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There is something missing from your question: what about the two consecutive integers? Is the proof required that one of them is divisible by 2? Or that their product is (which amounts to the same thing)?
Showing that one of two consecutive number is divisible by 2:
Suppose your two numbers are n and (n+1).
If n is divisible by 2, ie n = 2k, the result is shown.
Otherwise assume n is not divisible by 2. In this case n = 2m+1. Then:
(n+1) = ((2m+1)+1)
= 2m + 2
= 2(m+1)
which is a multiple of 2 and so divisible by 2. QED.
Showing that exactly one of two consecutive integers is divisible by two is shown above with the addition to the first part:
"as (n+1) = 2k+1 is not divisible by two and so only n is divisible by 2."
To show the product is divisible by 2, show either n is divisible by 2 or (n+1) is as above, then the result follows as one of n and (n+1) is divisible by 2 and so their product is.
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How do you prove that the product of two consecutive even integers is divisible by 8?
Because 6*8 = 48 and 48/8 = 6
Prove that one of every 3 consecutive positive integers is divisible by 3?
Let three consecutive integers be n, n+1 and n+2. If n is divisible by 3 then n+1 and n+2 cannot be divisible by 3 as these numbers will respectively leave remainders of 1 and 2. If n is not divisible by 3 then it will leave a remainder of 1 or 2. If n leaves a remainder of 1, then n+1 leaves a remainder of 2 and n+2 is therefore divisible by 3. If n leaves a remainder of 2, then n+1 is divisible by 3 and n+2 is not divisible by 3 as it leaves a remainder of 1.
The sum of 25 consecutive integers is divisible by?
The sum of n consecutive integers is divisible by n when n is odd. It is not divisible by n when n is even. So in this case the answer is it is divisible by 25! Proof: Case I - n is odd: We can substitute 2m+1 (where m is an integer) for n. This lets us produce absolutely any odd integer. Let's look at the sum of any 2m+1 consecutive integers. a + a+d + a+2d + a+3d + ... + a+(n-1)d = n(first+last)/2 (In our problem, the common difference is 1 and this is an arithmetic series.) a + (a+1) + (a+2) + ... + (a+2m) = (2m+1)(2a+2m)/2 = (2m+1)(a+m) It is obvious that this is divisible by 2m+1, our original odd number. That proves case I when n is odd, not for case when it is even. Case II - n is even: We can substitute 2m for n. We have another arithmetic series: a + (a+1) + (a+2) + ... + (a+2m-1) = (2m)(2a+2m-1)/2 = m(2a+2m-1) It is not too hard to prove that this is not divisible by 2m... try it!
How do you prove a number is divisible by 24 if it is divisible by 3 and 8?
Becuz 8 mutiplied by 3 is 24
Is the square root of a 143 a irrational number?
Yes. The square root of a positive integer can ONLY be either:* An integer (in this case, it isn't), OR * An irrational number. The proof is basically the same as the proof used in high school algebra, to prove that the square root of 2 is irrational.
When you double the first and quadruple the second integer their sum is 30 find the consecutive integers?
There are no such consecutive integer as is so simple to prove! Suppose the first integer is x. Then the next (consecutive) integer is x+1. Then 2*x + 4*(x+1) = 30 So that 2x + 4x + 4 = 30 6x + 4 = 30 6x = 30 - 4 = 26 x = 26/6 which is NOT an integer.
How do you prove that the product of two consecutive even integers is divisible by 8?
Because 6*8 = 48 and 48/8 = 6
Prove that one of every 3 consecutive positive integers is divisible by 3?
Let three consecutive integers be n, n+1 and n+2. If n is divisible by 3 then n+1 and n+2 cannot be divisible by 3 as these numbers will respectively leave remainders of 1 and 2. If n is not divisible by 3 then it will leave a remainder of 1 or 2. If n leaves a remainder of 1, then n+1 leaves a remainder of 2 and n+2 is therefore divisible by 3. If n leaves a remainder of 2, then n+1 is divisible by 3 and n+2 is not divisible by 3 as it leaves a remainder of 1.
How do you prove that an integer is an odd integer?
It leaves a remainder of 1 when divided by 2. Or, if the number is n, then n-1 or n+1 is even (divisible by 2).
Prove that one of every three consecutive positive integers is divisible by 3?
No
The sum of 25 consecutive integers is divisible by?
The sum of n consecutive integers is divisible by n when n is odd. It is not divisible by n when n is even. So in this case the answer is it is divisible by 25! Proof: Case I - n is odd: We can substitute 2m+1 (where m is an integer) for n. This lets us produce absolutely any odd integer. Let's look at the sum of any 2m+1 consecutive integers. a + a+d + a+2d + a+3d + ... + a+(n-1)d = n(first+last)/2 (In our problem, the common difference is 1 and this is an arithmetic series.) a + (a+1) + (a+2) + ... + (a+2m) = (2m+1)(2a+2m)/2 = (2m+1)(a+m) It is obvious that this is divisible by 2m+1, our original odd number. That proves case I when n is odd, not for case when it is even. Case II - n is even: We can substitute 2m for n. We have another arithmetic series: a + (a+1) + (a+2) + ... + (a+2m-1) = (2m)(2a+2m-1)/2 = m(2a+2m-1) It is not too hard to prove that this is not divisible by 2m... try it!
What is proof by induction?
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.
Do Prove that the product of an integer and arbitrary integer is even?
The statement is not true. Disprove by counter-example: 3 is an integer and 5 is an integer, their product is 15 which is odd.
How do you prove a number is divisible by 40 and 4?
4
Can you prove If m n and p are three consectuive integers then mnp is even?
If m, n, and p are three consecutive integers, then one of them must be even. Let's say the even number is m. Since m is even, it is divisible by two, and so can be written as 2*k, where k is some integer. This means that m*n*p = 2*k*n*p. Since we are multiplying the quantity k*n*p by 2, it must be divisible by two, and therefore must be even.
Why is the sum of consecutive odd integers always a multiple of 4?
It is quite easy to prove this using algebra. Suppose x is the smaller of the two odd integer. The fact that x is odd means that it is of the form 2m + 1 where m is an integer. [m integer => 2m is an even integer => 2m + 1 is odd] The next odd integer will be x + 2, which is (2m + 1) + 2 = 2m + 3 The sum of these two consecutive odd integers is, therefore, 2m + 1 + 2m + 3 = 4m + 4 = 4(m + 1) Since m is an integer, m+1 is an integer and so 4(m + 1) represents a factorisation of the answer which implies that 4 is a factor of the sum. In other words, the sum is a multiple of 4.
How do you prove a number is divisible by 24 if it is divisible by 3 and 8?
Becuz 8 mutiplied by 3 is 24
