Proof of $ x^2 + y = y^2 + x$ when $ x+ y =1$ and x is larger than y
$begingroup$
I know that whatever numbers you choose for x and y and their sum equals to 1 will satisfy the equation $x^2 + y = y^2 + x$
Algebraic proof:
Given: $x + y = 1$
$$LS = x^2+ y
= (1-y)^2 + y
= 1 - 2y+y^2 + y
= y^2 - y + 1$$
$$RS = y^2 + x
= y^2 + (1-y)
= y^2 - y + 1$$
Therefore,$$ LS = RS $$
How can this be proved geometrically? (Ex. in a diagram of rectangular areas)
I tried to add a square piece with side lengths y with a rectangle with side lengths x and x+y but I can't seem to prove it geometrically.
Can someone help?
proof-verification conjectures proof-theory
asked Jan 11 at 15:44
WadeWade
188211
$endgroup$
add a comment |
$begingroup$
I know that whatever numbers you choose for x and y and their sum equals to 1 will satisfy the equation $x^2 + y = y^2 + x$
Algebraic proof:
Given: $x + y = 1$
$$LS = x^2+ y
= (1-y)^2 + y
= 1 - 2y+y^2 + y
= y^2 - y + 1$$
$$RS = y^2 + x
= y^2 + (1-y)
= y^2 - y + 1$$
Therefore,$$ LS = RS $$
How can this be proved geometrically? (Ex. in a diagram of rectangular areas)
I tried to add a square piece with side lengths y with a rectangle with side lengths x and x+y but I can't seem to prove it geometrically.
Can someone help?
proof-verification conjectures proof-theory
asked Jan 11 at 15:44
WadeWade
188211
$endgroup$
3
$begingroup$
Hint: $x^2-y^2=(x-y)(x+y)$
$endgroup$
– robjohn♦
Jan 11 at 15:48
add a comment |
$begingroup$
I know that whatever numbers you choose for x and y and their sum equals to 1 will satisfy the equation $x^2 + y = y^2 + x$
Algebraic proof:
Given: $x + y = 1$
$$LS = x^2+ y
= (1-y)^2 + y
= 1 - 2y+y^2 + y
= y^2 - y + 1$$
$$RS = y^2 + x
= y^2 + (1-y)
= y^2 - y + 1$$
Therefore,$$ LS = RS $$
How can this be proved geometrically? (Ex. in a diagram of rectangular areas)
I tried to add a square piece with side lengths y with a rectangle with side lengths x and x+y but I can't seem to prove it geometrically.
Can someone help?
proof-verification conjectures proof-theory
asked Jan 11 at 15:44
WadeWade
188211
$endgroup$
I know that whatever numbers you choose for x and y and their sum equals to 1 will satisfy the equation $x^2 + y = y^2 + x$
Algebraic proof:
Given: $x + y = 1$
$$LS = x^2+ y
= (1-y)^2 + y
= 1 - 2y+y^2 + y
= y^2 - y + 1$$
$$RS = y^2 + x
= y^2 + (1-y)
= y^2 - y + 1$$
Therefore,$$ LS = RS $$
How can this be proved geometrically? (Ex. in a diagram of rectangular areas)
I tried to add a square piece with side lengths y with a rectangle with side lengths x and x+y but I can't seem to prove it geometrically.
Can someone help?
proof-verification conjectures proof-theory
proof-verification conjectures proof-theory
asked Jan 11 at 15:44
WadeWade
188211
asked Jan 11 at 15:44
WadeWade
188211
edited Jan 11 at 15:53
Wade
asked Jan 11 at 15:44
WadeWade
188211
asked Jan 11 at 15:44
WadeWade
188211
asked Jan 11 at 15:44
WadeWade
188211
188211
3
$begingroup$
Hint: $x^2-y^2=(x-y)(x+y)$
$endgroup$
– robjohn♦
Jan 11 at 15:48
add a comment |
3
$begingroup$
Hint: $x^2-y^2=(x-y)(x+y)$
$endgroup$
– robjohn♦
Jan 11 at 15:48
3
3
$begingroup$
Hint: $x^2-y^2=(x-y)(x+y)$
$endgroup$
– robjohn♦
Jan 11 at 15:48
$begingroup$
Hint: $x^2-y^2=(x-y)(x+y)$
$endgroup$
– robjohn♦
Jan 11 at 15:48
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here is a picture. The left shows $y^2+x$, the right $x^2+y$.
answered Jan 11 at 16:04
Jaap ScherphuisJaap Scherphuis
4,084617
$endgroup$
add a comment |
$begingroup$
Sorry for the description of the lacking shapes...
You have to consider for the LHS a square of side $x$ and a rectangle of sides $y$ and $x+y=1$.
This can be decomposed into the "big" square $x^2$, the "little" square $y^2$ and the remaining rectangle $xy$.
For the RHS, a square of side $y$ and a rectangle of sides $x$ and $x+y=1$.
In turn, this can be decomposed into the "little" square $y^2$, the "big" square $x^2$ and the remaining rectangle $yx$.
Then rotate $yx$.
answered Jan 11 at 15:51
Mauro ALLEGRANZAMauro ALLEGRANZA
65.2k448112
$endgroup$
add a comment |
$begingroup$
Because $$0=x^2+y-(y^2+x)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1).$$
answered Jan 11 at 16:35
Michael RozenbergMichael Rozenberg
99.1k1590189
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a picture. The left shows $y^2+x$, the right $x^2+y$.
answered Jan 11 at 16:04
Jaap ScherphuisJaap Scherphuis
4,084617
$endgroup$
add a comment |
$begingroup$
Here is a picture. The left shows $y^2+x$, the right $x^2+y$.
answered Jan 11 at 16:04
Jaap ScherphuisJaap Scherphuis
4,084617
$endgroup$
add a comment |
$begingroup$
Here is a picture. The left shows $y^2+x$, the right $x^2+y$.
answered Jan 11 at 16:04
Jaap ScherphuisJaap Scherphuis
4,084617
$endgroup$
Here is a picture. The left shows $y^2+x$, the right $x^2+y$.
answered Jan 11 at 16:04
Jaap ScherphuisJaap Scherphuis
4,084617
answered Jan 11 at 16:04
Jaap ScherphuisJaap Scherphuis
4,084617
answered Jan 11 at 16:04
Jaap ScherphuisJaap Scherphuis
4,084617
answered Jan 11 at 16:04
Jaap ScherphuisJaap Scherphuis
4,084617
4,084617
add a comment |
add a comment |
$begingroup$
Sorry for the description of the lacking shapes...
You have to consider for the LHS a square of side $x$ and a rectangle of sides $y$ and $x+y=1$.
This can be decomposed into the "big" square $x^2$, the "little" square $y^2$ and the remaining rectangle $xy$.
For the RHS, a square of side $y$ and a rectangle of sides $x$ and $x+y=1$.
In turn, this can be decomposed into the "little" square $y^2$, the "big" square $x^2$ and the remaining rectangle $yx$.
Then rotate $yx$.
answered Jan 11 at 15:51
Mauro ALLEGRANZAMauro ALLEGRANZA
65.2k448112
$endgroup$
add a comment |
$begingroup$
Sorry for the description of the lacking shapes...
You have to consider for the LHS a square of side $x$ and a rectangle of sides $y$ and $x+y=1$.
This can be decomposed into the "big" square $x^2$, the "little" square $y^2$ and the remaining rectangle $xy$.
For the RHS, a square of side $y$ and a rectangle of sides $x$ and $x+y=1$.
In turn, this can be decomposed into the "little" square $y^2$, the "big" square $x^2$ and the remaining rectangle $yx$.
Then rotate $yx$.
answered Jan 11 at 15:51
Mauro ALLEGRANZAMauro ALLEGRANZA
65.2k448112
$endgroup$
add a comment |
$begingroup$
Sorry for the description of the lacking shapes...
You have to consider for the LHS a square of side $x$ and a rectangle of sides $y$ and $x+y=1$.
This can be decomposed into the "big" square $x^2$, the "little" square $y^2$ and the remaining rectangle $xy$.
For the RHS, a square of side $y$ and a rectangle of sides $x$ and $x+y=1$.
In turn, this can be decomposed into the "little" square $y^2$, the "big" square $x^2$ and the remaining rectangle $yx$.
Then rotate $yx$.
answered Jan 11 at 15:51
Mauro ALLEGRANZAMauro ALLEGRANZA
65.2k448112
$endgroup$
Sorry for the description of the lacking shapes...
You have to consider for the LHS a square of side $x$ and a rectangle of sides $y$ and $x+y=1$.
This can be decomposed into the "big" square $x^2$, the "little" square $y^2$ and the remaining rectangle $xy$.
For the RHS, a square of side $y$ and a rectangle of sides $x$ and $x+y=1$.
In turn, this can be decomposed into the "little" square $y^2$, the "big" square $x^2$ and the remaining rectangle $yx$.
Then rotate $yx$.
answered Jan 11 at 15:51
Mauro ALLEGRANZAMauro ALLEGRANZA
65.2k448112
answered Jan 11 at 15:51
Mauro ALLEGRANZAMauro ALLEGRANZA
65.2k448112
answered Jan 11 at 15:51
Mauro ALLEGRANZAMauro ALLEGRANZA
65.2k448112
answered Jan 11 at 15:51
Mauro ALLEGRANZAMauro ALLEGRANZA
65.2k448112
65.2k448112
add a comment |
add a comment |
$begingroup$
Because $$0=x^2+y-(y^2+x)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1).$$
answered Jan 11 at 16:35
Michael RozenbergMichael Rozenberg
99.1k1590189
$endgroup$
add a comment |
$begingroup$
Because $$0=x^2+y-(y^2+x)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1).$$
answered Jan 11 at 16:35
Michael RozenbergMichael Rozenberg
99.1k1590189
$endgroup$
add a comment |
$begingroup$
Because $$0=x^2+y-(y^2+x)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1).$$
answered Jan 11 at 16:35
Michael RozenbergMichael Rozenberg
99.1k1590189
$endgroup$
Because $$0=x^2+y-(y^2+x)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1).$$
answered Jan 11 at 16:35
Michael RozenbergMichael Rozenberg
99.1k1590189
answered Jan 11 at 16:35
Michael RozenbergMichael Rozenberg
99.1k1590189
answered Jan 11 at 16:35
Michael RozenbergMichael Rozenberg
99.1k1590189
answered Jan 11 at 16:35
Michael RozenbergMichael Rozenberg
99.1k1590189
99.1k1590189
add a comment |
add a comment |
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3
$begingroup$
Hint: $x^2-y^2=(x-y)(x+y)$
$endgroup$
– robjohn♦
Jan 11 at 15:48