Which statement is NOT TRUE?
1) for all x E Z: if x^(squared) - 5x + 6 = 0, then x > 1 and x < 4.
2) there exists an integer x such that x < 0 and x^(squared) + 4 = 0.
3) there exists an integer x E {10, 100, 1000} such that x^(squared) E {10, 100, 1000}.
4) for all integers x E {10, 100, 1000}, x^(squared) is an even number.
My answer is 3 because x squared is not 10, 100 and 1000. Am I right???
Jump to PostNo, 10 squared is 100. And using E instead of in symbol is little confusing, 'in' whould be clearer. Check again.
Jump to PostAnd what is solution for number for number 2?
Jump to PostThen its 4, because I know that 1 and 2 are true
Are you guessing, you're the author of this thread so i believe you should have the answer
Jump to Post2 could be true if you include Gaussian integers. The joys of complex numbers :)
Jump to Post2 could be true if you include Gaussian integers. The joys of complex numbers
Could we have a breakdown so we can verify and work towards an answer
No, 10 squared is 100. And using E instead of in symbol is little confusing, 'in' whould be clearer. Check again.
Then its 4, because I know that 1 and 2 are true
And what is solution for number for number 2?
There exists an integer x, take for instance -2, muliplied into the equation gives 0 and it is less that 0. I'm a bit confused, this is my first level
(-2)*(-2)+ 4 = 4 + 4 = 8 Didn't learn multiplication yet? ;)
Oops my mistake! Feel silly now
Then its 4, because I know that 1 and 2 are true
Are you guessing, you're the author of this thread so i believe you should have the answer
I was confused about the answer, if I was sure about the answer already I woulndt have asked the question
2 could be true if you include Gaussian integers. The joys of complex numbers :)
2 could be true if you include Gaussian integers. The joys of complex numbers
Could we have a breakdown so we can verify and work towards an answer
Gaussian integers are complex numbers in which each part is an integer, for example: 3+2i.
For #2, if you used 0+2i the result would be -4 + 4 = 0. You can also use 0-2i.
One "small thing" though, that joice does not specify integer type (like Z or N), but also it says exist integer x < 0. So Momerath, I would be interested on your proof that -2i < 0
One "small thing" though, that joice does not specify integer type (like Z or N), but also it says exist integer x < 0. So Momerath, I would be interested on your proof that -2i < 0
Numbers are either positive, negative or zero.
Neither 2i nor -2i is zero.
Negative numbers are less than zero.
When multiplying two numbers, if both are negative it results in a positive number.
When multiplying two numbers, if one is negative it results in a negative number.
If 2i is positive, then -2i must be negative.
If 2i is negative, then -2i must be positive.
Therefor one of {2i, -2i} must be negative and is a solution for the problem x < 0, x^2 + 4 = 0.
Momerath:Numbers are either positive, negative or zero.
Sorry not True, numbers can be negative only if the imaginary part == 0
Numbers are either zero or non-zero, that is True. 'Negative' does not make sense with numbers with imaginary parts.
Only thing that we can say comparing -2i, 0 and +2i, is that their real part is 0. But abs(-2i) == abs(2i) == 2 == abs(2) == abs(-2).
No you can't compare imaginary and real numbers like that. #2 and #3 are both false. And #1 and #4 are true.
@OP, can you first show that 1 and 4 are true? Then show that 2 and 3 are false.
can you first show that 1 and 4 are true?
1 is false. Try using the value x = 1.1
There is no restriction that the numbers must be integers for question 1.
Sorry not True, numbers can be negative only if the imaginary part == 0
Proof please.
But abs(-2i) == abs(2i) == 2 == abs(2) == abs(-2).
This is just wrong. If it were true then i would equal 1 or -1.
Python 2.7.2 (default, Jun 12 2011, 15:08:59) [MSC v.1500 32 bit (Intel)] on win32
Type "copyright", "credits" or "license()" for more information.
>>> 2j < 0
Traceback (most recent call last):
File "<pyshell#0>", line 1, in <module>
2j < 0
TypeError: no ordering relation is defined for complex numbers
>>>## let's try one definition for negative not with <
>>> def negative(x): return (x*x)**0.5 != x
>>> negative(-3)
True
>>> negative(-2j)
True
>>> negative(2j)
True
>>> negative(4)
False
>>> negative(4+3j)
False
>>> negative(4-3j)
False
>>> negative(-4-3j)
True
>>> negative(-4+3j)
True
# this was in doubt
>>> abs(-2j) == abs(2j) == 2 == abs(2) == abs(-2)
True
There is no restriction that the numbers must be integers for question 1.
Proof please.
That's the definition of "negative". FFS what did you think negative meant?
This is just wrong. If it were true then i would equal 1 or -1.
Instead of talking about stuff you don't have a clue about, you should try not talking about stuff you don't have a clue about. It's a good idea.
pyTony, on the other hand, manages to talk about mathematics without just making stuff up and then arguing with standard definitions.
And no it would simply mean that abs(i) equals 1.
1 is false. Try using the value x = 1.1
There is no restriction that the numbers must be integers for question 1.
I see a lot of people attacking you. But rather than saying you don't know what you are talking about,I will assume the question wasn't clearly stated to you and so I would like to explain why #1 is true. Its all in the wording. So this is question #1.
1) for all x E Z: if x^(squared) - 5x + 6 = 0, then x > 1 and x < 4.
In English words it is read as follows, for all values x in Z, if the value x^2 - 5x + 6 equals 0, then x has to be in between 1 and 4. Notice that it first needs to satisfy the equation x^2 - 5x + 6 first. And if it does, then it says that x has to be withing 1 and 4, exclusively. So lets see if this is true.
First we need to see what values of x satisfies the equation y(x) = x^2 - 5x + 6. We can reformulate the above equation as, y(x) = (x-3)(x-2) = 0. And we can instantly see that x = 3 and x = 2 is a solution. Hence if we plug-in x = 3 or x = 2, in y(x) we get the result y(x) = 0 for x = 3 or x = 2.
And now we check if x = 2 or x = 3 satisfies the second clause of the original statement 1). And it does since 1 < 2,3 < 4. Hence the the statement 1) is true.
So in all, I guess the mistake was to think that you can evaluate the second clause without the first clause being satisfied.
Thanks, firstPerson, additionally Z traditionally denotes integers including negatives.
Here more exactly the statements in pyTonyc way, in Python:
""" executable formulation in Python for post:
http://www.daniweb.com/software-development/computer-science/threads/384061/1653965#post1653965
"""
# point 3)
print next((x, x*x) for x in {10, 100, 1000} if x*x in {10, 100, 1000})
""" Output:
(10, 100)
"""
# point 4)
print all(not(x*x & 1) for x in {10, 100, 1000})
""" Output:
True
"""
# kind of proof without direct statement that x*x >= 0
# for point 2)
print any(x**2 + 4 == 0 for x in range(-2,0))
""" Output, therefore -> this is the answer
False
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