USE A PROOF BY CONTRADICTION TO PROVE: Please show and explain every single step

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USE A PROOF BY CONTRADICTION TO PROVE: Please show and explain every single step since I'm really struggling in this class. QUESTION 6&7

1. Use a proof by contradiction to prove: For all integers m and n, if m n is odd, then m is odd or n is odd. 2. Use a proof by contradiction to prove: For all real numbers r1,r2, ...,rk and m, if then ri 2 m or r2 2 m or or rk m. 4. se a proof by contradiction to prove: For all positive real numbers ar, a, and b, if r2 z ab, then r a or r b. 5. Use a proof by contradiction to prove: For all real numbers r and y, if r y is rational but is irrational, then y is irrational. 6. Use a proof by contradiction to prove: For all sets A, B, and C, if A g B and A Z B C, then An 0. 7. Use a proof by contradiction to prove V3 2 is irrational

Explanation / Answer

6) for the all the sets A,B,C

Given A is subset or equal to B and A is not subset of B-c then we have to to prove that A intersection c is not empty set.

Use proof by contradiction.

Let us take A intersection C = Null set. ----------------(1)

that means A and C are disjoint sets also no common elements in A and C.

i) Given A is subset of B

that means at least one element X which is in B but not in A.

that is X belongs to B and X doesnot belongs to A

ii) GIven A is not subset of B-C.

that is at least one element present in A but not present B-C.

implies let Y belongs to A and Y does not belongs to B-C,

i.e Y belongs to A and (Y does not belongs to B and Y belongs to C)

implies Y belongs to A and Y belongs to C,

that means Y is an element in A and C,

but we considered A and C are disjoint sets, no common elements in A and C,

So this is a contradiction, Our assumption is wrongs.

Hence by proof by contradiction A intersection C not empty.

7) We have to Square root 3 and Square root 2 is irrational.

First let us prove Square root 3 is irrational.

case(1) Let us take Square root 3 is rational.

then square root 3 = a/b, where a and b are integers with no common factor other than 1.

squaring on both sides,

implies 3 = a2/b2   

implies 3b2=a2, since 3 is a factor, implies a2 is divisible by 3.

If a2 is divisible by 3, then a is also divisble by 3. -------------------(1)

since a is divisble by 3, 3 is a factor of a,

that is a = 3k, where k is some integer.

now 3b2=a2, implies 3b2=(3k)2

implies 3b2=9k2,

dividing both sides by 3, we have b2=3k2,

which means b2 is divisible by 3

implies b is also divisible by 3 -----------------------(2)

from (1) and (2) , there is a contradiction that a and b have common factor except 1.

therefore square root 3 is not rational,

our assumption is wrong.

therefore square root 3 is irrational

case(2) similarly square root 2 is also irrational

case (3) we have a result that difference of two irrational numbers is also a irrational number,

hence from case (1) ,(2) and (3) ,

square root of 3 - square root of 2 is irrational.

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