Question

prove that 4+3√ 5 is an irrational number​

Answer 1

Step-by-step explanation:

Let us assume that 4+3√5 is a rational number.

4+3√5=a/b [ Here a and b are co prime]

3√5=a/b-4

√5=a-4b/3b

Since, a,b,4b,3b are integers.

a-4b/3b is a rational number and √5 is an irrational number.

So, 4+3√5 is an irrational number....

Proved!!!!

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Answer 2

CONCEPT:

rational and irrational numbers: rational numbers can be written in the form of fraction(a/b) and irrational numbers cannot be written in the form of fractions.

Here to prove 4+3√5 is an irrational number we use the method of contradiction.

Contradiction method is a method in which we assume the opposite to the statement and proves that the opposite of statement is wrong thereby proving that the original statement is right

GIVEN:

a statement is given

4+3√5 is an irrational number

FIND:

we want to prove that 4+3√5 is an irrational number.

SOLUTION:

let the statement be P:4+3√5 is an irrational number.

let us assume a CONTRADICTION to the given statement and check it's validity

CONTRADICTION IS 4+3√5 is rational.

let's check it's validity

if 4+3√5 is rational we can write it in fractional form

so 4+3√5 =a/b

3√5=a/b-4

3√5=(a-4b)/b

√5=(a-4b)/3b

in the RHS a,b,4,3 are integers.

so (a-4b)/3b is rational

in LHS it's √5.

we know that  SQUARE ROOT OF ANY INTEGERS IS IRRATIONAL.

so here it comes the situation that irrational =rational

so the contradiction 4+3√5 is rational is wrong

so the number 4+3√5 is irrational and hence proved

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