Question

prove that 5-2√3 is an number​

Answer 1

Answer:

Step-by-step explanation:

let,

5 - 2√3 be a rational number.

.°. 5 - 2√3 = p/q [ where p and q are integer , q ≠ 0 and q and p are co- prime number ]

=> - 2√3 = p/q -5

=>- 2√3 = p - 5q / q

=> √3 = p - 5q / - 2q

we know that p/q is a rational number.

.°. √3 is also a rational number.

This contradicts our assumption

5 - 2√3 is an irrational number

Answer 2

Answer:

Rational numbers are a field and it is close to addition, subtraction, multiplication and division (except division by zero).

So let's suppose that 3+25–√ is rational.

If 3+25–√ is rational, then (3+25–√)−3=25–√ is rational. And if 25–√ is rational, then 25–√÷2=5–√ would be rational.

Now, if 5–√ is rational, then there would exist to integers p,q such as 5–√=pq. If q is negative, then 5–√=−p−q for a positive integer −q. And, among all possible values of pq=5–√ with positive q there is one pair with the less possible positive q.

So let's take this value: pq=5–√ for the less possible positive integer q, which does exist if 5–√ is rational.

Then (pq)2=5, meaning p2q2=5 and therefore p2=5q5.

Right side is divisible by 5. If 5 does not divide p then p2 won't be divisible by 5. So p=5k and p2=25k2.

Now we have 25k2=5q2. We can simplify and we get that 5k2=q2. Left side is divisible by 5 so q must be divisible by 5. Let q=5h. We know that 0<h<q.

Let's replace and we get 5k2=25h2 from which we get that k2h2=5 and kh=5–√.

With h positive and less than q. So there is no minimum positive integer q such as 5–√=pq (with p integer). So 5–√ is not rational. So 25–√ is not rational either. So 3+25–√ is irrational.

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