Question

prove that (7-5√3) is irrational.​

Answer 1

Answer:

In x = 7 - 3√5, both sides are rational as assumed. ⇒ Here, as both sides of x = 7 - 3√5 are rational, then so should be √5 = (7 - x)/3. But this contradicts our earlier assumption that 7 - 3√5 is rational, because the RHS of √5 = (7 - x)/3 is rational while the LHS is irrational. ⇒ Hence proved!!!

Answer 2

Answer:

Let us assume 7 - 5√3 is rational

Then we can find such numbers a and b where a and b are coprimes and is not equal to 0.

7-5√3 = a/b

7 - a/b = 5√3

7b - a/b = 5√3

7b - a/5b = √3

Now, a and b are integers

Therefore 7b - a/5b is rational

and therefore √3 is rational

But this contradicts the fact that √3 is irrational

This contradiction has arose because of our wrong assumption that 7- 5√3 is rational

Therefore 7-5√3 is irrational

Hence Proved

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