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130 The Nuts and Bolts of Proof, Third Edition
7. Assume that (xo,yo) is a solution of Si. Is it a solution of S2I By
definition of solution, (xo,)^o) satisfies both equations of the system Si.
So (xo,}^o) satisfies the first equation of S2. Thus, we only need to prove
that it satisfies the second equation of S2. By rearranging the terms we
can write {ax -h ^^2)^0 + (^i + bb2)y{) = (aixo + fci jo) + ^fe^o + biyo).
Because (xo,yo) is a solution of Si, a\XQ + biyo = ci, ^2X0 H- biyo = ci.
Thus, {a\ + ba'i)X(^ + (fti + bb2)y{s = ci + bci. So (xo, yo) is a solution of
S2. Assume now that (xo,}^o) is a solution of S2. Is it a solution of Si?
By definition of solution, (xo,yo) satisfies both equations of system
S2. Thus, (xo,}^o) satisfies the first equation of Si as well. Therefore, we
must prove that it satisfies the second equation of Si. By hypothesis,
(a\ + bai)%{) + (/?i + ^^2)^0 = ci + bci. We can rewrite the left-hand
side of the second equation of S2 as (^i + ^^2)^0 + (^i + bb2)y^ =
(flixo+ fci}^o)+ K^2^o + ^2}^o)- Because (xo,}^o) is a solution of S2,
the left-hand side of the equation is equal to c\ + bc^, and the first
expression in the parentheses on the right-hand side of the equation
is equal to c\. Therefore, we obtain ci + fcc2 = ci + fc(a2Xo + fo2yo)-
This equahty impHes that a2Xo + biy^ = C2, as fois a nonzero number.
This proves that (XQ, jo) is a solution of Si.
USE OF COUNTEREXAMPLES
1. (Let us consider the statement. It seems to suggest that the "growth" of
/ should be cancelled by the "drop" of g. But the two functions could
increase and decrease at different rates. Therefore, the statement does
not seem to be true.) Let us construct a counterexample, using simple
functions. We could try to use linear functions. Consider/(x) = x -f 1
and ^(x) = — 3x. Clearly, / is increasing and g is decreasing. (If you
wish to do so, prove these claims.) Their sum is h(x) = —2x + 1, which
is decreasing.
2. There might be an angle in the first quadrant for which 2 sin t = sin It.
But the equahty is not true for all the angles in the first quadrant.
Consider t — n/A. The left-hand side of the equation equals
2sin(;r/4) = 2(l/\/2) = V2. The right-hand side equals sin(7r/2) = L
Thus, the statement is false.
3. (It might seem plausible that y = P(x) is always negative, because
its leading coefficient is negative [it is —1]. But, when the values of
the variable x are not too large, the value of the monomial —x^ can
be smaller than the value of the monomial 2x. This remark seems
to suggest that for positive values of x that are not too large the
variable y = P(x) might be positive [or at least nonnegative].) We
will look for a value of the variable x that makes the polynomial