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1. all prime numbers are odd

2. the sum off two numbers is always greater than the larger number

3. if the product of two numbers is even, then the two numbers must be even

4. if the product of two numbers is positive, then the two numbers must both be positive

5. the square root of a number "x" is always less then "x"

6. if "m" is a nonzero integer, then m+1 over m is always greater than 1

2007-08-25 16:19:17 · 6 answers · asked by shauna0126 1 in Science & Mathematics Mathematics

6 answers

For #2, use the numbers 1 and -1. The sum is zero, which is less than the larger number.

For #5, use 1/4. The square root of 1/4 is 1/2, which is larger than 1/4.

For #6, use m = -2. Then (-2 + 1)/-2 = 1/2 < 1.

2007-08-25 16:37:04 · answer #1 · answered by Anonymous · 0 0

===== Counterexample to Statement 1 =====

2 is even as it is divisible by 2, so it is not odd.

2 is prime since its only divisors are 2 (itself) and 1.

===== Counterexample to Statement 2 =====

(3) + (-3) = 0

0 < 3 (which means that 0 > 3 is not true)

The sum of the two numbers, 3 and -3 is not greater than the larger number in the sum, which is 3.

===== Counterexample to Statement 3 =====

3 * 2 = 6

3 ÷ 2 = 3/2 or 1 remainder 1 (meaning that 3 is not divisible by 2, so 3 is odd, and not even)

6 is the product of two numbers and is even, but 3 (one of the two numbers in the product is not even) as 3/2 is not an integer.

===== Counterexample to Statement 4 =====

(-1) * (-1) = 1

-1 < 0 (so -1 is negative, and therefore, not positive)

1 > 0 (so 1 is positive)

The product of two numbers is positive, but neither of those two numbers themselves are positive.

===== Counterexample to Statement 5 =====

The square root of 1/4 = 1/2.

1/4 > 1/2 (so 1/2 is not less than 1/4)

So let x = 1/4, then the square root of "x" is not less than "x"

===== Counterexample to Statement 6 =====

(-1) = 1 / (-1) = 0 / (-1) = 0

0 < 1 (so 0 is not greater than 1)

Let m = -1, which is a nonzero integer.

m+1 over m = 0 which is not greater than 1.

2007-08-25 16:38:20 · answer #2 · answered by darthsherwin 3 · 0 0

1. 2 is the one and only example of an even prime number. Therefore, the conjecture is proved false.

2. Let -2 and 2 be the two numbers.
It can be proved that -2 + 2 < 2. Therefore, the conjecture is false.

3. Let 3 and 6 be the two numbers.
The product of 3 and 6 = 18 is even, but 3 is not an even number. Therefore, the conjecture is false.

4. Let -4 and -5 be the two numbers.
The product of -4 and -5 = 20 is positive, but -4 and -5 are both not positive numbers. Therefore, the conjecture is false.

5. Let the number x be 1.
The square root of 1 which is 1 is not less than 1 itself. Therefore, the conjecture is false.

6. Let the non-zero integer m be -1.
The value of (-1 + 1)/-1 = 0 is not greater than 1. Therefore, the conjecture is false.

2007-08-25 16:23:14 · answer #3 · answered by Anonymous · 3 1

1. 2 is prime and even.
2. If both numbers are negative, this is false.
For example -1 + -2 = -3 and -3 < -1.
3. Only one need be even. For example 1*2=2.
4. The numbers could be negative. For example
-1 * -2 = 2.
5. This is false if 0 and .8 > .64
6. False. Let m = -1. Then (m+1)/m = 0.

2007-08-26 03:04:53 · answer #4 · answered by steiner1745 7 · 0 0

1. 9
2. 3+(-2)=1
3. 6 times 1/2
4. -2+(-2)
5. square root of 1/2 is less then 1/2
6. 1/2 is always greater then 1

2007-08-29 11:25:07 · answer #5 · answered by Anonymous · 0 0

1) . . . false . . . . 2 is also prime
2) . . . true
3) . . . false .. . . at least one is even
4) . . .false . . . . 2 negatives can be positive
5) . . true
6) . . . true

2007-08-25 16:29:50 · answer #6 · answered by CPUcate 6 · 0 2

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