In the last statement, can i prove the contrapositive of it to be true to prove the statement?
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I have worked out the problem A B C, my question is in the last statement " If the numbers e, π, π^2, e^2 and eπ are irrational, prove that at most one of the numbers π+e, π−e, π^2−e^2, π^2+e^2 is rational." can i prove the contrapositive of it (i.e.at least two of the numbers π+e, π−e, π^2−e^2, π^2+e^2 is rational, then at least one of the e, π, π^2, e^2 and eπ are rational) to be true to prove the statement?
What does it mean to say that a number x is irrational?
Prove by contradiction statements A and B below, where p and q are real numbers.
A: If pq is irrational, then at least one of p and q is irrational.
B: If p + q is irrational, then at least one of p and q is irrational.
Disprove by means of a counterexample statement C below, where p and q are real numbers.
C: If p and q are irrational, then p + q is irrational.
If the numbers e, π, π^2, e^2 and eπ are irrational, prove that at most one of the numbers π+e,
π−e, π^2−e^2, π^2+e^2 is rational.
logic propositional-calculus
asked Jan 2 at 13:54
KevinKevin
143
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add a comment |
$begingroup$
I have worked out the problem A B C, my question is in the last statement " If the numbers e, π, π^2, e^2 and eπ are irrational, prove that at most one of the numbers π+e, π−e, π^2−e^2, π^2+e^2 is rational." can i prove the contrapositive of it (i.e.at least two of the numbers π+e, π−e, π^2−e^2, π^2+e^2 is rational, then at least one of the e, π, π^2, e^2 and eπ are rational) to be true to prove the statement?
What does it mean to say that a number x is irrational?
Prove by contradiction statements A and B below, where p and q are real numbers.
A: If pq is irrational, then at least one of p and q is irrational.
B: If p + q is irrational, then at least one of p and q is irrational.
Disprove by means of a counterexample statement C below, where p and q are real numbers.
C: If p and q are irrational, then p + q is irrational.
If the numbers e, π, π^2, e^2 and eπ are irrational, prove that at most one of the numbers π+e,
π−e, π^2−e^2, π^2+e^2 is rational.
logic propositional-calculus
asked Jan 2 at 13:54
KevinKevin
143
$endgroup$
$begingroup$
What have you tried? Surely, for part A, you can show that $p,qin mathbb Qimplies pqin mathbb Q$, no?
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– lulu
Jan 2 at 13:59
$begingroup$
That comment looks like it should be part of the question. The title looks like it should be in the question too. And what's all the other stuff in the question? Are we supposed to say something about A, B, and C, or just the last paragraph?
$endgroup$
– David K
Jan 2 at 13:59
add a comment |
$begingroup$
I have worked out the problem A B C, my question is in the last statement " If the numbers e, π, π^2, e^2 and eπ are irrational, prove that at most one of the numbers π+e, π−e, π^2−e^2, π^2+e^2 is rational." can i prove the contrapositive of it (i.e.at least two of the numbers π+e, π−e, π^2−e^2, π^2+e^2 is rational, then at least one of the e, π, π^2, e^2 and eπ are rational) to be true to prove the statement?
What does it mean to say that a number x is irrational?
Prove by contradiction statements A and B below, where p and q are real numbers.
A: If pq is irrational, then at least one of p and q is irrational.
B: If p + q is irrational, then at least one of p and q is irrational.
Disprove by means of a counterexample statement C below, where p and q are real numbers.
C: If p and q are irrational, then p + q is irrational.
If the numbers e, π, π^2, e^2 and eπ are irrational, prove that at most one of the numbers π+e,
π−e, π^2−e^2, π^2+e^2 is rational.
logic propositional-calculus
asked Jan 2 at 13:54
KevinKevin
143
$endgroup$
I have worked out the problem A B C, my question is in the last statement " If the numbers e, π, π^2, e^2 and eπ are irrational, prove that at most one of the numbers π+e, π−e, π^2−e^2, π^2+e^2 is rational." can i prove the contrapositive of it (i.e.at least two of the numbers π+e, π−e, π^2−e^2, π^2+e^2 is rational, then at least one of the e, π, π^2, e^2 and eπ are rational) to be true to prove the statement?
What does it mean to say that a number x is irrational?
Prove by contradiction statements A and B below, where p and q are real numbers.
A: If pq is irrational, then at least one of p and q is irrational.
B: If p + q is irrational, then at least one of p and q is irrational.
Disprove by means of a counterexample statement C below, where p and q are real numbers.
C: If p and q are irrational, then p + q is irrational.
If the numbers e, π, π^2, e^2 and eπ are irrational, prove that at most one of the numbers π+e,
π−e, π^2−e^2, π^2+e^2 is rational.
logic propositional-calculus
logic propositional-calculus
asked Jan 2 at 13:54
KevinKevin
143
asked Jan 2 at 13:54
KevinKevin
143
edited Jan 2 at 14:10
Kevin
asked Jan 2 at 13:54
KevinKevin
143
asked Jan 2 at 13:54
KevinKevin
143
asked Jan 2 at 13:54
KevinKevin
143
143
$begingroup$
What have you tried? Surely, for part A, you can show that $p,qin mathbb Qimplies pqin mathbb Q$, no?
$endgroup$
– lulu
Jan 2 at 13:59
$begingroup$
That comment looks like it should be part of the question. The title looks like it should be in the question too. And what's all the other stuff in the question? Are we supposed to say something about A, B, and C, or just the last paragraph?
$endgroup$
– David K
Jan 2 at 13:59
add a comment |
$begingroup$
What have you tried? Surely, for part A, you can show that $p,qin mathbb Qimplies pqin mathbb Q$, no?
$endgroup$
– lulu
Jan 2 at 13:59
$begingroup$
That comment looks like it should be part of the question. The title looks like it should be in the question too. And what's all the other stuff in the question? Are we supposed to say something about A, B, and C, or just the last paragraph?
$endgroup$
– David K
Jan 2 at 13:59
$begingroup$
What have you tried? Surely, for part A, you can show that $p,qin mathbb Qimplies pqin mathbb Q$, no?
$endgroup$
– lulu
Jan 2 at 13:59
$begingroup$
What have you tried? Surely, for part A, you can show that $p,qin mathbb Qimplies pqin mathbb Q$, no?
$endgroup$
– lulu
Jan 2 at 13:59
$begingroup$
That comment looks like it should be part of the question. The title looks like it should be in the question too. And what's all the other stuff in the question? Are we supposed to say something about A, B, and C, or just the last paragraph?
$endgroup$
– David K
Jan 2 at 13:59
$begingroup$
That comment looks like it should be part of the question. The title looks like it should be in the question too. And what's all the other stuff in the question? Are we supposed to say something about A, B, and C, or just the last paragraph?
$endgroup$
– David K
Jan 2 at 13:59
add a comment |
1 Answer
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$begingroup$
Yes, proving the contrapositive also proves the statement.
We need to be sure the contrapositive is really a contrapositive.
"At least two" is the exact negation of "at most one,"
and "at least one is rational" is the exact negation of "all are irrational,"
so I think you're doing OK so far.
answered Jan 2 at 14:07
David KDavid K
55.4k345120
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$begingroup$
Yes, proving the contrapositive also proves the statement.
We need to be sure the contrapositive is really a contrapositive.
"At least two" is the exact negation of "at most one,"
and "at least one is rational" is the exact negation of "all are irrational,"
so I think you're doing OK so far.
answered Jan 2 at 14:07
David KDavid K
55.4k345120
$endgroup$
add a comment |
$begingroup$
Yes, proving the contrapositive also proves the statement.
We need to be sure the contrapositive is really a contrapositive.
"At least two" is the exact negation of "at most one,"
and "at least one is rational" is the exact negation of "all are irrational,"
so I think you're doing OK so far.
answered Jan 2 at 14:07
David KDavid K
55.4k345120
$endgroup$
add a comment |
$begingroup$
Yes, proving the contrapositive also proves the statement.
We need to be sure the contrapositive is really a contrapositive.
"At least two" is the exact negation of "at most one,"
and "at least one is rational" is the exact negation of "all are irrational,"
so I think you're doing OK so far.
answered Jan 2 at 14:07
David KDavid K
55.4k345120
$endgroup$
Yes, proving the contrapositive also proves the statement.
We need to be sure the contrapositive is really a contrapositive.
"At least two" is the exact negation of "at most one,"
and "at least one is rational" is the exact negation of "all are irrational,"
so I think you're doing OK so far.
answered Jan 2 at 14:07
David KDavid K
55.4k345120
answered Jan 2 at 14:07
David KDavid K
55.4k345120
answered Jan 2 at 14:07
David KDavid K
55.4k345120
answered Jan 2 at 14:07
David KDavid K
55.4k345120
55.4k345120
add a comment |
add a comment |
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$begingroup$
What have you tried? Surely, for part A, you can show that $p,qin mathbb Qimplies pqin mathbb Q$, no?
$endgroup$
– lulu
Jan 2 at 13:59
$begingroup$
That comment looks like it should be part of the question. The title looks like it should be in the question too. And what's all the other stuff in the question? Are we supposed to say something about A, B, and C, or just the last paragraph?
$endgroup$
– David K
Jan 2 at 13:59