prove that 3 - root 5 is irrational
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it's totally same just replace 2√5 with -√5
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proshan2004:
thnx
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Hi Friend !!!
Here is ur answer !!
Let's assume that 3-√5 is rational number a
Every rational number is in the form of p/q where p, q are integers
3-√5 = p/q
√5 = 3-p/q
√5 = 3q-p/q
If p, q are integers then 3q-p/q is a rational number
Then √5 is also an rational number
But it contradicts the fact that √5 is irrational number
So, our assumption is wrong
So, 3-√5 is irrational number
Hope it helps u : )
Here is ur answer !!
Let's assume that 3-√5 is rational number a
Every rational number is in the form of p/q where p, q are integers
3-√5 = p/q
√5 = 3-p/q
√5 = 3q-p/q
If p, q are integers then 3q-p/q is a rational number
Then √5 is also an rational number
But it contradicts the fact that √5 is irrational number
So, our assumption is wrong
So, 3-√5 is irrational number
Hope it helps u : )
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