What are the methods of discrete mathematics proof, discrete mathematics proof problems

Direct method: Since m is an even integer, it can be set to m 2n, then m+7 2n+7 2(n+3)+1 is an odd integer.

Indirect method: (Counter-argument method) Assuming that m+7 is not an odd integer, then m+7 is an even integer, which can be set to m+7 2n, then m 2n-7 2(n-3)-1 is also an odd integer, which contradicts the problem, so m+7 is an odd integer.

1) ara<==>a -1*a=1 h, 2) if arb, then braIn fact, (b -1*a) -1 = a -1*b h, h is a subgroup of g, b -1*a h

3) If ARB, BRC, then.

a -1*b h, b -1*c h, (a -1*b)(b -1*c) = a -1*c h, then arc

In summary, r is an equivalence relation.

Reflexivity: Clearly true.

Symmetry: If a b, then there is a nonsingular matrix p,q such that b = paq, then p inverse bq inverse = a, so there is b a

Transitivity: If a b, b c , there are nonsingular matrices p1, q1, p2, q2 such that.

b=p1aq1，c=p2bq2

So b=p1aq1, p2 inverse, cq2 inverse=b

So there is p1aq1=p2 inverse cq2 inverse, so p2p1aq1q2=c, i.e. a c

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Reflexivity: u+v=u+v=> r

Symmetry: If r then there is u+y=x+v => x+v=u+y => r transmissibility: if r then there is u+y=x+v (1).

If r then there is x+t=s+y (2).

1)+(2) u+t=s+v =>(u,v)r(s,t) relation r is reflexive, symmetrical and transitive on a a, and is therefore equivalence.

If a n=e, the order of a is k, a k=e

n k, you might as well let n=mk+b, if b≠0, then 0 a b=e, and k is the order of a, k b

This is the same as ba n=a (mk)=(a k) m=e m=e.

As shown in the figure below, two mathematical inductions were used.

5.(a) Let f(n)=[1 (-1 3)+2 (-1 3)+3 (-1 3)+....n^(-1/3)]-n+1)^(2/3),f(n+1)-f(n)=(n+1)^(1/3)+(n+1)^(2/3)-(n+2)^(2/3)

n+2-(n+1) (1 3)*(n+2) (2 3)] n+1) (1 3)>0, so f(n) is incremental, f(4) = 1 (-1 3)+2 (-1 3)+3 (-1 3)+4 (-1 3)-5 (2 3) 1+

0, so f(n)>f(4)>0, the proposition holds.

b) (i) n = 1 (7n + 1) * 6 n + (-1) (n + 1) = 49, divisible by 49.

ii) Assuming that n=k(7k+1)*6 k+(-1) (k+1) is divisible by 49, then n=k+1.

7k+8)*6^(k+1)+(1)^(k+2)

7k+1)*6^(k+1)+7*6^(k+1)+(1)^(k+2)

6[(7k+1)*6^k+(-1)^(k+1)]+7*6^(k+1)+(1)^(k+2)-6*(-1)^(k+1),①

6 (k+1)=7m+(-1) (k+1), where m is an integer, so 7*6 (k+1)+(1) (k+2)-6*(-1) (k+1).

7[7m+(-1)^(k+1)]+1)^(k+2)-6*(-1)^(k+1)

49m, hypothesized by induction, is divisible by 49.

By mathematical induction, the proposition is true.