What's wrong with this bogus proof?
$begingroup$
What is the mistake here? Is it matter of the unit?
discrete-mathematics proof-verification
asked Mar 12 at 0:18
王文军 or Wenjun Wang王文军 or Wenjun Wang
514
$endgroup$
add a comment |
$begingroup$
What is the mistake here? Is it matter of the unit?
discrete-mathematics proof-verification
asked Mar 12 at 0:18
王文军 or Wenjun Wang王文军 or Wenjun Wang
514
$endgroup$
5
$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
Mar 12 at 0:20
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
Mar 12 at 0:20
add a comment |
$begingroup$
What is the mistake here? Is it matter of the unit?
discrete-mathematics proof-verification
asked Mar 12 at 0:18
王文军 or Wenjun Wang王文军 or Wenjun Wang
514
$endgroup$
What is the mistake here? Is it matter of the unit?
discrete-mathematics proof-verification
discrete-mathematics proof-verification
asked Mar 12 at 0:18
王文军 or Wenjun Wang王文军 or Wenjun Wang
514
asked Mar 12 at 0:18
王文军 or Wenjun Wang王文军 or Wenjun Wang
514
asked Mar 12 at 0:18
王文军 or Wenjun Wang王文军 or Wenjun Wang
514
asked Mar 12 at 0:18
王文军 or Wenjun Wang王文军 or Wenjun Wang
514
asked Mar 12 at 0:18
王文军 or Wenjun Wang王文军 or Wenjun Wang
514
514
5
$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
Mar 12 at 0:20
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
Mar 12 at 0:20
add a comment |
5
$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
Mar 12 at 0:20
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
Mar 12 at 0:20
5
5
$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
Mar 12 at 0:20
$begingroup$
Yes, the units don’t match across the 2nd equals sign
$endgroup$
– Alex
Mar 12 at 0:20
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
Mar 12 at 0:20
$begingroup$
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
Mar 12 at 0:20
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
$$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.
answered Mar 12 at 0:24
ArthurArthur
119k7118202
$endgroup$
add a comment |
$begingroup$
You can clearly see the fallacy if you keep track of the units:
In the second equality, $$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 Mar 12 at 0:43
J. W. Tanner
3,4751320
answered Mar 12 at 0:25
Martin ArgeramiMartin Argerami
128k1184184
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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$begingroup$
$$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.
answered Mar 12 at 0:24
ArthurArthur
119k7118202
$endgroup$
add a comment |
$begingroup$
$$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.
answered Mar 12 at 0:24
ArthurArthur
119k7118202
$endgroup$
add a comment |
$begingroup$
$$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.
answered Mar 12 at 0:24
ArthurArthur
119k7118202
$endgroup$
$$0.01=(sqrt{$}0.1)^2$, not $($0.1)^2$.
answered Mar 12 at 0:24
ArthurArthur
119k7118202
answered Mar 12 at 0:24
ArthurArthur
119k7118202
answered Mar 12 at 0:24
ArthurArthur
119k7118202
answered Mar 12 at 0:24
ArthurArthur
119k7118202
119k7118202
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}.
$$
This is not $(10c)^2=100c^2$.
In conclusion, two equalities are bogus, and so is the argument.
edited Mar 12 at 0:43
J. W. Tanner
3,4751320
answered Mar 12 at 0:25
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
add a comment |
$begingroup$
You can clearly see the fallacy if you keep track of the units:
In the second equality, $$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 Mar 12 at 0:43
J. W. Tanner
3,4751320
answered Mar 12 at 0:25
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}.
$$
This is not $(10c)^2=100c^2$.
In conclusion, two equalities are bogus, and so is the argument.
edited Mar 12 at 0:43
J. W. Tanner
3,4751320
answered Mar 12 at 0:25
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
You can clearly see the fallacy if you keep track of the units:
In the second equality, $$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 Mar 12 at 0:43
J. W. Tanner
3,4751320
answered Mar 12 at 0:25
Martin ArgeramiMartin Argerami
128k1184184
edited Mar 12 at 0:43
J. W. Tanner
3,4751320
edited Mar 12 at 0:43
J. W. Tanner
3,4751320
edited Mar 12 at 0:43
J. W. Tanner
3,4751320
3,4751320
answered Mar 12 at 0:25
Martin ArgeramiMartin Argerami
128k1184184
answered Mar 12 at 0:25
Martin ArgeramiMartin Argerami
128k1184184
answered Mar 12 at 0:25
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
Mar 12 at 0:20
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
Mar 12 at 0:20