why in induction problems we assume that f(n) is true and then proceed.....this is only the thing to prove
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Shubham Gupta
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3 years, 3 months ago

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@Shubham Gupta
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I would like to explain the whole principle:
First we check for \(f(1)\), if it is true we proceed.Then we assume that it is true for a natural number k.Then we check whether it is true for k+1.If it is also true then our given expression is true for all natural numbers.We assumed that it is true for k then we proceed, but previously we checked that it is true for 1.So in this case k=1.Now we prove that it is true for k+1,so it is true for k=2.So now we proved that it is true for 2.Now ,let k=2 as k=2 satisfies the condition,so we again prove that it is true for k+1 i.e 3.Again we let k=3 and prove for k=4.The cycle goes on and on which shows that expression is true for all natural numbers.Let me know where I am wrong.Yes I accept my weak english.Sorry for it.
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Kishan K
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3 years, 3 months ago

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@Kishan K
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thnks kevin ,and about your english ,its absolutely flawless
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Shubham Gupta
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3 years, 3 months ago

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TopNewestYou are referring to the First Principle of Finite Induction, which can be proved by contradiction. – Calvin Lin Staff · 3 years, 3 months ago

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why in induction problems we assume that f(n) is true and then proceed.....this is only the thing to prove – Shubham Gupta · 3 years, 3 months ago

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– Kishan K · 3 years, 3 months ago

I would like to explain the whole principle: First we check for \(f(1)\), if it is true we proceed.Then we assume that it is true for a natural number k.Then we check whether it is true for k+1.If it is also true then our given expression is true for all natural numbers.We assumed that it is true for k then we proceed, but previously we checked that it is true for 1.So in this case k=1.Now we prove that it is true for k+1,so it is true for k=2.So now we proved that it is true for 2.Now ,let k=2 as k=2 satisfies the condition,so we again prove that it is true for k+1 i.e 3.Again we let k=3 and prove for k=4.The cycle goes on and on which shows that expression is true for all natural numbers.Let me know where I am wrong.Yes I accept my weak english.Sorry for it.Log in to reply

– Shubham Gupta · 3 years, 3 months ago

thnks kevin ,and about your english ,its absolutely flawlessLog in to reply