What's wrong with this bogus proof?What's wrong with this random variable proof?What's wrong with this proof...
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What's wrong with this bogus proof?
What's wrong with this random variable proof?What's wrong with this proof that all UFDs are Bezout?What's wrong with this proof by contradiction?A bogus proof of countable power setWhat's wrong with this proof $1=i^2=-1$What is wrong with this proofWhat's wrong with this proof of symmetry of equality?What's wrong with this 1 = -1 proof?What's wrong with this proof? (Regular languages)What's wrong in this proof of 1=2?
$begingroup$
What is the mistake here? Is it matter of the unit?
discrete-mathematics proof-verification
asked 1 hour ago
ShinobuIsMyWifeShinobuIsMyWife
313
$endgroup$
add a comment |
$begingroup$
What is the mistake here? Is it matter of the unit?
discrete-mathematics proof-verification
asked 1 hour ago
ShinobuIsMyWifeShinobuIsMyWife
313
$endgroup$
5
$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
1 hour ago
2
$begingroup$
Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
$endgroup$
– Dean Young
1 hour ago
add a comment |
$begingroup$
What is the mistake here? Is it matter of the unit?
discrete-mathematics proof-verification
asked 1 hour ago
ShinobuIsMyWifeShinobuIsMyWife
313
$endgroup$
What is the mistake here? Is it matter of the unit?
discrete-mathematics proof-verification
discrete-mathematics proof-verification
asked 1 hour ago
ShinobuIsMyWifeShinobuIsMyWife
313
asked 1 hour ago
ShinobuIsMyWifeShinobuIsMyWife
313
asked 1 hour ago
ShinobuIsMyWifeShinobuIsMyWife
313
asked 1 hour ago
ShinobuIsMyWifeShinobuIsMyWife
313
asked 1 hour ago
ShinobuIsMyWifeShinobuIsMyWife
313
313
5
$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
1 hour ago
2
$begingroup$
Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
$endgroup$
– Dean Young
1 hour ago
add a comment |
5
$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
1 hour ago
2
$begingroup$
Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
$endgroup$
– Dean Young
1 hour ago
5
5
$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
1 hour ago
$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
1 hour ago
2
2
$begingroup$
Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
$endgroup$
– Dean Young
1 hour ago
$begingroup$
Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
$endgroup$
– Dean Young
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.
answered 1 hour ago
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
You can clearly see the fallacy if you keep track of the units:
In the second equality, $$0.01 = $0.1times $0.1$ is not true, if you are doing units.
Even if the second equality were true, the third one gives problems: since $c=$/100$, you have
$$
($0.1)^2=left(frac c{100},0.1right)^2=frac{c^2}{100}timesfrac1{10}=frac{c^2}{1000}.
$$
This is not $(10c)^2=100c^2$.
In conclusion, two equalities are bogus, and so is the argument.
edited 1 hour ago
J. W. Tanner
3,0981320
answered 1 hour ago
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.
answered 1 hour ago
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
$$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.
answered 1 hour ago
ArthurArthur
117k7116200
$endgroup$
add a comment |
$begingroup$
$$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.
answered 1 hour ago
ArthurArthur
117k7116200
$endgroup$
$$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.
answered 1 hour ago
ArthurArthur
117k7116200
answered 1 hour ago
ArthurArthur
117k7116200
answered 1 hour ago
ArthurArthur
117k7116200
answered 1 hour ago
ArthurArthur
117k7116200
117k7116200
add a comment |
add a comment |
$begingroup$
You can clearly see the fallacy if you keep track of the units:
In the second equality, $$0.01 = $0.1times $0.1$ is not true, if you are doing units.
Even if the second equality were true, the third one gives problems: since $c=$/100$, you have
$$
($0.1)^2=left(frac c{100},0.1right)^2=frac{c^2}{100}timesfrac1{10}=frac{c^2}{1000}.
$$
This is not $(10c)^2=100c^2$.
In conclusion, two equalities are bogus, and so is the argument.
edited 1 hour ago
J. W. Tanner
3,0981320
answered 1 hour ago
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
$begingroup$
You can clearly see the fallacy if you keep track of the units:
In the second equality, $$0.01 = $0.1times $0.1$ is not true, if you are doing units.
Even if the second equality were true, the third one gives problems: since $c=$/100$, you have
$$
($0.1)^2=left(frac c{100},0.1right)^2=frac{c^2}{100}timesfrac1{10}=frac{c^2}{1000}.
$$
This is not $(10c)^2=100c^2$.
In conclusion, two equalities are bogus, and so is the argument.
edited 1 hour ago
J. W. Tanner
3,0981320
answered 1 hour ago
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
$begingroup$
You can clearly see the fallacy if you keep track of the units:
In the second equality, $$0.01 = $0.1times $0.1$ is not true, if you are doing units.
Even if the second equality were true, the third one gives problems: since $c=$/100$, you have
$$
($0.1)^2=left(frac c{100},0.1right)^2=frac{c^2}{100}timesfrac1{10}=frac{c^2}{1000}.
$$
This is not $(10c)^2=100c^2$.
In conclusion, two equalities are bogus, and so is the argument.
edited 1 hour ago
J. W. Tanner
3,0981320
answered 1 hour ago
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
You can clearly see the fallacy if you keep track of the units:
In the second equality, $$0.01 = $0.1times $0.1$ is not true, if you are doing units.
Even if the second equality were true, the third one gives problems: since $c=$/100$, you have
$$
($0.1)^2=left(frac c{100},0.1right)^2=frac{c^2}{100}timesfrac1{10}=frac{c^2}{1000}.
$$
This is not $(10c)^2=100c^2$.
In conclusion, two equalities are bogus, and so is the argument.
edited 1 hour ago
J. W. Tanner
3,0981320
answered 1 hour ago
Martin ArgeramiMartin Argerami
128k1184184
edited 1 hour ago
J. W. Tanner
3,0981320
edited 1 hour ago
J. W. Tanner
3,0981320
edited 1 hour ago
J. W. Tanner
3,0981320
3,0981320
answered 1 hour ago
Martin ArgeramiMartin Argerami
128k1184184
answered 1 hour ago
Martin ArgeramiMartin Argerami
128k1184184
answered 1 hour ago
Martin ArgeramiMartin Argerami
128k1184184
128k1184184
add a comment |
add a comment |
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5
$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
1 hour ago
2
$begingroup$
Yes, you have to square the unit. Conversion of squared units is different: if 100 cents is a dollar, then $100^2$ cents squared is a dollar squred.
$endgroup$
– Dean Young
1 hour ago