Proof verification: Prove that the product of $2$ real negative numbers is positive.
$begingroup$
Can someone please verify any mistakes in my proof of this theorem. Is this a correct proof of the corollary of $(-1)times (-1) = 1$ that the product of two negative numbers is positive?
Theorem: The product of two real negative numbers is positive
Proof: Let $x,yinmathbb{R^+}$. We have
begin{equation}
begin{alignedat}{2}
(-1) times0=0quad
Rightarrowquad &&
(-1) times (-xy+xy) &= 0
\
Rightarrowquad &&
(-1)times((-1)times xy+xy)&= 0
\
Rightarrowquad &&
(-1)times(-1)times xy+(-1)times xy &= 0 \ Rightarrowquad && (-1)times x times (-1) times y +(-1)times xy&=0 \
Rightarrowquad &&(-x)times (-y)- xy&=0 \
Rightarrowquad &&(-x)times (-y)&=xy.
end{alignedat}
end{equation}
proof-verification proof-writing
edited Jan 8 at 20:44
greedoid
38.7k114797
asked Jan 8 at 19:31
ImranImran
30715
$endgroup$
add a comment |
$begingroup$
Can someone please verify any mistakes in my proof of this theorem. Is this a correct proof of the corollary of $(-1)times (-1) = 1$ that the product of two negative numbers is positive?
Theorem: The product of two real negative numbers is positive
Proof: Let $x,yinmathbb{R^+}$. We have
begin{equation}
begin{alignedat}{2}
(-1) times0=0quad
Rightarrowquad &&
(-1) times (-xy+xy) &= 0
\
Rightarrowquad &&
(-1)times((-1)times xy+xy)&= 0
\
Rightarrowquad &&
(-1)times(-1)times xy+(-1)times xy &= 0 \ Rightarrowquad && (-1)times x times (-1) times y +(-1)times xy&=0 \
Rightarrowquad &&(-x)times (-y)- xy&=0 \
Rightarrowquad &&(-x)times (-y)&=xy.
end{alignedat}
end{equation}
proof-verification proof-writing
edited Jan 8 at 20:44
greedoid
38.7k114797
asked Jan 8 at 19:31
ImranImran
30715
$endgroup$
add a comment |
$begingroup$
Can someone please verify any mistakes in my proof of this theorem. Is this a correct proof of the corollary of $(-1)times (-1) = 1$ that the product of two negative numbers is positive?
Theorem: The product of two real negative numbers is positive
Proof: Let $x,yinmathbb{R^+}$. We have
begin{equation}
begin{alignedat}{2}
(-1) times0=0quad
Rightarrowquad &&
(-1) times (-xy+xy) &= 0
\
Rightarrowquad &&
(-1)times((-1)times xy+xy)&= 0
\
Rightarrowquad &&
(-1)times(-1)times xy+(-1)times xy &= 0 \ Rightarrowquad && (-1)times x times (-1) times y +(-1)times xy&=0 \
Rightarrowquad &&(-x)times (-y)- xy&=0 \
Rightarrowquad &&(-x)times (-y)&=xy.
end{alignedat}
end{equation}
proof-verification proof-writing
edited Jan 8 at 20:44
greedoid
38.7k114797
asked Jan 8 at 19:31
ImranImran
30715
$endgroup$
Can someone please verify any mistakes in my proof of this theorem. Is this a correct proof of the corollary of $(-1)times (-1) = 1$ that the product of two negative numbers is positive?
Theorem: The product of two real negative numbers is positive
Proof: Let $x,yinmathbb{R^+}$. We have
begin{equation}
begin{alignedat}{2}
(-1) times0=0quad
Rightarrowquad &&
(-1) times (-xy+xy) &= 0
\
Rightarrowquad &&
(-1)times((-1)times xy+xy)&= 0
\
Rightarrowquad &&
(-1)times(-1)times xy+(-1)times xy &= 0 \ Rightarrowquad && (-1)times x times (-1) times y +(-1)times xy&=0 \
Rightarrowquad &&(-x)times (-y)- xy&=0 \
Rightarrowquad &&(-x)times (-y)&=xy.
end{alignedat}
end{equation}
proof-verification proof-writing
proof-verification proof-writing
edited Jan 8 at 20:44
greedoid
38.7k114797
asked Jan 8 at 19:31
ImranImran
30715
edited Jan 8 at 20:44
greedoid
38.7k114797
asked Jan 8 at 19:31
ImranImran
30715
edited Jan 8 at 20:44
greedoid
38.7k114797
edited Jan 8 at 20:44
greedoid
38.7k114797
edited Jan 8 at 20:44
greedoid
38.7k114797
38.7k114797
asked Jan 8 at 19:31
ImranImran
30715
asked Jan 8 at 19:31
ImranImran
30715
asked Jan 8 at 19:31
ImranImran
30715
30715
add a comment |
add a comment |
1 Answer
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Yes it is correct, but perhaps this would be faster. Take any two negative number $x$ and $y$. Then $x=-a$ and $y=-b$ where $a,b$ are positive. Now we have:
begin{eqnarray} xy &=& (-a)(-b)\ &=& (-1)a(-1)b\
&=& (-1)(-1)ab \ &=& 1cdot ab \&=& ab>0end{eqnarray}
answered Jan 8 at 19:36
greedoidgreedoid
38.7k114797
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add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Yes it is correct, but perhaps this would be faster. Take any two negative number $x$ and $y$. Then $x=-a$ and $y=-b$ where $a,b$ are positive. Now we have:
begin{eqnarray} xy &=& (-a)(-b)\ &=& (-1)a(-1)b\
&=& (-1)(-1)ab \ &=& 1cdot ab \&=& ab>0end{eqnarray}
answered Jan 8 at 19:36
greedoidgreedoid
38.7k114797
$endgroup$
add a comment |
$begingroup$
Yes it is correct, but perhaps this would be faster. Take any two negative number $x$ and $y$. Then $x=-a$ and $y=-b$ where $a,b$ are positive. Now we have:
begin{eqnarray} xy &=& (-a)(-b)\ &=& (-1)a(-1)b\
&=& (-1)(-1)ab \ &=& 1cdot ab \&=& ab>0end{eqnarray}
answered Jan 8 at 19:36
greedoidgreedoid
38.7k114797
$endgroup$
add a comment |
$begingroup$
Yes it is correct, but perhaps this would be faster. Take any two negative number $x$ and $y$. Then $x=-a$ and $y=-b$ where $a,b$ are positive. Now we have:
begin{eqnarray} xy &=& (-a)(-b)\ &=& (-1)a(-1)b\
&=& (-1)(-1)ab \ &=& 1cdot ab \&=& ab>0end{eqnarray}
answered Jan 8 at 19:36
greedoidgreedoid
38.7k114797
$endgroup$
Yes it is correct, but perhaps this would be faster. Take any two negative number $x$ and $y$. Then $x=-a$ and $y=-b$ where $a,b$ are positive. Now we have:
begin{eqnarray} xy &=& (-a)(-b)\ &=& (-1)a(-1)b\
&=& (-1)(-1)ab \ &=& 1cdot ab \&=& ab>0end{eqnarray}
answered Jan 8 at 19:36
greedoidgreedoid
38.7k114797
answered Jan 8 at 19:36
greedoidgreedoid
38.7k114797
answered Jan 8 at 19:36
greedoidgreedoid
38.7k114797
answered Jan 8 at 19:36
greedoidgreedoid
38.7k114797
38.7k114797
add a comment |
add a comment |
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