# Translating $\forall$ and $\exists$ in a statement

The following statement has two versions – one where $d$ is quantified by $\forall$ and the second where it's quantified by $\exists$. The task here is to find a counterexample where the statements below are false. The domain is all integers.

1. $\forall a \forall b \forall c \forall d(a^d + b^d = c^d$)
2. $\forall a \forall b \forall c \exists d(a^d + b^d = c^d$)

The first statement is false for at least some values of the variables. When $a=1, b=2, c=3, d=4$, the statement does not hold for all variables $a$, $b$, $c$, and $d$. For instance:

$$1^4 + 2^4 \ne 3^4$$

The second statement is false when $a = 1, b = 2, c = 10$ because there doesn't exist a $d$ where the statement would be true. For instance, if $d$ was $5$:

$$1^5 + 2^5 \ne 10^5$$

Am I interpreting this correctly and do my counterxamples make sense?

• Looks good to me. $\checkmark$ – Clarinetist Sep 6 '16 at 16:35
• Yes, but you can remove the for all variables a, b, c, and d part for the first statement, and you should remove the example given for the second statement (giving an example in this case is simply not enough for proving $\not\exists{d}$). Thinking about it, you need to prove your counterexample of $a=1,b=2,c=10$, since it is not so trivial. – barak manos Sep 6 '16 at 16:35
• "For instance, if d was 5..." So what? That only shows it isn't true for 5. You have to show it is never true no matter what $d$ is . $3^5 + 4^5 \ne 5^5$ for example. But I can not conclude that $3^d + 4^d \ne 5^d$ ever. (What if d = 2). – fleablood Sep 6 '16 at 16:41
• Your first counterexample makes sense. Your second counter example would make sense if you showed $1^d + 2^d \ne 10^d$. You did not show that at all. You only showed $1^5 + 2^5 \ne 10^5$ which is one case. Now you have to show for n=1,2,3,4,6,7,8,9,10,.......... – fleablood Sep 6 '16 at 16:43
• Assuming you are not allowed to quote Fermat's Last Theorem unless you can prove it yourself, a good counter example would be any c < a. Then $a^d + b^d > a^d > c^d$ for all $d$ so no $d$ exists where $a^d + b^d = c^d$. Example $a = 5,327$ and $b = 65,439$ and $c = 1$. Then $5,327^d + 65,439^d = 1^d$ is never true. – fleablood Sep 6 '16 at 16:49

Turning a comment into an answer:

For the first statement, you can remove the for all variables a, b, c, and d part, as the counterexample of $[a=1,b=2,c=3,d=4]$ already states this fact.

For the second statement, you need to prove that your counterexample of $[a=1,b=2,c=10]$ is true for ALL values of $d$.

I'm not sure whether or not it is indeed true for all values of $d$, but it is certainly not so trivial.

Alternatively, you can use the counterexample of $[a=0,b=0,c=1]$ which is a lot more trivial.

• Just to clarify, is $0$ a part of the set of all integers? – Shrey Sep 6 '16 at 16:53
• $1^d + 2^d \ne 10^d$ because $1^d + 2^d$ is odd and $10^d$ is even. But the OP never gave a reason why. S/he only gave a single example it wasn't true for $d = 5$. – fleablood Sep 6 '16 at 16:54
• @Shrey: Yes, though your question doesn't state anything about integers. You need to prove it for all (real) numbers. – barak manos Sep 6 '16 at 16:55
• @fleablood: So I suggested a simpler (much more obvious) counterexample. – barak manos Sep 6 '16 at 16:55
• Yes, 0 and negatives. So you also have to show $\frac 1{a^k} + \frac 1{b^k} \ne \frac 1{c^k}$ ... which blows my last comment out of the water. Oops. – fleablood Sep 6 '16 at 16:56

The counterexample you gave for the first statement is correct.

To prove the second statement is false, you need to prove that its negation is true. This negation is the statement $\exists a, \exists b, \exists c, \forall d, a^d+b^d \ne c^d$. Take $a=b=0$, $c=1$. Observe that for all integers $d$, $0^d + 0^d \ne 1^d$.