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Proof by Induction or Contradiction By Ayman D’Souza
Very often in Mathematics we find ourselves needing to prove something: a formula, a fact, a theorem. But what do we do when we can no longer use a 3 ଶ as 9x= 3*3*x? Very often, we can use the method of Proof by Induction, or Proof by Contradiction. In this article I will explain both methods by employing the use of examples. Proof by Induction ଵ ଶ n (n +1) If we already know that to find the value of this some we should take the average simple proof, e.g. 9x must be divisible by We are given the arithmetic progression: 1 + 2 + 3 + ..... + n = sense. But how could prove it by induction? Let us assume that the arithmetic progression is true for n = k, rewriting the formula as: ଵ ଶ k (k + 1 ) If it is true for all values, we can fill the formula in with k + 1 as well as k, reading: {Formula A}: 1 + 2 + 3 + ..... + k = ( ଵ ଶ k + 1 ) (k + 1 ) = ଵ ଶ (k + 1 ) (k + 2 ) We can manipulate the right hand side of the formula to resemble the left hand side ଵ ଶ (k + 1 ) not affecting the overall value. Since we have proved that the original statement to prove works for n + 1 (remember n represents k ), it logically works for any integer equal to or larger than one, and so we have proved the formula to work for any value of n . Proof by Contradiction Often, is something is untrue, we can prove this by contradicting it. For example, let is cancelled down as far as possible) which we can contradict to prove the square root of two is an irrational number, similarly to pi. if we halve (k + 2 ), and double us take the example √2 ൌ (assuming that and multiply it by the number of terms, the formula ଵ ଶ n (n +1) makes perfect {Formula B}: 1 + 2 + 3 + ..... + k + (k + 1 ) = ଵ ଶ (k + 1 ) ((k + 1 ) + 1 ) Since we already know that 1 + 2 + 3 + ..... + k = ଵ ଶ k (k + 1 ) {Formula A} , we can rewrite the {Formula B} above as: ଵ ଶ k (k + 1 ) + (k + 1 ) = ଵ ଶ (k + 1 ) (k + 2 )
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