Is my proof that the square of any real number is non-negative correct?
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Lemma: Let $ k $ be a real number. Then $ k^2 geq 0 $.
Proof: I will do this by considering cases and using the following axioms:
Axiom 1: Every $ xinmathbbR $ has a negative $ -x $ such that $ x + (-x) = 0 $.
Axiom 2: The real numbers are closed under addition (and hence subtraction).
Axiom 3: Every non-integer lies between two consecutive integers.
Axiom 4: Multiplication of two real numbers is commutative.
Axiom 5: Addition of two non-negative numbers yields a non-negative number.
Axiom 6: A line having positive gradient implies it is increasing from left to right.
We will now begin the proof.
Case 1: $ k = 0 $
Since $ 0^2 = 0 $, we have that $ k^2 = 0 geq 0 $. Therefore true for $ k = 0 $.
Case 2: $ k neq 0 $
Fix $ k neq 0 $. By Axiom 1, $ exists-kinmathbbR $ such that
$ k + (-k) = 0 $
Multiplying both sides by $ k $:
$ k^2 + k(-k) = 0 $ $ (1) $
Multiplying both sides by $ -k $:
$ (-k)k + (-k)^2 = 0 $ $ (2) $
Equating $ (1) $ and $ (2) $:
$ k^2 + k(-k) = (-k)k + (-k)^2 $
But $ k(-k) = (-k)k $ by Axiom 4, so by cancellation we are left with $ k^2 = (-k)^2 $. Therefore, it suffices to show that the square of any positive real number is non-negative.
Fix $ k > 0 $. If $ kinmathbbN $, then, by definition, $ k^2 = ktimesk = underbracek+k+dots+k_k textit times $. But, using Axiom 5,
$ k > 0 $
$ Longrightarrow k+k > k > 0 $
$ Longrightarrow k+k+k > k+k > k > 0 $
.
.
.
$ Longrightarrow $ $ underbracek+k+dots+k_k text times $ $ > ... > k > 0 geq 0 $.
$ therefore $ $ k^2 geq 0 $.
If $ knotinmathbbN $, then by Axiom 3 there exists two integers $ [k] $ and $ [k] + 1 $ such that $ [k] < k < [k]+1 $. Note that $ [k] geq 0 $.
Now consider the points $ ([k], [k]^2) $ and $ (k, k^2) $ in the plane. Let $ m $ denote the gradient of the line segment joining these points. Then
$ m = frack^2-[k]^2k-[k] = frac(k-[k])(k+[k])k-[k] = k+[k] geq k > 0 $.
Hence, the line joining these two points has positive gradient. So by Axiom 6 the $ y $-values of this line are increasing as $ x $ increases. $ therefore $ $ k^2 > [k]^2 $ since $ k > [k] $. But $ [k]^2 = underbrace[k]+[k]+dots+[k]_[k] text times geq 0 $ by repeated application of Axiom 5.
$ therefore k^2 geq 0 $ for $ knotinmathbbN $ as well as for $ kinmathbbN $, which completes the proof. $ square $
proof-verification inequality
asked Aug 14 at 5:56
A.Abbas
266
add a comment |Â
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1
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Lemma: Let $ k $ be a real number. Then $ k^2 geq 0 $.
Proof: I will do this by considering cases and using the following axioms:
Axiom 1: Every $ xinmathbbR $ has a negative $ -x $ such that $ x + (-x) = 0 $.
Axiom 2: The real numbers are closed under addition (and hence subtraction).
Axiom 3: Every non-integer lies between two consecutive integers.
Axiom 4: Multiplication of two real numbers is commutative.
Axiom 5: Addition of two non-negative numbers yields a non-negative number.
Axiom 6: A line having positive gradient implies it is increasing from left to right.
We will now begin the proof.
Case 1: $ k = 0 $
Since $ 0^2 = 0 $, we have that $ k^2 = 0 geq 0 $. Therefore true for $ k = 0 $.
Case 2: $ k neq 0 $
Fix $ k neq 0 $. By Axiom 1, $ exists-kinmathbbR $ such that
$ k + (-k) = 0 $
Multiplying both sides by $ k $:
$ k^2 + k(-k) = 0 $ $ (1) $
Multiplying both sides by $ -k $:
$ (-k)k + (-k)^2 = 0 $ $ (2) $
Equating $ (1) $ and $ (2) $:
$ k^2 + k(-k) = (-k)k + (-k)^2 $
But $ k(-k) = (-k)k $ by Axiom 4, so by cancellation we are left with $ k^2 = (-k)^2 $. Therefore, it suffices to show that the square of any positive real number is non-negative.
Fix $ k > 0 $. If $ kinmathbbN $, then, by definition, $ k^2 = ktimesk = underbracek+k+dots+k_k textit times $. But, using Axiom 5,
$ k > 0 $
$ Longrightarrow k+k > k > 0 $
$ Longrightarrow k+k+k > k+k > k > 0 $
.
.
.
$ Longrightarrow $ $ underbracek+k+dots+k_k text times $ $ > ... > k > 0 geq 0 $.
$ therefore $ $ k^2 geq 0 $.
If $ knotinmathbbN $, then by Axiom 3 there exists two integers $ [k] $ and $ [k] + 1 $ such that $ [k] < k < [k]+1 $. Note that $ [k] geq 0 $.
Now consider the points $ ([k], [k]^2) $ and $ (k, k^2) $ in the plane. Let $ m $ denote the gradient of the line segment joining these points. Then
$ m = frack^2-[k]^2k-[k] = frac(k-[k])(k+[k])k-[k] = k+[k] geq k > 0 $.
Hence, the line joining these two points has positive gradient. So by Axiom 6 the $ y $-values of this line are increasing as $ x $ increases. $ therefore $ $ k^2 > [k]^2 $ since $ k > [k] $. But $ [k]^2 = underbrace[k]+[k]+dots+[k]_[k] text times geq 0 $ by repeated application of Axiom 5.
$ therefore k^2 geq 0 $ for $ knotinmathbbN $ as well as for $ kinmathbbN $, which completes the proof. $ square $
proof-verification inequality
asked Aug 14 at 5:56
A.Abbas
266
7
You cannot define multiplication as repeated addition for $k$ that isn't a non-negative integer. How do you add something $2.3$ times, or $pi$ times? (ETA: Sorry, I see how you dealt with negative numbers. But you still have to deal with positive non-integers.)
– Brian Tung
Aug 14 at 5:58
2
If you know (or can prove) that $,1 gt 0,$ and $,(-1)cdot(-1)=1,$ then the rest follows.
– dxiv
Aug 14 at 6:07
2
You also have to mention what axioms of the real numbers you are allowing answerers to use without proof. For example, is $1 > 0$ obvious, or is it to be proved using some definition of order on the real numbers that you have?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 14 at 6:33
1
Prove it for $k<0$, then $k=0$, then $k>0$ and you're done.
– David G. Stork
Aug 14 at 6:39
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Lemma: Let $ k $ be a real number. Then $ k^2 geq 0 $.
Proof: I will do this by considering cases and using the following axioms:
Axiom 1: Every $ xinmathbbR $ has a negative $ -x $ such that $ x + (-x) = 0 $.
Axiom 2: The real numbers are closed under addition (and hence subtraction).
Axiom 3: Every non-integer lies between two consecutive integers.
Axiom 4: Multiplication of two real numbers is commutative.
Axiom 5: Addition of two non-negative numbers yields a non-negative number.
Axiom 6: A line having positive gradient implies it is increasing from left to right.
We will now begin the proof.
Case 1: $ k = 0 $
Since $ 0^2 = 0 $, we have that $ k^2 = 0 geq 0 $. Therefore true for $ k = 0 $.
Case 2: $ k neq 0 $
Fix $ k neq 0 $. By Axiom 1, $ exists-kinmathbbR $ such that
$ k + (-k) = 0 $
Multiplying both sides by $ k $:
$ k^2 + k(-k) = 0 $ $ (1) $
Multiplying both sides by $ -k $:
$ (-k)k + (-k)^2 = 0 $ $ (2) $
Equating $ (1) $ and $ (2) $:
$ k^2 + k(-k) = (-k)k + (-k)^2 $
But $ k(-k) = (-k)k $ by Axiom 4, so by cancellation we are left with $ k^2 = (-k)^2 $. Therefore, it suffices to show that the square of any positive real number is non-negative.
Fix $ k > 0 $. If $ kinmathbbN $, then, by definition, $ k^2 = ktimesk = underbracek+k+dots+k_k textit times $. But, using Axiom 5,
$ k > 0 $
$ Longrightarrow k+k > k > 0 $
$ Longrightarrow k+k+k > k+k > k > 0 $
.
.
.
$ Longrightarrow $ $ underbracek+k+dots+k_k text times $ $ > ... > k > 0 geq 0 $.
$ therefore $ $ k^2 geq 0 $.
If $ knotinmathbbN $, then by Axiom 3 there exists two integers $ [k] $ and $ [k] + 1 $ such that $ [k] < k < [k]+1 $. Note that $ [k] geq 0 $.
Now consider the points $ ([k], [k]^2) $ and $ (k, k^2) $ in the plane. Let $ m $ denote the gradient of the line segment joining these points. Then
$ m = frack^2-[k]^2k-[k] = frac(k-[k])(k+[k])k-[k] = k+[k] geq k > 0 $.
Hence, the line joining these two points has positive gradient. So by Axiom 6 the $ y $-values of this line are increasing as $ x $ increases. $ therefore $ $ k^2 > [k]^2 $ since $ k > [k] $. But $ [k]^2 = underbrace[k]+[k]+dots+[k]_[k] text times geq 0 $ by repeated application of Axiom 5.
$ therefore k^2 geq 0 $ for $ knotinmathbbN $ as well as for $ kinmathbbN $, which completes the proof. $ square $
proof-verification inequality
asked Aug 14 at 5:56
A.Abbas
266
Lemma: Let $ k $ be a real number. Then $ k^2 geq 0 $.
Proof: I will do this by considering cases and using the following axioms:
Axiom 1: Every $ xinmathbbR $ has a negative $ -x $ such that $ x + (-x) = 0 $.
Axiom 2: The real numbers are closed under addition (and hence subtraction).
Axiom 3: Every non-integer lies between two consecutive integers.
Axiom 4: Multiplication of two real numbers is commutative.
Axiom 5: Addition of two non-negative numbers yields a non-negative number.
Axiom 6: A line having positive gradient implies it is increasing from left to right.
We will now begin the proof.
Case 1: $ k = 0 $
Since $ 0^2 = 0 $, we have that $ k^2 = 0 geq 0 $. Therefore true for $ k = 0 $.
Case 2: $ k neq 0 $
Fix $ k neq 0 $. By Axiom 1, $ exists-kinmathbbR $ such that
$ k + (-k) = 0 $
Multiplying both sides by $ k $:
$ k^2 + k(-k) = 0 $ $ (1) $
Multiplying both sides by $ -k $:
$ (-k)k + (-k)^2 = 0 $ $ (2) $
Equating $ (1) $ and $ (2) $:
$ k^2 + k(-k) = (-k)k + (-k)^2 $
But $ k(-k) = (-k)k $ by Axiom 4, so by cancellation we are left with $ k^2 = (-k)^2 $. Therefore, it suffices to show that the square of any positive real number is non-negative.
Fix $ k > 0 $. If $ kinmathbbN $, then, by definition, $ k^2 = ktimesk = underbracek+k+dots+k_k textit times $. But, using Axiom 5,
$ k > 0 $
$ Longrightarrow k+k > k > 0 $
$ Longrightarrow k+k+k > k+k > k > 0 $
.
.
.
$ Longrightarrow $ $ underbracek+k+dots+k_k text times $ $ > ... > k > 0 geq 0 $.
$ therefore $ $ k^2 geq 0 $.
If $ knotinmathbbN $, then by Axiom 3 there exists two integers $ [k] $ and $ [k] + 1 $ such that $ [k] < k < [k]+1 $. Note that $ [k] geq 0 $.
Now consider the points $ ([k], [k]^2) $ and $ (k, k^2) $ in the plane. Let $ m $ denote the gradient of the line segment joining these points. Then
$ m = frack^2-[k]^2k-[k] = frac(k-[k])(k+[k])k-[k] = k+[k] geq k > 0 $.
Hence, the line joining these two points has positive gradient. So by Axiom 6 the $ y $-values of this line are increasing as $ x $ increases. $ therefore $ $ k^2 > [k]^2 $ since $ k > [k] $. But $ [k]^2 = underbrace[k]+[k]+dots+[k]_[k] text times geq 0 $ by repeated application of Axiom 5.
$ therefore k^2 geq 0 $ for $ knotinmathbbN $ as well as for $ kinmathbbN $, which completes the proof. $ square $
proof-verification inequality
asked Aug 14 at 5:56
A.Abbas
266
edited Aug 14 at 11:23
asked Aug 14 at 5:56
A.Abbas
266
asked Aug 14 at 5:56
A.Abbas
266
asked Aug 14 at 5:56
A.Abbas
266
266
7
You cannot define multiplication as repeated addition for $k$ that isn't a non-negative integer. How do you add something $2.3$ times, or $pi$ times? (ETA: Sorry, I see how you dealt with negative numbers. But you still have to deal with positive non-integers.)
– Brian Tung
Aug 14 at 5:58
2
If you know (or can prove) that $,1 gt 0,$ and $,(-1)cdot(-1)=1,$ then the rest follows.
– dxiv
Aug 14 at 6:07
2
You also have to mention what axioms of the real numbers you are allowing answerers to use without proof. For example, is $1 > 0$ obvious, or is it to be proved using some definition of order on the real numbers that you have?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 14 at 6:33
1
Prove it for $k<0$, then $k=0$, then $k>0$ and you're done.
– David G. Stork
Aug 14 at 6:39
add a comment |Â
7
You cannot define multiplication as repeated addition for $k$ that isn't a non-negative integer. How do you add something $2.3$ times, or $pi$ times? (ETA: Sorry, I see how you dealt with negative numbers. But you still have to deal with positive non-integers.)
– Brian Tung
Aug 14 at 5:58
2
If you know (or can prove) that $,1 gt 0,$ and $,(-1)cdot(-1)=1,$ then the rest follows.
– dxiv
Aug 14 at 6:07
2
You also have to mention what axioms of the real numbers you are allowing answerers to use without proof. For example, is $1 > 0$ obvious, or is it to be proved using some definition of order on the real numbers that you have?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 14 at 6:33
1
Prove it for $k<0$, then $k=0$, then $k>0$ and you're done.
– David G. Stork
Aug 14 at 6:39
7
7
You cannot define multiplication as repeated addition for $k$ that isn't a non-negative integer. How do you add something $2.3$ times, or $pi$ times? (ETA: Sorry, I see how you dealt with negative numbers. But you still have to deal with positive non-integers.)
– Brian Tung
Aug 14 at 5:58
You cannot define multiplication as repeated addition for $k$ that isn't a non-negative integer. How do you add something $2.3$ times, or $pi$ times? (ETA: Sorry, I see how you dealt with negative numbers. But you still have to deal with positive non-integers.)
– Brian Tung
Aug 14 at 5:58
2
2
If you know (or can prove) that $,1 gt 0,$ and $,(-1)cdot(-1)=1,$ then the rest follows.
– dxiv
Aug 14 at 6:07
If you know (or can prove) that $,1 gt 0,$ and $,(-1)cdot(-1)=1,$ then the rest follows.
– dxiv
Aug 14 at 6:07
2
2
You also have to mention what axioms of the real numbers you are allowing answerers to use without proof. For example, is $1 > 0$ obvious, or is it to be proved using some definition of order on the real numbers that you have?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 14 at 6:33
You also have to mention what axioms of the real numbers you are allowing answerers to use without proof. For example, is $1 > 0$ obvious, or is it to be proved using some definition of order on the real numbers that you have?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 14 at 6:33
1
1
Prove it for $k<0$, then $k=0$, then $k>0$ and you're done.
– David G. Stork
Aug 14 at 6:39
Prove it for $k<0$, then $k=0$, then $k>0$ and you're done.
– David G. Stork
Aug 14 at 6:39
add a comment |Â
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7
You cannot define multiplication as repeated addition for $k$ that isn't a non-negative integer. How do you add something $2.3$ times, or $pi$ times? (ETA: Sorry, I see how you dealt with negative numbers. But you still have to deal with positive non-integers.)
– Brian Tung
Aug 14 at 5:58
2
If you know (or can prove) that $,1 gt 0,$ and $,(-1)cdot(-1)=1,$ then the rest follows.
– dxiv
Aug 14 at 6:07
2
You also have to mention what axioms of the real numbers you are allowing answerers to use without proof. For example, is $1 > 0$ obvious, or is it to be proved using some definition of order on the real numbers that you have?
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Aug 14 at 6:33
1
Prove it for $k<0$, then $k=0$, then $k>0$ and you're done.
– David G. Stork
Aug 14 at 6:39