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Prove that if x and y are positive integers then x square plus y square is even but not divisible by 4?
Wiki User
∙ 13y ago
The statement cannot be proven because it is FALSE.
If one of x and y is odd and the other is even then x2 + y2 MUST be odd.
Also if x and y are even then x2 + y2 MUST be divisible by 4.
The statement is only true if x and y are odd integers.
Whether or not they are positive makes little difference.
Wiki User
∙ 13y ago
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Is the square root of 6 rational?
No; you can prove the square root of any positive number that's not a perfect square is irrational, using a similar method to showing the square root of 2 is irrational.
Prove the square root of 2 not irrational?
This is impossible to prove, as the square root of 2 is irrational.
What is the prime factorization of the number 223?
223 is a prime number. You can prove it by showing it is not divisible by 2, 3, 5, 7, 9, 11, 13, and 15 (2 and all the odd numbers up to the square root of 223)
What is the smallest set of numbers closed under subtraction?
Any one of the sets of the form: {kz : where k is any fixed integer and z belongs to the set of all integers} Thus, k = 1 gives the set of all integers, k = 2 is the set of all even integers, k = 3 is the set of all multiples of 3, and so on. You might think that as k gets larger the sets become smaller because the gaps between numbers in the set increases. However, it is easy to prove that the cardinality of each of these infinite sets is the same.
Prove that root of 5 is irrational?
Suppose we start with the opposite view. That is, the square root of 5 is rational. This means that it can be expressed as a ratio of two integers. Let the square root of 5 in its simplest terms be a/b so that a and b are integers with no common factors and b > 0.sqrt(5) = a/b so 5 = a2/b2that is, 5b2= a25 is a factor of the left hand side (LHS) so 5 must be a factor of the right hand side (RHS).Since 5 is a prime and a is an integer, 5 must be a factor of a. That is, a = 5c for some integer c and since a and b are coprime b and c must also be coprime.But now,5 = (5c)2/b2so that 5 = 25c2/b2and so b2= 5c2. As before, this implies that 5 must be a factor of b. But that contradicts the supposition that a and b are coprime.The contradiction implies that the original assumption was incorrect. That is, sqrt(5) cannot be rational.
How do you prove that the product of two consecutive even integers is divisible by 8?
Because 6*8 = 48 and 48/8 = 6
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.
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.
Is it true that the square root of 2 cannot be expressed by a ratio of two integers?
yes. But I'm too lazy to show you how to prove it.
Prove that one of every three consecutive positive integers is divisible by 3?
No
Is the square root of 6 rational?
No; you can prove the square root of any positive number that's not a perfect square is irrational, using a similar method to showing the square root of 2 is irrational.
Prove that the two consecutive positive integer is divisible by 2?
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.
How do you prove a number is divisible by 40 and 4?
4
How can properties be used to prove rules for multiplying integers?
The answer depends on which properties are being used to prove which rules.
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 -0.15 a rational number?
Yes, because you can express it as a ratio of two integers, for example, -15 / 100. This can be simplified, but that's not necessary to prove it is rational.Yes, because you can express it as a ratio of two integers, for example, -15 / 100. This can be simplified, but that's not necessary to prove it is rational.Yes, because you can express it as a ratio of two integers, for example, -15 / 100. This can be simplified, but that's not necessary to prove it is rational.Yes, because you can express it as a ratio of two integers, for example, -15 / 100. This can be simplified, but that's not necessary to prove it is rational.
How can you Prove that if v is multiple of 3 then z is also a multiple of three?
With the information you gave in the question you cannot prove it unequivocally.If Z/V has no remainder then z is also divisible by whatever v is divisible by.
