When proving a problem by reduction to absurdity, we must use "counter-hypothesis" for reasoning, otherwise it is not reduction to absurdity. When proving a problem by reducing to absurdity, if there is only one proposition that we want to prove, we can just refute this situation, which is also called reducing to absurdity. If the conclusion is multifaceted, then all the negative situations must be refuted one by one in order to infer the original conclusion. This method of proof is also called "exhaustive method".

Example 1 Prove that when both p and q are odd numbers,? Curve? y=？ x？ The abscissa of the intersection of 2-2px+2q and the X axis is an irrational number. (2009 Tsinghua University Summer Camp Selection Examination)?

Thinking analysis?

At present, there is no direct judgment method to explain the ignorance of quadratic equation, so the reduction to absurdity is adopted.

Proof?

The abscissa of the inverse intersection is a rational number. That is, the abscissa of the intersection point is x=uv ((u, v)= 1), then uv? 2-2puv+2q=0, that is, u? 2-2puv+2qv？ 2=0，u？ 2=2(puv-qv？ 2)① is an even number. So u is even.

(u，v)= 1，？ V is odd.

In addition, there are v|u from ①? 2, so that v | u. and (u, v)=? 1，v= 1。

Let u=2s, then 4s? 2-4ps+2q=0, that is? 2s？ 2-? 2ps+q=0，q=2(ps-s？ 2) Even number, which contradicts the parity of known conditions.

Therefore, the inverse hypothesis is not established, which shows that the conclusion is established.

That is, the curve y=x? The abscissa of the intersection of 2-2px+2q and the X axis is an irrational number.

Problem solving review?

In simple integer theory, reduction to absurdity is a common method. The main applicable situation is that when it can't be dealt with positively, the assumption conclusion is not established, taking the assumption as a condition, and finally negating the assumption by deducing contradictions. In the simple integer theory, many contradictions are even and odd, for example, the most classical reduction to absurdity proves that 2 is an irrational number.

Example 2 It is known that there are 1 9 (different) positive integers between1and 90. Are there necessarily three equal differences? (1990 Hungarian Mathematics Competition)?

Analysis?

It is very difficult to deal with such a problem from the front. We can consider starting from the opposite side: there are no three equal situations, and at most two are equal, so what information can we get? If arranged in order of size, there are 18 differences, and at most two of these differences are equal, so there are some overlaps, so there are at least 9 different numbers. Let's try to find out the existence or contradiction.

Proof?

Let the number of 19 be 1≤a? 10。 ?

What about series a? 2- 1m，a3m+2？ - 1m，a6m+2？ - 1m，a9m+2？ -1m is not a geometric series.

So, a series? 2- 1m，a3m+2？ - 1m，a6m+2？ - 1m，a9m+2？ -1m in geometric series,? d=3m。 ?

Problem solving review?

This is the law of solving mathematical problems. When solving the problem head-on, it is necessary for us to adjust our direction. Starting from the opposite side of the problem is equivalent to adding a condition. In this problem, d≥3m+ 1 is much smaller than d=3m, and the progression increases slowly, so the original d=3m is just enough. Now that we have regressed, there should naturally be contradictions. At this time, intuitive qualitative analysis is also helpful.

Example 4 proves that if the absolute value of the integer coefficient polynomial of variable X is 1 when taking three different integer values, then this polynomial has no integer root.

Proof?

Let the integer coefficient polynomial f(x) have?

|f(a)|=|f(b)|=|f(c)|= 1。 ? ( 1)?

Suppose f(x) has an integer root x? 0,? So f(x)=(x-x? 0)Q(x)。 ? (2) (where Q(x) is an integer coefficient? Project type)?

From (1)(2),? |(a-x？ 0)Q(a)|=？ |(a-x？ 0)|? |Q(a)|= 1。 ?

Since Q(a) is an integer, Then |a-x? 0|= 1, similarly |b-x? 0|= 1，|c-x？ 0|= 1.？

So those three numbers a-x 0, b-x? 0，c-x？ Two of 0 must be equal. Therefore, two of A, B and C are equal.

This contradicts what is known, so f(x) has no integer root.

Problem solving review?

(1) has the attribute: Polynomial? f(x)？ ,? For a, b∈R, a≠b, a-b must be? f(a)-？ The factor of f(b); ?

(2) When studying related propositions with negative words such as "non-existence", "no", "inequality" and "impossible", our common strategy is to consider the problem from the opposite side, that is, if it is difficult, it will be reversed.

Example 5? Known function f(x)=ax? 2+bx+？ c(a≦？ 0), f(x)=x has no real root. Q: Does f(f(x))=x have a real root? And prove your conclusion. (2009 Shanghai Jiaotong University Self-enrollment Examination Questions)?

Analysis?

Reduction to absurdity. If there is f(f(x? 0))=x？ 0, make f(x? 0)=t, then f(t)=x? 0, that is, (t, x? 0) is a point on the image of y=f(x). 0)=t, that is, (x? 0, t) is also a point on the image of y=f(x). Obviously, these two points do not coincide. They are symmetrical about the straight line y=x, and y=f(x)=ax? 2+bx+c is a continuous function, so y=f(x)=ax? 2+bx+c and y=x must intersect, so f(x)=x has a real number solution. Contradiction!

Problem solving review?

Using the reduction to absurdity, the solution of the problem is intuitive and clear. At the same time, the conclusion of this problem is also valid for the general continuous function f(x), and its processing method can be used for reference.

Example 6 (peking university independent recruitment test questions in 2008)? Real number a? i(i= 1，2，3)，b？ I(i= 1, 2,3) satisfies? Answer? 1+a？ 2+a？ 3=b？ 1+b？ 2+b？ 3，a？ 1a？ 2+a？ 2a？ 3+a？ 3a？ 1=? b？ 1b？ 2+b？ 2b？ 3+b？ 3b？ 1, ? Min? (a？ 1，a？ 2，a？ 3)≤? Min? (b？ 1，b？ 2，b？ 3).？

Proof: max? (a？ 1，a？ 2，a？ 3)≤? Max. (b？ 1，b？ 2，b？ 3).？

Thinking analysis?

This problem is difficult to prove directly, so we think it is difficult to prove directly, and the proof is completed by reduction to absurdity and function construction.

Analysis?

Why not set an a? 1≤a？ 2≤a？ 3，b？ 1≤b？ 2≤b？ 3. What about A? 1≤b？ 1. A? 3≤b？ 3. Use reduction to absurdity. What if? Answer? 3>? b？ 3. Construct two functions f(x)=(x-a? 1)(x-a？ 2)(x-a？ 3)，g(x)=(x-b？ 1)(x-b？ 2)(x-b？ 3). From the known condition A? 1+a？ 2+a？ 3=b？ 1+b？ 2+b？ 3，a？ 1a？ 2+a？ 2a？ 3+a？ 3a？ 1=b？ 1b？ 2+b？ 2b？ 3+b？ 3b？ 1，f(x)=g(x)+b？ 1b？ 2b？ 3-a？ 1a？ 2a？ 3. On the one hand, f(a? 1)=g(a？ 1)+b？ 1b？ 2b？ 3-a？ 1a？ 2a？ 3=0, ? f(a？ 3)=? g(a？ 3)+b？ 1b？ 2b？ 3-a？ 1a？ 2a？ 3=0, so g(a? 1)=g(a？ 3). On the other hand, g(a? 1)=(a？ 1-b？ 1)(a？ 1-b？ 2)(a？ 1-b？ 3)、a？ 1-b？ 1≤0，a？ 1-? b？ 2≤? 0，a？ 1-b？ 3≤0, so g(a? 1)≤0; And then what? g(a？ 3)=? (a？ 3-b？ 1)(a？ 3-b？ 2)(a？ 3-b？ 3)、a？ 3-? b？ 1>? 0，a？ 3-b？ 2>0，a？ 3-b？ 3 > 0, so g(a? 3) > 0, and g(a? 1)=g(a？ 3) contradiction. So a? 3≤b？ 3, ? Max. (a？ 1，a？ 2，a？ 3)≤? Max. (b？ 1，b？ 2，b？ 3).？

Problem solving review?

Mathematics competition test is a contest of wisdom, especially how to get rid of the predicament. At present, there are bound to be "new questions" in mathematics competitions, independent entrance examinations and college entrance examinations, which may make candidates unable to grasp them for a while and make them think hard and stop solving problems. Unilateral blind attack on these strategic highlands is not the best policy. To learn to attack from the flank, there is often a sense of "strategic detour" and a breakthrough from the side or the opposite point.

Newton once said: "Reduction to absurdity is one of the best weapons for mathematicians". Generally speaking, the types of questions commonly proved by reduction to absurdity are: propositions whose conclusions appear in the form of "negative form", "at least" or "at most", "unique" and "infinite"; Or use more obvious, concrete and simple conclusions to deny the proposition; Or directly prove the difficult proposition, change its thinking direction, and think negatively from the conclusion, and the problem may be solved very simply.

Consolidate training?

1? It is proved that if f(f(x)) has a unique fixed point, then F (x) also has a unique fixed point. (20 10 Zhejiang university independent enrollment examination paper adaptation)

2 Known function f(x)= 13x? 3-2 times? 2+3x？ (x∈？ The image of r) is curve c,? It is proved that there is no straight line and curve C tangent to two different points at the same time.

3 it is known that there is an integral coefficient a? 1，a？ 2、…、a？ Polynomial f (x) with n = x? n+a？ 1xn- 1？ +…+an- 1？ x+a？ N, for four different integers a, b, c, d, f(a)=f(b)=f(c)=f(d)=5. It is proved that there is no integer k and f(k)=8. (Name of preliminary contest of Sichuan Division in 2009)

Let f(x)=ax? 2+bx+c，？ It is known that F (1), F (2), F (3), F (4) and F (5) are all prime numbers. Proof: F(x) No? Can be decomposed into the product of two integral coefficients in linear form. (20 10 preliminary contest of Fujian Mathematics Competition)

1? Proof: Why not set X? 0 is the only fixed point of f(f(x)). That is, f(f(x? 0))=x？ 0, make f(x? 0)=t, then f(t)=x? 0, then, f(f(t))=f(x? 0), and f(x? 0)=t, that is, f(f(t))=t, which means that t is also the fixed point of f(f(x)). If f(f(x)) has a unique fixed point, do you know X? 0=t, so f(t)=t,? Does that mean t is too? F(x) does not move? Point, proof of existence.

The uniqueness of the next proof. Suppose f(x) has another fixed point t? 0,? That's f(t? 0)=t？ 0 (t≠t？ 0), then f(f(t? 0))=f(t？ 0)=t？ 0, which means that f(f(x)) still has a fixed point t? 0,? Contradicting with the topic.

Problem solving review? When f(x? 0)=x？ At 0 o'clock? We call it x? 0 is the fixed point of the function f(x). Some mathematical problems can be solved by using the fixed point principle, which is a hot issue in the independent entrance examination.

2? Proved that A(x? 1，y？ The tangent of 1 is tangent to two points of curve c at the same time. The other vertex is B(x? 2，y？ 2)、x？ 1≠x？ 2.？

Then the tangent equation is:? y- 13x？ 3 1? -2x？ 2 1? +3x 1？ =? (x？ 2 1? -? 4x 1？ +3)(x-x 1？ ),?

Simplify: y=(x？ 2 1? -4x 1？ +3)x+-23x？ 3 1? +2x？ 2 1? . ?

And B(x? 2，y？ The tangent equation of 2) is y=(x? 22? -4x2？ +3)x+-23x？ 32? +2x？ 22? ,?

Because the two tangents are the same straight line,

Have:? x？ 2 1? -4x 1？ +3=x？ 22? -4x2？ +3，x？ 1+? x？ 2=? 4.？

Again -23x? 3 1? +2x？ 2 1? =-23x？ 32? +2x？ 22? ,?

That is -23(x 1? -x2？ )(x？ 2 1? +x 1？ x2？ +x？ 22? )+2(x 1？ -x2？ )(x 1？ +x2？ )=0,?

- 13(x？ 2 1? +x 1？ x2？ +x？ 22? ) +4=0, which means x 1? (x 1？ +? x2？ )+x？ 22? - 12=0,?

That is, (4-x2? )×4+x？ 22? - 12=0，x？ 22? -4x2？ +4=0，x？ 2=2.？

But when x? When 2=2, x? 1+x？ 2=4 x？ 1=2, which is different from x? 1≠x？ 2 contradiction.

So there is no straight line and curve c tangent to two points at the same time.

3? Analysis: Note that A, B, C and D are the roots of polynomial f(x)-5, so we can construct a polynomial f(x)-5, and then prove it by factorial theorem and reduction to absurdity.

Prove: from what is known, Should f(x)-5 be equal to (x-a) (x-b) (x-c) (x-d) g (x)? Where g(x) is an integer coefficient polynomial.

If there is an integer k that makes f(k)=8, that is, (k-a)(k-b)(k-c)(k-d)g(k)=3.

But prime number 3 cannot have more than four different factors, which is contradictory.

So there is no integer k so that f(k)=8.

3 inverse f(x)=g(x)h(x),? Where g (x) and h (x) are linear expressions of integer coefficients.

Then f (1) = g (1) h (1), f(2)=g(2)h(2),? f(3)=？ g(3)h(3)，f(4)=g(4)h(4)，f(5)=g(5)h(5)，？

The left ends of these five equations are all prime numbers, so g (1), g (2), g (3), g (4), g (5), h (1), h (2), h (3),? h(4)，h(5)？ At least five of them are +/- 1. Since g(x) is a linear expression of integral coefficient, G (1), G (2), G (3), G (4) and G (5) are different numbers, that is, at most one is 1 and the other is -66. Similarly, h( 1),? h(2)？ , h(3), h(4), h(5), at most one is 1, and the other is-1, which is contradictory.

Therefore, the reverse hypothesis does not hold, so the original proposition holds.