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jhonyy9
 5 years ago
 let a,b >=1 and n>=3 numbers from the set of natural numbers N ,
 how can prove it that always for any value of n exist numbers a and b such that this equation
a+b+1=n is true ?
 is this prove with reductio ad absurdum correct ?
in a reducto ad absurdum argument, we assume the opposite is true. in this case, that would be:
for some natural number n >=3 there are no natural numbers a >=1, b >=1 with a+b+1 = n.
we don't know what n may be (except that it is greater or equal 3).
since n >=3, n  1 >=2.
since n  1 >=2, (n  1)/2 >=1.
(n  1)/2 + (n  1)/2 + 1 = (n  1) + 1 = n.
jhonyy9
 5 years ago
 let a,b >=1 and n>=3 numbers from the set of natural numbers N ,  how can prove it that always for any value of n exist numbers a and b such that this equation a+b+1=n is true ?  is this prove with reductio ad absurdum correct ? in a reducto ad absurdum argument, we assume the opposite is true. in this case, that would be: for some natural number n >=3 there are no natural numbers a >=1, b >=1 with a+b+1 = n. we don't know what n may be (except that it is greater or equal 3). since n >=3, n  1 >=2. since n  1 >=2, (n  1)/2 >=1. (n  1)/2 + (n  1)/2 + 1 = (n  1) + 1 = n.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Unfortunately, (n1)/2 is not a natural number for every n. If it were, you would have demonstrated that every number greater than or equal to three is odd. You'll do much better using induction rather than reduction.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in a reducto ad absurdum argument, we assume the opposite is true. in this case, that would be: for some natural number n >=3 there are no natural numbers a >=1, b >=1 with a+b+1 = n. we don't know what n may be (except that it is greater or equal 3). since n >=3, n  1 >= 2. since n  1 >=2, (n  1)/2 >=1. (n  1)/2 + (n  1)/2 + 1 = (n  1) + 1 = n. but if n is a natural number, so is n  1, and so is (n  1)/2. therefore we have proved that for this n >=3, the natural numbers (n  1)/2 don't exist. this is absurd. (the logical conclusion therefore being that no such natural number >=3 actually exists, so our assumption was wrong, so EVERY natural number >=3 has such an a and b >=1 with a+b+1 = n).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let n be 4. Then n1 is 3. (n1)/2 is 3/2 which is not a natural number.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There are two reasons to want to do this by induction. The first is that it's trivially easy. The second is that induction is how most theorems about natural numbers are proved.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0 can you writing please one prove with induction ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Normally induction would start with n=1 but the theorem concerns n>=3 so that's our first point. Is the theorem true for n=3? Can you write a+b+1 =3 for some values of a and b?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Don't overthink this. I said it was trivial.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exactly. Now we assume that the theorem is true for every value up to n, and we test whether it's true for n+1. So assuming we can write n = a+b+1, can we write n+1 as the sum of two numbers plus 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exactly. And the right hand side can be grouped so that it is two numbers plus 1 rather than three plus one as you've written it. Right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like n+1=a+(b+1)+1  ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Again correct. You can bludgeon this one by examining cases, but it's much more useful and aesthetic as an exercise in induction.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you help me please how ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What further help would you like?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one complete math induction prove what will may be acceptebilly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You've done it, but let me repeat. To prove that the theorem is true by induction, we prove that it's true for the lowest value (3) and then prove that if it's true for n, it is also true for n+1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For n=3, we see that n=1+1+1 so it works for a=1 and b=1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now assume that it's true for all values up to. This means that n=a+b+1 for some numbers a and b. Then n+1 = a+b+1+1 = a+(b+1)+1 and we have our two numbers, a and b+1 satisfying the equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since the set of natural numbers is generated by starting with 1 and adding 1 repeatedly, this process covers every possible natural number. QED by induction.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I said it was trivial.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can also do it by the two possible cases. Any number n greater or equal to three is either even or odd. If even, there is some number a, such that n=a+a. Since n>3, a>1 so a1>=1. so N=a+(a1)+1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If n is odd, then n1 is even, so n1 = a+a, so n=a+a+1 and the two values are a and a.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thank you very much  but i will come back again tomorrow
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