Math, asked by kannananand293, 1 year ago

prove that 2√3-7 is an irrational number

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Answered by CamilaaCabello
45
Hey.!



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Let us assume in contradiction that 2√3-7 is rational number.

Let 2√3-7 =a/b where a and b are co-primes.

2√3-7=a/b


2√3 =a/b+7

√3 =a/b +7/2

√3 =2a+7b /2b

√3 = (integer) + (integer) /2(integer)

√3 = p/q

But , we know that √3 is irrational

So , our assumption is wrong.

2√3-7 is irrational number.


# HOPE IT HELPS #

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Answered by anitakumar15021976
4

Answer:

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23–7

Now, let us assume 23–7 be a rational number, ‘r’

So,23–7=r

23=r+7

3=(r+7)/2

We know that ‘r’ is rational, ‘(r+7)/2’ is rational, so ‘3’ is also rational.

This contradicts the statement that 3 is irrational.

So, 23–7 is an irrational number.

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