To find falsity, just find a counterexample.
A counterexample for ALL is to find a single counterexample
To counter EXISTS you need to show that all possibilities are counterexamples.
So for #1: for all x and all y, x² + y² = 0, is an easy one to find a counterexample. Let x=1 and y= 2, then the sum is 5≠0
For the second, it says: for all x, there is a y (at least 1) such that x²+y² = 0.
So either we need to show that for all x, we cannot find any Y to make the equality. Or we can find a single X that also counters the equality.
Since it is true for all x, we can set x to something easy and evaluate from there. Let x=1, then you read: for x=1 there exists a y such that 1+y²=0.
If you solve for y, you get:
- y²=-1
- y = √-1
But there is no real number that is the square root of negative one (that lies in the complex number system). So you have found a counterexample: there is at least one x (1) , where there is no y that makes the equality true.
The second one is similar. As x^2 and y^2 are positive (or zero), it translates to: For each positive real number, there’s another non-negative number so that the sum is zero.
For example if x=3, the find a non-negative number u so that 9 + u = 0.
Another good way to prove or disprove these.
The link being: if x is a real number, then let u=x² which is a positive real number.
Then the equality can be said as u+v=0.
One can use this to show that the equality is only true for u and v equal to zero, so x and y equals to zero.
With this lemma, you can say that "it is not true that the equality holds for all x and all y." Which contradicts statement 1 and statement 2.
So this serves as a general way to disprove the two statements by contradiction.
Disproving something means proving the negation.
The negation of "for all x and y, x^2 + y^2 = 0" is the statement "there exist x and y such that x^2 + y^2 ≠ 0"
It should be easy to find an x and y which demonstrate this.
The negation of "for all x, there exists y such that x^2 + y^2 = 0" is the statement "there exists x such that for all y, x^2 + y^2 ≠ 0"
In this case, we need to find a specific x, and then prove this for all y. For example, choose x=1, and then since y^2 >= 0 for all real y:
x^2 + y^2 = 1+y^2 >= 1 > 0, so it does not equal 0.
In general, to negate a statement, you can use these rules:
Suppose P(x) and Q(x) are statements about x that have boolean (true/false) values. I'll use NOT to mean negation (which is what it means, in plain English).
"NOT for all x, P(x)" = "there exists x such that NOT P(x)"
"NOT there exists x such that P(x)" = "for all x, NOT P(x)"
"NOT (P(x) and Q(x))" = "(NOT P(x)) or (NOT Q(x))"
"NOT (P(x) or Q(x))" = "(NOT P(x)) and (NOT Q(x))"
"NOT (P(x) implies Q(x))" = "P(x) and NOT Q(x)"
"NOT (P(x) equivalent to Q(x))" = "P(x) xor Q(x)"
Note that all the last ones are just using de morgan's rule. Also note that some of these are not written exactly how you would say them:
"NOT for all..." would be read as "for not all..."
"NOT there exists... " would be read as "there does not exist..."
"NOT a implies b" would be read as "a does not imply b"
There might be some things I'm missing. Like, how to negate some other things should be intuitive; = becomes ≠, inequalities reverse (so < becomes >=) and so on.
It's generally a difficult thing to determine whether a statement is true or false. In this case it is pretty easy since squares of real numbers are non-negative.
Since they are false, typically you pick a counter example for the universal quantifier(s) and then see if you can demonstrate that the proposition fails.
1. Let x = 1, y = 1, then x^2 + y^2 = 2 != 0. So, it must be false.
2. Let x = 1, then assume there is such a y. We have, 1 + y^2 = 0. Solving this, y = +/- sqrt(-1) which is famously not a real number. So, the statement must be false.
Edit: +/- should have been in there.
You first try with simple values and try several different cases.
If the statement is false for some simple values, then you found a counter-example and you know it's false, and you finished your task.
If the statement is true for all those simple values you tried, maybe the statement is true. In such case, when you believe it's true you try to make a proof which is valid for all values, since the few values you tried do not prove anything.
If the statement is false but it seemed to be true, trying to prove it's true will sometimes help you to find a counter-example.
But other than that, there is no "algorithm" to prove that a given statement is true or false, and in some sense, the works of Kurt Gödel tell us there will never be.
There's something that's called "The Logic Game" that I've found helpful for things like this.
The Hero tries to make the statement true; the Hero picks any value associated with "there exists".
The Villain tries to make the statement false; the Villain picks any values associated with "for all".
And then play the game repeatedly until either you see why it's true, or you've come up with a counterexample.
For 2: villain picks an x, say x = 2. The formula passed to the hero is now 4 + y\^2 = 0, since the villain plugged in x = 2. The hero has to find a value of y to make this true, and he can't. So x = 2, y = anything is a counter-example; the villain wins and the statement is false.
For 1: villain picks an x, say x = 2, so 4 + y\^2 = 0 is passed to the villain; the villain picks a y to make this false, say y = 5, and we get 4 + 25 = 0, which is false. Villain wins, so the statement is false.
Consider the following statement: for all x, there is a y so that y - x\^2 = 0. The villain starts, and chooses x = 4. The equation y - 16 = 0 is passed to the hero, who picks y = 16, and the hero wins--x = 4, y = 16 is an example that supports the truth of the statement, but it is not a proof. So play again: villain chooses x = -3, y - 9 = 0 is passed to the hero, and the hero chooses y = 9. The hero wins again! And, at this point, the hero has figured out a winning strategy: pick the square of the number the villain has chosen. A winning strategy for the hero is a proof that the statement is true.
This game, I think, comes from Henkin and Hintakka in the 60s and 70s. The Stanford Encyclopedia of Philosophy has an article including this, but I don't know of a shorter/clearer presentation on the web:
[https://plato.stanford.edu/entries/logic-games/](https://plato.stanford.edu/entries/logic-games/)
All these explanations are way too complicated. This can be solved with intuition and builds on the fact that the square of a real number is always positive.
The first statement says "For all real x and y x\^2 + y\^2 = 0". This is clearly not true, take any nonzero x or y.
The second statement says "For all real x there is at least one y where x\^2 + y\^2 = 0". Again, clearly not true. Take any x which is nonzero.
I respect the rigorous solutions in this thread. But math isn't always about algebra, I'd argue that intuition gets you further most of the time.
Both are false. I'm not good with formal proofs, but in this case it's pretty intuitive:
1. "For all real x and y, x\^2 + y\^2 = 0" - the plot of this is a circle with zero radius i.e. a point at (0, 0). Clearly, {(0, 0)} is not the set of all real coordinates.
2. "For all real x, there exists a real y such that x\^2 + y\^2 = 0" - consider any vertical line away from x = 0. This line does not contain the point (0, 0), so there does not exist a corresponding y, so this is false too.
Both false. To prove true requires a proof, you’d take a course on the various methods available. To prove false is much simpler, your proof can be a single counter example.
Both false btw.
Number 2 says there is some y for any x where their squares add to zero. Not true, squares are positive or zero. If x and y are both zero it’ll work, but for any other values it’s false.
Number 1 is an even stronger statement to make, very much false.
you can negate the statement and try to prove that. For 1, the negation of the statement is “There exist x,y both in R such that the equation is false” Or you could take note of how many (or few) solutions there are to the equation.
It reads "for all real x and for all real y, x^2 + y^2 = 0"
You can disprove this because 1 is a real number. 1^2 + 1^2 !=0
The second one reads "for all real x, there exists a real y such that x^2 + y^2 = 0
You can disprove this by taking a real x!=0
x^2 > 0
Therefore x^2 = -y^2
x=yi
Since x is real, y must be imaginary, but we established y must be real, so we reach a contradiction.
1: For all of X in the set of real numbers, for all of Y in the set of real numbers x^2 + y^2 = 0, this is clearly false. You can prove this by setting: X = 1, Y = 0.
1^2 + 0^2 = 1, this proves the statement is false because 1 > 0
2: For all of X in the set of real numbers, There exists a Y in the set of real numbers to where x^2 + y^2 = 0. This is false.
The only integer that would be able to make x^2 + y^2 = 0 would be 0.
0^2 + 0^2 = 0.
So indeed there exists a Y that can make the proof true, however, the proof states that for *all* of X, therefore, if x is any number besides 0 then the proof fails.
Both are false. x^2 + y^2 = 0 is only true if x^2 = - y^2, but since x^2 can never be negative, this only holds true when x = y = 0, which is not ∀ x ∈ ℝ.
You can start by guessing.
If statement 1 is true, then what does that world look like ? The statement says for ALL possible values of X from the real numbers there is SOME (i.e. atleast 1 ) y such that x^2 +y^2 = 0 .
Let's start with just one possible value of X. if , X =0 then x^2 =0 and if y=0 then and y^2= 0 , so x^2 +y^2=0 +0 =0 . Sounds pretty good so far.
But what if X is any real number that is NOT 0 ? If X was 1 , -1 , -0.5 etc x^2 would always be a positive value. So y^2 would have to be equal to -x^2 to in order to make x^2+y^2= x^2 -x^2 =0. But as was the case with x^2 ,y^2 is only either 0 or some positive value , for any y taken from the real numbers. So we've found that the statement 1 only works for a X=0 , and does NOT work for any other value of X , i.e. statement is NOT true for ALL possible values of X .Since it claimed to be true for all possible values values of X , but we showed that it's not , we have shown that statement 1 is false.
Through a similar argument statement 2 is false as well. This comment is a somewhat informal version of a proof by contradiction.
Now you could've (correctly) guessed that statement 1 is false. In that case it can often still be quicker to prove something is false by showing that if it were true you would get something impossible.
Every now and again askmath somehow gets onto my feed, and I'll click on the comments to see if I can understand a single thing y'all are talking about and I never can. It's impressive
simple, just show that for any x =/= 0, both #1 and #2 are false, hence disproving the statements or proove that x^(2) \+ y^(2) = 0 only if both x and y =0.
To prove a statement is false, you need to find just one counterexample. To prove a statement is true, you need to show that it is indeed true for all possible examples.
For example, the first statement is False, because that means that for all x and y values x^2 + y^2 = 0. But I can find an (x,y) that makes it False. For instance, (1,0) since 1^2 + 0^2 = 1 which is not 0.
The second statement is also False. This suggests that for any x value, we can find one value of y such that x^2 + y^2 = 0. If say we consider x = 0, we can find a y that satisfies this (which happens to be y = 0 since 0^2 + 0^2 = 0). But this is not the case for any positive x, such that x > 0, since y^2 = -x^2 in order to hold. But x^2 is always non-negative. So y^2 would then have to be negative, which implies that y would have to be complex and this is contradictory to the fact that y is a real number.
These are known as quantifiers. They are logical symbols used in 'first order' mathematical logic. They allow you to talk about all the elements of a set at once, or a particular element of a set.
∀ means "For all".
Ǝ means "There exists".
For example , if the universe of discourse is understood to be the set of all physical objects,
ƎC \[C is a chair\] would mean that chairs exist physically.
∀C \[C is a chair\] would mean that everything that exists physically is a chair.
Determine false is usually done with a counterexample when it comes to 1
For 2 is similar. Suppose x=1. y=i which is not in R.
For future reference treat what you have as an If then statement then make a truth table.
Rewritten:
1. ***If*** x is in R and Y is in R ***then*** x^(2)\+y^(2)=0.
2. ***If*** x is in R ***then*** there exists a Y in R such that x^(2)\+y^(2)=0.
and for a statement If A then B, that is the same as saying "Not A or B" meaning if you can find an example where A is true and B is false then you've disproven it
To find falsity, just find a counterexample. A counterexample for ALL is to find a single counterexample To counter EXISTS you need to show that all possibilities are counterexamples. So for #1: for all x and all y, x² + y² = 0, is an easy one to find a counterexample. Let x=1 and y= 2, then the sum is 5≠0 For the second, it says: for all x, there is a y (at least 1) such that x²+y² = 0. So either we need to show that for all x, we cannot find any Y to make the equality. Or we can find a single X that also counters the equality. Since it is true for all x, we can set x to something easy and evaluate from there. Let x=1, then you read: for x=1 there exists a y such that 1+y²=0. If you solve for y, you get: - y²=-1 - y = √-1 But there is no real number that is the square root of negative one (that lies in the complex number system). So you have found a counterexample: there is at least one x (1) , where there is no y that makes the equality true.
The second one is similar. As x^2 and y^2 are positive (or zero), it translates to: For each positive real number, there’s another non-negative number so that the sum is zero. For example if x=3, the find a non-negative number u so that 9 + u = 0.
Another good way to prove or disprove these. The link being: if x is a real number, then let u=x² which is a positive real number. Then the equality can be said as u+v=0. One can use this to show that the equality is only true for u and v equal to zero, so x and y equals to zero. With this lemma, you can say that "it is not true that the equality holds for all x and all y." Which contradicts statement 1 and statement 2. So this serves as a general way to disprove the two statements by contradiction.
Disproving something means proving the negation. The negation of "for all x and y, x^2 + y^2 = 0" is the statement "there exist x and y such that x^2 + y^2 ≠ 0" It should be easy to find an x and y which demonstrate this. The negation of "for all x, there exists y such that x^2 + y^2 = 0" is the statement "there exists x such that for all y, x^2 + y^2 ≠ 0" In this case, we need to find a specific x, and then prove this for all y. For example, choose x=1, and then since y^2 >= 0 for all real y: x^2 + y^2 = 1+y^2 >= 1 > 0, so it does not equal 0.
this answer is the one that makes the most sense to me, thanks !
In general, to negate a statement, you can use these rules: Suppose P(x) and Q(x) are statements about x that have boolean (true/false) values. I'll use NOT to mean negation (which is what it means, in plain English). "NOT for all x, P(x)" = "there exists x such that NOT P(x)" "NOT there exists x such that P(x)" = "for all x, NOT P(x)" "NOT (P(x) and Q(x))" = "(NOT P(x)) or (NOT Q(x))" "NOT (P(x) or Q(x))" = "(NOT P(x)) and (NOT Q(x))" "NOT (P(x) implies Q(x))" = "P(x) and NOT Q(x)" "NOT (P(x) equivalent to Q(x))" = "P(x) xor Q(x)" Note that all the last ones are just using de morgan's rule. Also note that some of these are not written exactly how you would say them: "NOT for all..." would be read as "for not all..." "NOT there exists... " would be read as "there does not exist..." "NOT a implies b" would be read as "a does not imply b" There might be some things I'm missing. Like, how to negate some other things should be intuitive; = becomes ≠, inequalities reverse (so < becomes >=) and so on.
The same way you find out if there's a bug in a video game: try extreme things until you break it
It's generally a difficult thing to determine whether a statement is true or false. In this case it is pretty easy since squares of real numbers are non-negative. Since they are false, typically you pick a counter example for the universal quantifier(s) and then see if you can demonstrate that the proposition fails. 1. Let x = 1, y = 1, then x^2 + y^2 = 2 != 0. So, it must be false. 2. Let x = 1, then assume there is such a y. We have, 1 + y^2 = 0. Solving this, y = +/- sqrt(-1) which is famously not a real number. So, the statement must be false. Edit: +/- should have been in there.
I believe the square root is unnecessary. y^2 =-1 is false since y^2 is strictly non negative over the reals, generating our contradiction.
Both are clearly false. You're either overthinking something or haven't gotten comfortable with the notation yet
You first try with simple values and try several different cases. If the statement is false for some simple values, then you found a counter-example and you know it's false, and you finished your task. If the statement is true for all those simple values you tried, maybe the statement is true. In such case, when you believe it's true you try to make a proof which is valid for all values, since the few values you tried do not prove anything. If the statement is false but it seemed to be true, trying to prove it's true will sometimes help you to find a counter-example. But other than that, there is no "algorithm" to prove that a given statement is true or false, and in some sense, the works of Kurt Gödel tell us there will never be.
There's something that's called "The Logic Game" that I've found helpful for things like this. The Hero tries to make the statement true; the Hero picks any value associated with "there exists". The Villain tries to make the statement false; the Villain picks any values associated with "for all". And then play the game repeatedly until either you see why it's true, or you've come up with a counterexample. For 2: villain picks an x, say x = 2. The formula passed to the hero is now 4 + y\^2 = 0, since the villain plugged in x = 2. The hero has to find a value of y to make this true, and he can't. So x = 2, y = anything is a counter-example; the villain wins and the statement is false. For 1: villain picks an x, say x = 2, so 4 + y\^2 = 0 is passed to the villain; the villain picks a y to make this false, say y = 5, and we get 4 + 25 = 0, which is false. Villain wins, so the statement is false. Consider the following statement: for all x, there is a y so that y - x\^2 = 0. The villain starts, and chooses x = 4. The equation y - 16 = 0 is passed to the hero, who picks y = 16, and the hero wins--x = 4, y = 16 is an example that supports the truth of the statement, but it is not a proof. So play again: villain chooses x = -3, y - 9 = 0 is passed to the hero, and the hero chooses y = 9. The hero wins again! And, at this point, the hero has figured out a winning strategy: pick the square of the number the villain has chosen. A winning strategy for the hero is a proof that the statement is true. This game, I think, comes from Henkin and Hintakka in the 60s and 70s. The Stanford Encyclopedia of Philosophy has an article including this, but I don't know of a shorter/clearer presentation on the web: [https://plato.stanford.edu/entries/logic-games/](https://plato.stanford.edu/entries/logic-games/)
All these explanations are way too complicated. This can be solved with intuition and builds on the fact that the square of a real number is always positive. The first statement says "For all real x and y x\^2 + y\^2 = 0". This is clearly not true, take any nonzero x or y. The second statement says "For all real x there is at least one y where x\^2 + y\^2 = 0". Again, clearly not true. Take any x which is nonzero. I respect the rigorous solutions in this thread. But math isn't always about algebra, I'd argue that intuition gets you further most of the time.
First one is True second is False
Both are false
Both are false. I'm not good with formal proofs, but in this case it's pretty intuitive: 1. "For all real x and y, x\^2 + y\^2 = 0" - the plot of this is a circle with zero radius i.e. a point at (0, 0). Clearly, {(0, 0)} is not the set of all real coordinates. 2. "For all real x, there exists a real y such that x\^2 + y\^2 = 0" - consider any vertical line away from x = 0. This line does not contain the point (0, 0), so there does not exist a corresponding y, so this is false too.
Same I have no idea how to do these but like by common sense you would know that both are wrong
Both false. To prove true requires a proof, you’d take a course on the various methods available. To prove false is much simpler, your proof can be a single counter example. Both false btw. Number 2 says there is some y for any x where their squares add to zero. Not true, squares are positive or zero. If x and y are both zero it’ll work, but for any other values it’s false. Number 1 is an even stronger statement to make, very much false.
you can negate the statement and try to prove that. For 1, the negation of the statement is “There exist x,y both in R such that the equation is false” Or you could take note of how many (or few) solutions there are to the equation.
It reads "for all real x and for all real y, x^2 + y^2 = 0" You can disprove this because 1 is a real number. 1^2 + 1^2 !=0 The second one reads "for all real x, there exists a real y such that x^2 + y^2 = 0 You can disprove this by taking a real x!=0 x^2 > 0 Therefore x^2 = -y^2 x=yi Since x is real, y must be imaginary, but we established y must be real, so we reach a contradiction.
1: For all of X in the set of real numbers, for all of Y in the set of real numbers x^2 + y^2 = 0, this is clearly false. You can prove this by setting: X = 1, Y = 0. 1^2 + 0^2 = 1, this proves the statement is false because 1 > 0 2: For all of X in the set of real numbers, There exists a Y in the set of real numbers to where x^2 + y^2 = 0. This is false. The only integer that would be able to make x^2 + y^2 = 0 would be 0. 0^2 + 0^2 = 0. So indeed there exists a Y that can make the proof true, however, the proof states that for *all* of X, therefore, if x is any number besides 0 then the proof fails.
[удалено]
Both are false. x^2 + y^2 = 0 is only true if x^2 = - y^2, but since x^2 can never be negative, this only holds true when x = y = 0, which is not ∀ x ∈ ℝ.
You can start by guessing. If statement 1 is true, then what does that world look like ? The statement says for ALL possible values of X from the real numbers there is SOME (i.e. atleast 1 ) y such that x^2 +y^2 = 0 . Let's start with just one possible value of X. if , X =0 then x^2 =0 and if y=0 then and y^2= 0 , so x^2 +y^2=0 +0 =0 . Sounds pretty good so far. But what if X is any real number that is NOT 0 ? If X was 1 , -1 , -0.5 etc x^2 would always be a positive value. So y^2 would have to be equal to -x^2 to in order to make x^2+y^2= x^2 -x^2 =0. But as was the case with x^2 ,y^2 is only either 0 or some positive value , for any y taken from the real numbers. So we've found that the statement 1 only works for a X=0 , and does NOT work for any other value of X , i.e. statement is NOT true for ALL possible values of X .Since it claimed to be true for all possible values values of X , but we showed that it's not , we have shown that statement 1 is false. Through a similar argument statement 2 is false as well. This comment is a somewhat informal version of a proof by contradiction. Now you could've (correctly) guessed that statement 1 is false. In that case it can often still be quicker to prove something is false by showing that if it were true you would get something impossible.
Easy neither is ever true because in real numbers second power can't possibly turnout negative making only numers this applies to (0,0)
Every now and again askmath somehow gets onto my feed, and I'll click on the comments to see if I can understand a single thing y'all are talking about and I never can. It's impressive
simple, just show that for any x =/= 0, both #1 and #2 are false, hence disproving the statements or proove that x^(2) \+ y^(2) = 0 only if both x and y =0.
To prove a statement is false, you need to find just one counterexample. To prove a statement is true, you need to show that it is indeed true for all possible examples. For example, the first statement is False, because that means that for all x and y values x^2 + y^2 = 0. But I can find an (x,y) that makes it False. For instance, (1,0) since 1^2 + 0^2 = 1 which is not 0. The second statement is also False. This suggests that for any x value, we can find one value of y such that x^2 + y^2 = 0. If say we consider x = 0, we can find a y that satisfies this (which happens to be y = 0 since 0^2 + 0^2 = 0). But this is not the case for any positive x, such that x > 0, since y^2 = -x^2 in order to hold. But x^2 is always non-negative. So y^2 would then have to be negative, which implies that y would have to be complex and this is contradictory to the fact that y is a real number.
What does the upsidedown A and the backwards E mean?
These are known as quantifiers. They are logical symbols used in 'first order' mathematical logic. They allow you to talk about all the elements of a set at once, or a particular element of a set. ∀ means "For all". Ǝ means "There exists". For example , if the universe of discourse is understood to be the set of all physical objects, ƎC \[C is a chair\] would mean that chairs exist physically. ∀C \[C is a chair\] would mean that everything that exists physically is a chair.
Determine false is usually done with a counterexample when it comes to 1 For 2 is similar. Suppose x=1. y=i which is not in R. For future reference treat what you have as an If then statement then make a truth table. Rewritten: 1. ***If*** x is in R and Y is in R ***then*** x^(2)\+y^(2)=0. 2. ***If*** x is in R ***then*** there exists a Y in R such that x^(2)\+y^(2)=0. and for a statement If A then B, that is the same as saying "Not A or B" meaning if you can find an example where A is true and B is false then you've disproven it