
q^2 = 2k^2 (ii)
From (ii),
2 divides q^2.
Therefore, 2 divides q also. (b)
From statements (a) and (b) , we can say that :-
p and q have a common factor namely 2.
Hence, our assumption that p and q are co-prime is wrong. Hence , √2 is an irrational no.
Hence proved.
This method is called contradiction method
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