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Unfortunately, (n-1)/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.
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. 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).
Let n be 4. Then n-1 is 3. (n-1)/2 is 3/2 which is not a natural number.
There 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.
- can you writing please one prove with induction ?
Normally 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?
Don't overthink this. I said it was trivial.
Exactly. 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?
Exactly. 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?
like n+1=a+(b+1)+1 - ?
Again correct. You can bludgeon this one by examining cases, but it's much more useful and aesthetic as an exercise in induction.
can you help me please how ?
What further help would you like?
one complete math induction prove what will may be acceptebilly
You'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.
For n=3, we see that n=1+1+1 so it works for a=1 and b=1.
Now 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.
Since 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.
- so and thats all ?
I said it was trivial.
You 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 a-1>=1. so N=a+(a-1)+1.
If n is odd, then n-1 is even, so n-1 = a+a, so n=a+a+1 and the two values are a and a.
ok thank you very much - but i will come back again tomorrow
I'll watch for you.