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Prove that for all positive integers n 10 power n equals 1 in modulo 9?
Wiki User
∙ 14y ago
Mmm, this is a really good quality question! I mean, once you've thought about it for a second, it seems so obvious. 10^3-1 = 999, 10^5-1 = 99999, and in general anything of the form of 10^n - 1 will be divisible by 9 (because the number is made up of nothing but 9's!)
This means that all numbers of the form 10^n (where n is an integer) will be equal to 1 mod 9 (i.e., they will equal 9k + 1, where k is an integer.)
However, such an argument is far from constituting a mathematical proof. So how are we to go about proving this? Well, can we manipulate the equation to show that
10^n - 1 = 9k (k,n are integers) ???????
While this may be possible, such manipulations are not easy. Another option is to try to find a proof by contradiction: lets say that 10^n is not congruent to 1 mod 9 for some value of n. Maybe it is congruent to 2 mod 9. Does this lead to a contradiction? If it does, we still aren't done the proof: we would have to show that 10^n being congruent to 3 mod 9, 4 mod 9, etc. all lead to contradictions. Once again, this is possiblly possible, but good maths is in general about being lazy, so rather than all that work, we'd rather try and find something simpler.
And then we finally run upon the idea of mathematical induction. If you don't know what that is, see here: http://en.wikipedia.org/wiki/Mathematical_induction
Mathematical induction is an exceptionally powerful tool one can use, in particular when trying to find proofs related to the integers. We basically assume that what we are trying to prove is correct, and then see if we can find a proof that this assumption necessarily leads to the statement being correct for the next greatest number. So, we assume a formula works for any integer n, and then try and show it works for n + 1. If we are successful, then all we need to show is that the formula works for n = 1, and we will have proven it for 1,2,3,4,5,6... indeed for every integer.
So can we do that here? Well, lets firstly assume that 10^n = 1 mod 9. This means that 10^n = 9k + 1 (k element of z). Even though we want to prove that, lets just for the time being assume it's true.
Now we need to prove that 10^(n+1) = 9s + 1 (where n and s are integers).
10^(n+1) = 10^n * 10 = 9s + 1
therefore (9s + 1)/10 = 10^n
and we have assumed that 10^n = 9k + 1
therefore (9s + 1)/10 = 9k + 1
9s + 1 = 90k + 10 = 90k + 9 + 1
9s + 1 = 9 (10k + 1) + 1
as 10k + 1 is an integer, s is an integer, so therefore
10^(n+1) = 9s + 1 (where n and s are integers)
so therefore, if 10^n is congruent 1mod9, 10^(n+1) is also congruent 1mod9
All we have left to do is show that the formula works for n = 1
10^1 = 10 = 9*1 + 1. It is obvious that 10 is congruent 1mod9. And that's the proof. End it with a statement like: therefore 10^n is congruent to 1 mod 9 for all n greater than or equal to 1, by mathematical induction
Hope that helps someone out there
Wiki User
∙ 14y ago
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Mathematical induction is just a way of proving a statement to be true for all positive integers: prove the statement to be true about 1; then assume it to be true for a generic integer x, and prove it to be true for x + 1; it therefore must be true for all positive integers.
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By using modulo (also known as clock) arithmetic. In this type of arithmetic, when the modulus is reached, the counting restarts back at zero - it is the same as the remainder when the number is divided by the modulus. In the case of this question, the modulus is 12: (11 + 2) MOD 12 = 13 MOD 12 13 ÷ 12 = 1 r 1 → 13 MOD 12 = 1 → (11 + 2) MOD 12= 1 (9 + 5) MOD 12 = 14 MOD 14 14 ÷ 12 = 1 r 2 → 14 MOD 12 = 2 → (9 + 5) MOD 12 = 2
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9
What is proof by induction?
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The answer depends on which properties are being used to prove which rules.
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Prove that 2 divided by10 equals 2?
Cannot prove that 2 divided by 10 equals 2 because it is not true.
How do you prove that eleven plus two equals one and nine plus five equals two?
By using modulo (also known as clock) arithmetic. In this type of arithmetic, when the modulus is reached, the counting restarts back at zero - it is the same as the remainder when the number is divided by the modulus. In the case of this question, the modulus is 12: (11 + 2) MOD 12 = 13 MOD 12 13 ÷ 12 = 1 r 1 → 13 MOD 12 = 1 → (11 + 2) MOD 12= 1 (9 + 5) MOD 12 = 14 MOD 14 14 ÷ 12 = 1 r 2 → 14 MOD 12 = 2 → (9 + 5) MOD 12 = 2
How do you prove integers aren't closed under subtraction?
Take any two integers, and subtract one from another, you will have another integer. If there was a situation where you could show that this statement is not true, then that would prove your hypothesis, but I cannot think of any.
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9
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