Question

prove that √3+√5 is irrational. don't use P, q use only and b​

Answer 1

Answer:

we know that, √3 and √5 are irrational numbers ... a rational no. => √ 3-√ 5 = p/q , where p & q are integers , but q not= 0… ... This is only possible when √3-√5 is an irrational number.

Answer 2

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How can you prove that √3-√5 is irrational?

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Let √3-√5 be any rational number x

x=√3-√5

squaring both sides

x²=(√3-√5)²

x²=3+5-2√15

x²=8-2√15

x²+8=2√15

(x²+8)/2=√15

as x is a rational number so x²is also a rational number, 8 and 2 are rational nos. , so √15 must also be a rational number as quotient of two rational numbers is rational

but, √15 is an irrational number

so we arrive at a contradiction t

this shows that our supposition was wrong

so √3-√5 is not a rational number

OR U CAN DO IT LIKE THIS :

we know that, √3 and √5 are irrational numbers

so we know that sum of two irrational numbers is also irrational

√3-√5 is also irrational

hope this helps

How do you prove that √2+√3 is irrational?

How can we prove that √5-√3 is an irrational number?

Is √5 - √3 irrational?

How can I prove that 1+√2 and √3+√2 are irrational numbers?

How can I prove that 2 + √3 is irrational?

Let

a=3–√−5–√ be rational

Then

1a=13–√−5–√ will also be rational.

Now,

13–√−5–√

=13–√−5–√∗3–√+5–√3–√+5–√

=3–√+5–√−2

Sum of two rational is again rational.

a is rational as assumed , 2∗1a is also rational.

So a−2∗1a must be a rational(As sum or difference of two rational is again a rational) .

BUT,

a−2∗1a

=(3–√−5–√)+(3–√+5–√)

=2∗3–√

But we know 3–√ is irrational , hence a contradiction, hence

(3–√−5–√)

Is irrati

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We can prove it by contradictory method…

Let √3 -√5 be a rational no

=> √ 3-√ 5 = p/q , where p & q are integers , but q not= 0…………… (1)

On squaring both sides

3 + 5 - 2 √15 = p^2 / q^2

=> -2√ 15 = (p^2/q^2) - 8

=> -2 √15 = (p^2 -8q^2)/q^2

=> √15 = ( 8q^2 - p^2) / 2q^2

Now, in the above expression LHS is an irrational . We have a proof for √15 is an irrational.

RHS is a rational because numerator (8q^2 - p^2)is an integer (as product of 2 integers also square of an integer is integer) & denominator is also an integer. & denominator not=0

But LHS is irrational

=> LHS not= RHS

Hence equation (1) is incorrec

For a instant let us assume that √3-√5 is a rational

then let r=√3-√5 ( where r is a rational number)

r+√5=√3

squaring

(r+√5)^2=(√3)^2

r^2+5+2√5=3

2√5=-r^2–2

√5=-1/2*r^2–1

We found that in right side -1/2*r^2 is rational

so -1/2*r^2–1 is a rational

This indicates that

(i) left side √5 is a rational

(ii)But we know that √5 is a irrational number

The statements in (i) and (ii) are contradictory

This contradiction appeared because we assumed that √3-√5 is a rational number

This proves that our assumption is not correct and √3-√5 is an irrational number

How can you prove that 3+√2 is an irrational number?

How can you prove that 3-√5 is irrational?

Is 3 +√5 an irrational number?

How can we prove that 2+√3 is an irrational number?

Can you prove that 1/√2 is an irrational number?

It is a must that

The Sum of a rational number and an irrational number is always an irrational number.

The sum of any two rational numbers is always an rational number.

However the sum of any two irrational numbers may or may not be a rational number.

Now let's solve this problem:-

We know that

-√3 is an irrational number and so is √5 .

The sum of -√3 and √5 or -√3+√5 is also an irrational number.

When we add this number to √3-√5 ,we get

sum=0 =0/1 which is a rational number. This is only possible when √3-√5 is an irrational number.

Therefore, √3-√5 is irrational number.

I hope it helps.