Question

PROVE 3 Is An IRRATIONAL NUMBER,
$proof.if root 3 is a rational number. let root 3 is equal to m/n take it has 1 at equation. here m and n are positive integers with no common factor greater than 1. root 3 is equal to m/n. squaring on both the sides. root 3square is equal to bracket of m /n square. 3 equal to m square /n square. 3 n square equal to m square. m square eqaul to 3n square take it has equation 2 m square is divisible by 3 and hence m is divisible by 3. let m equal to 3k. substitute m equal to ek in equation 2 m square is equal to 3n square 3k square is equal to 3n square. 9k square is equal to 3n square. n sqaure is equal to 3 k square c n square is divisible by 3 and hence n is divisible by 3 we arrived at a contradiction because m and n are divisible by 3 and have common factor 3 hence root 3 ia an irrational number.2$
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Answer 1

Answer:

Let 8 be a rational number

Then 8=nm where m, n are integers and m, n are expenses and n=0

⇒m=8n

Squaring both sides we get

m2=8n2

⇒8m2=n2 ……………(iv)

⇒8 divides m2 i.e., 8 divides m

Then m can be written as

m=8k for some integer k.

Substituting value of m in (iv) we get

⇒3(8k)2=n2

⇒8k2=n2

⇒k2=8n2

⇒8 divides n2 i.e., 8 divides n

Thus we get that 8 is a common factor of m and n but m and n are co-primes which is a contradiction to our assumption.

Hence 8 is an irrational number.

Now consider 3+8 to be an rational number

Then 3+8=ba where a, b are integers, co-primes and b