摘要：The longer the end point segment, the higher the total fraction of itshighest form , the more inverse high shrinking fgsx ordershr

Chapter 6 4.

The relationship between the highest total fraction value and the end point segment fraction value.

After calculating the shrinkage and cut impairment generated by the previous inverse high shrinkage fgsx order , we find that the highest total fraction value is linked to its end point segment fraction value:

The longer the end point segment , the higher the total fraction of its highest form , the more inverse high shrinking fgsx order shrinkage factors are formed , the greater the shrinkage impairment generated, and vice versa.

set up ,

The highest total fraction is the sister order fraction because its ny=1 ,

If its inverse high shrinking fgsx order is reduced, since the sister order cannot be continuous, its end order is ny≥2 .

Comparison of the numeric fields of the two orders, the full numeric fields of the end order are greater than twice the highest total sister order.

In order to make the inverse high shrinking order , form the largest reduction amount for the highest whole formula, and cut the impairment amount, the B1 section of the end order, (suppose) when it reaches its extreme value ny= 2 , the end section must reach 1/3 of the end order number field . Only by inverting the high-reducing fgs1 order number 1 position can be cut to the center of the highest full-digit field, and cut impairment can be generated . Because the highest total order ny=1 , its stage is smaller than any stage higher than it, and it cannot cross any anti-high shrink fgsx stage between the same rankings . Only 1 cutting point G is generated, which is the sister order fraction of ny= 1 of the highest total formula , and the maximum inverse high shrink fgs1 impairment number. Only the B1 stage of the end pointTo lengthen, theB1 segment is the same vertically, the more orders of the inverse high shrink fgsx , the more inverse high shrink fgsxaccumulated shrink factors are generated for the highest total fraction , the larger the number of inverse high shrink fgsx order shrink fgs1 formed.

In the end point segment fraction, its value-added qxz accumulates , and is in the largest and most states of each fraction of the order (greater than the highest full-form fraction); the inverse high-definition fgsx order is in the state with the least inverse shrinking cumulative factor and the smallest shrinking rate (greater than the highest full-form fraction). Therefore, the factor condition for producing fractional values ​in the end point segment should be greater than the condition for producing fractional values ​​in the highest total formula .

Therefore, it is known that if the B1 segment of the end order is close to its extreme full order B segment, then the fractional value of the end point segment , based on the previous " estimated 8 ," is greater than the 2ny value of the highest full order .

From ny≥2 , we can obtain the fractional value of his end point segment:

>2ny=2*2=4。

To make the highest full expression, because the inverse high shrink fgsx cuts , only its integer value is reduced. According to the previous calculation, fgsx must be cut to the center point of its stage segment of the highest full expression . This state can only be formed when segment B1 reaches 1/3 of segment B ( segment B1 is in a folded state).

So,

If the end segment reaches 1/3 of the entire number field of the end segment, the fraction value of the end segment will be greater than 4/3>1. The end segment fraction can obtain at least 1 [conjecture] solution.

After the end point segment is lengthened , the inversely high-decreased fgsx order is the calculation result that the increased maximum total-decreased fgsx order reduces the impairment result. Since the value formation condition of the end point segment is greater than the value formation condition of the highest total-deformation fraction, it is smaller than the increase result of the value of the end point segment.

Obviously, the larger the ny value of the end-order all digit field , the larger its 2ny value . The more integer value of the end point segment fraction is generated.

calculate,

Let: the extreme value of the end point segment , B1 segment ≌ segment B ,

Because it is close to the extreme value of the end-order order all digit field, the most inverse high shrinking fgsx order shrink factor accumulation can be generated. The number of 2ny in segment B is in its extreme value.

Inverse high shrink fgsx cutting point: G≤n y value,

All the fractional values ​​of his end point segment are, 2ny -ny =ny

because,

ny=2 of the end point segment , the condition for generating the fractional value of is greater than the condition for generating the fractional value of the highest total formula,

To conclude,

The end segment fraction, after approaching its order all-digit field, has at least 2 [conjecture] solutions.

To conclude,

The highest total fraction value , after ≥5 orders, if it is impaired because it is cut, the sum of the fraction value with the end point segment: ≥1+2=3.

In order to prevent extreme phenomena, we still determine the end point segment fraction value caused by cutting impairment and lengthening as: ≥1.

However, when the end point segment fraction value is less than 1 , the first inverse high shrinkage fgs1 ( point c ), for the sister order fraction with the highest total expression ny=1 , there is no order shrinkage (integer value), and the effect of cutting impairment,

The highest total fraction of even-ordered equations , their fractional values ​​are still ≥2ny , so it is concluded that they all have at least ≥2 [conjecture] solutions.

To conclude,

"The Next Chapter", Chapter 5, Conclusion 15,

For any even-numbered highest total fraction, in the case of fgsx cut impairment, it has at least 1 [conjecture] solution. However, in the case of fgsx cutting impairment, its end point segment fraction obtains at least 1 [conjecture] solution.

1+1=2, which is the highest total fraction value of the sister order of ny=1, and the minimum extreme value of the sum of the two fraction values ​​with its end point segment fraction value.

Since ny≥2 the highest total fraction value: ≥2*2=4. The larger its order ny value, the larger the fraction value . The number of impairments caused by shrinking and cutting is increasingly smaller than the number of added value of its fractional value.

To conclude,

The highest total fraction value is related to the fraction value of its end point segment and the two order fraction values ​​they are linked:

The highest full formula value is greater than 2ny value without the inverse high shrinking fgsx order reduction and the inverse high shrinking fgsx cut impairment . Both have at least 2 [conjecture] solutions. With the maximum reduction of the order of inverse high shrink fgsx and the maximum cut impairment of inverse high shrink fgsx , the highest total fraction value is still greater than 1, and at least 1 [conjecture] solution is obtained. At this time, its end point segment fraction value obtains at least 1 [conjecture] solution.

To conclude,

The minimum number of extreme values ​​of the highest total formula, the sum of the final segment fraction value of the two fraction values, are: ≥2 groups.

In any even-number order calculation, as long as even numbers have order total expressions , they must have their highest total fraction . The sum of their highest total fraction and their end point segment fraction values: they are ≥2 groups , proving that they have at least 2 groups of solutions .

"73" has finished.

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