Math, asked by prakashsingh40p13xry, 8 days ago

# prove that root 5 is an irrational number

2
HEY BUDDY !!! ✍

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Let take √5 as rational number
If a and b are two co prime number and bis not equal to 0.
We can write √5 = a/b
Multiply by b both side we get
b√5 = a
To remove root, Squaring on both sides, we get
5b2 = a2 …  (i)

Therefore, 5 divides a2 and according to theorem of rational number, for any prime number p which is divides a2 then it will divide a also.
That means 5 will divide a. So we can write
a = 5c
Putting value of a in equation (i) we get
5b2 = (5c)2
5b2 = 25c2
Divide by 25 we get

b2/5 = c2

Similarly, we get that b will divide by 5
and we have already get that a is divide by 5
but a and b are co prime number. so it contradicts.
Hence √5 is not a rational number, it is irrational.
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HOPE IT HELP U !
0

Let us assume that √5 is a rational number.

we know that the rational numbers are in the form of p/q form where p,q are intezers.

so, √5 = p/q

p = √5q

we know that 'p' is a rational number. so √5 q must be rational since it equals to p

but it doesnt occurs with √5 since its not an intezer

therefore, p =/= √5q

this contradicts the fact that √5 is an irrational number

hence our assumption is wrong and √5 is an irrational number.

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