if x is not equal to y and x, y are real numbers and A=
, then
A<0
0<A<1
A>0
A=0
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Page 5. Problem 8. Prove that if x and y are real numbers, then
Proof. First we prove that if x is a real number, then x2 ≥ 0. The product of two positive numbers is always positive, i.e., if x ≥ 0 and y ≥ 0, then xy≥0. Inparticularifx≥0thenx2 =x·x≥0. Ifxisnegative,then−x is positive, hence (−x)2 ≥ 0. But we can conduct the following computation by the associativity and the commutativity of the product of real numbers:
2xy ≤ x + y .
0 ≥ (−x)2 = (−x)(−x) = ((−1)x)((−1)x) = (((−1)x))(−1))x 2
= (((−1)(x(−1)))x = (((−1)(−1))x)x = (1x)x = xx = x .
The above change in bracketting can be done in many ways. At any rate, this shows that the square of any real number is non-negaitive. Now if x and y are real numbers, then so is the difference, x − y which is defined to be x + (−y). Therefore we conclude that 0 ≤ (x + (−y))2 and compute:
0≤(x+(−y))2 =(x+(−y))(x+(−y))=x(x+(−y))+(−y)(x+(−y)) =x2 +x(−y)+(−y)x+(−y)2 =x2 +y2 +(−xy)+(−xy)
= x2 + y2 + 2(−xy);
adding 2xy to the both sides,
2xy=0+2xy≤(x2 +y2 +2(−xy))+2xy=(x2 +y2)+(2(−xy)+2xy)
2xy ≤ x2 + y2
for every pair of real numbers x and y. ♥ 1
22 22
=(x +y )+0=x +y . Therefore, we conclude the inequality: Page 5. Problem 11. If a and b are real numbers with a
Proof. First we prove that if x is a real number, then x2 ≥ 0. The product of two positive numbers is always positive, i.e., if x ≥ 0 and y ≥ 0, then xy≥0. Inparticularifx≥0thenx2 =x·x≥0. Ifxisnegative,then−x is positive, hence (−x)2 ≥ 0. But we can conduct the following computation by the associativity and the commutativity of the product of real numbers:
2xy ≤ x + y .
0 ≥ (−x)2 = (−x)(−x) = ((−1)x)((−1)x) = (((−1)x))(−1))x 2
= (((−1)(x(−1)))x = (((−1)(−1))x)x = (1x)x = xx = x .
The above change in bracketting can be done in many ways. At any rate, this shows that the square of any real number is non-negaitive. Now if x and y are real numbers, then so is the difference, x − y which is defined to be x + (−y). Therefore we conclude that 0 ≤ (x + (−y))2 and compute:
0≤(x+(−y))2 =(x+(−y))(x+(−y))=x(x+(−y))+(−y)(x+(−y)) =x2 +x(−y)+(−y)x+(−y)2 =x2 +y2 +(−xy)+(−xy)
= x2 + y2 + 2(−xy);
adding 2xy to the both sides,
2xy=0+2xy≤(x2 +y2 +2(−xy))+2xy=(x2 +y2)+(2(−xy)+2xy)
2xy ≤ x2 + y2
for every pair of real numbers x and y. ♥ 1
22 22
=(x +y )+0=x +y . Therefore, we conclude the inequality: Page 5. Problem 11. If a and b are real numbers with a
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