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KonstantinChe [14]
3 years ago
10

Prove that 7 + 3 root 2 is not a rational number​

Mathematics
2 answers:
Tanya [424]3 years ago
5 0

Answer:

Step-by-step explanation:

let 7 + 3√2 be an rational number where

7+3√2 = a/b [ a and b are coprime and b is not equal to zero]

3√2= a/b-7

3√2 =( a-7b) /b

√2 = (a-7b) /3b .....(i)

Now ,from equation (i) ,we get that √2 is rational but we know that √2 is irrational. so actually 7 + 3√2 is irrational not rational. thus our assumption is wrong. The number is irrational.

In-s [12.5K]3 years ago
4 0

Answer:

Prove by contradiction.

See below.

Step-by-step explanation:

Assume it is rational, then:

Let x = 7 + 3√2 = p/q      where p and q are integers and p/q is in simplest form.

x^2 = 49 + 42√2  + 18 = p^2/q^2

67q^2 = p^2  - 42√2q^2

67q^2 - p^2 = -42√2 q^2

67q^2 - p^2 is rational and q^2 is rational so that makes -42√2  rational.

but we know that -42√2  is irrational, so our original assumption is not true.

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