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Asked by Laxmi Last Modified
Answer To prove that x2−xx2−x is divisible by 2 for all positive integer xx, we can use mathematical induction.
Base Case: Let's start by checking the base case. When x=1x=1, x2−x=12−1=0x2−x=12−1=0. Since 0 is divisible by 2, the base case holds.
Inductive Hypothesis: Assume that x2−xx2−x is divisible by 2 for some positive integer kk, i.e., k2−kk2−k is divisible by 2.
Inductive Step: We need to show that if the statement holds for kk, then it also holds for k+1k+1.
(k+1)2−(k+1)=k2+2k+1−k−1=k2+k=(k2−k)+k(k+1)2−(k+1)=k2+2k+1−k−1=k2+k=(k2−k)+k
By the inductive hypothesis, we know that k2−kk2−k is divisible by 2. And we know that kk is a positive integer, so kk is also divisible by 2 or it is an odd number.
If kk is divisible by 2, then k2−kk2−k is divisible by 2, and kk is divisible by 2, so their sum (k2−k)+k(k2−k)+k is also divisible by 2.
If kk is an odd number, then k2−kk2−k is still divisible by 2, and when you add kk to it, you still get an even number, which is divisible by 2.
In both cases, (k+1)2−(k+1)(k+1)2−(k+1) is divisible by 2.
Therefore, by mathematical induction, we conclude that x2−xx2−x is divisible by 2 for all positive integers xx.
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