Question #3879

show if a,b ЄR then,
a. max (a,b) = (1/2)(a+b+|a-b|) and min (a,b)=(1/2)(a+b-|a-b|)
b. min (a,b,c)=min(min(a,b),c)

Expert's answer

a.& There are two possibilities: either a≥b or a<b.

1) max(a,b)=1/2 (a+b+|a-b|)

& If a≥b then (a-b)≥0 and thus |a-b|=(a-b). Let us see at the desired equality:

1/2 (a+b+|a-b|)=1/2 (a+b+(a-b))=1/2 (a+b+a-b)=1/2∙2a=a

On the other hand max(a,b)=a, because a≥b.

Now let’s check second case: a<b.

If a<b then (a-b)<0 and thus |a-b|=-(a-b)=(b-a). Let us look at the desired equality:

1/2 (a+b+|a-b|)=1/2 (a+b+(b-a))=1/2 (a+b+b-a)=1/2∙2b=b

On the other hand max(a,b)=b, because a<b.

So we’ve checked the equality for both cases, thus max(a,b)=1/2 (a+b+|a-b|) for any real numbers a,b ∈ R.

2)& min(a,b)=1/2 (a+b-|a-b|)

For second equality we can make almost the same proof.

If a≥b then (a-b)≥0 and thus |a-b|=(a-b). Let us see at the desired equality:

1/2 (a+b-|a-b|)=1/2 (a+b-(a-b))=1/2 (a+b-a+b)=1/2∙2b=b

On the other hand min(a,b)=b, because a≥b.

Now let’s check second case: a<b.

If a<b then (a-b)<0 and thus |a-b|=-(a-b)=(b-a). Let us look at the desired equality:

1/2 (a+b-|a-b|)=1/2 (a+b-(b-a))=1/2 (a+b-b+a)=1/2∙2a=a

On the other hand min (a,b) = a,& because a<b.

So we’ve checked the equality for both cases, thus min(a,b)=1/2 (a+b-|a-b|) for any real numbers a,b∈R.

b. min(a,b,c)=min(min(a,b),c)

Analogical to the part a. we can divide the problem into three cases:

1) If min(a,b,c)=a, then the left side of the equality is a. This means a≤b and at the same time a≤c. Let’s look at the right side:

min(a,b)=a, because a≤b. Then min(min(a,b),c)=min(a,c)=a, because a≤c. Thus the right side of the equality is a too.

2) If min(a,b,c)=b then the left side of the equality is b. In this case the proof is absolutely the same, just write a instead of b and write b instead of a. So b≤a and at the same time b≤c. Let’s look at the right side:

min(a,b)=b, because b≤a. Then min(min(a,b),c)=min(b,c)=b, because b≤c. Thus the right side of the equality is b too.

3) If min(a,b,c)=c then the left side of the equality is c. This means c≤a and at the same time c≤b. Then not knowing what is min(a,b) (it may be either a or b) we can conclude that c≤min(a,b), thus min(min(a,b),c)=c.

So we’ve checked the equality for each case and thus min(a,b,c)=min(min(a,b),c) is true for any real numbers a,b,c ∈ R.

1) max(a,b)=1/2 (a+b+|a-b|)

& If a≥b then (a-b)≥0 and thus |a-b|=(a-b). Let us see at the desired equality:

1/2 (a+b+|a-b|)=1/2 (a+b+(a-b))=1/2 (a+b+a-b)=1/2∙2a=a

On the other hand max(a,b)=a, because a≥b.

Now let’s check second case: a<b.

If a<b then (a-b)<0 and thus |a-b|=-(a-b)=(b-a). Let us look at the desired equality:

1/2 (a+b+|a-b|)=1/2 (a+b+(b-a))=1/2 (a+b+b-a)=1/2∙2b=b

On the other hand max(a,b)=b, because a<b.

So we’ve checked the equality for both cases, thus max(a,b)=1/2 (a+b+|a-b|) for any real numbers a,b ∈ R.

2)& min(a,b)=1/2 (a+b-|a-b|)

For second equality we can make almost the same proof.

If a≥b then (a-b)≥0 and thus |a-b|=(a-b). Let us see at the desired equality:

1/2 (a+b-|a-b|)=1/2 (a+b-(a-b))=1/2 (a+b-a+b)=1/2∙2b=b

On the other hand min(a,b)=b, because a≥b.

Now let’s check second case: a<b.

If a<b then (a-b)<0 and thus |a-b|=-(a-b)=(b-a). Let us look at the desired equality:

1/2 (a+b-|a-b|)=1/2 (a+b-(b-a))=1/2 (a+b-b+a)=1/2∙2a=a

On the other hand min (a,b) = a,& because a<b.

So we’ve checked the equality for both cases, thus min(a,b)=1/2 (a+b-|a-b|) for any real numbers a,b∈R.

b. min(a,b,c)=min(min(a,b),c)

Analogical to the part a. we can divide the problem into three cases:

1) If min(a,b,c)=a, then the left side of the equality is a. This means a≤b and at the same time a≤c. Let’s look at the right side:

min(a,b)=a, because a≤b. Then min(min(a,b),c)=min(a,c)=a, because a≤c. Thus the right side of the equality is a too.

2) If min(a,b,c)=b then the left side of the equality is b. In this case the proof is absolutely the same, just write a instead of b and write b instead of a. So b≤a and at the same time b≤c. Let’s look at the right side:

min(a,b)=b, because b≤a. Then min(min(a,b),c)=min(b,c)=b, because b≤c. Thus the right side of the equality is b too.

3) If min(a,b,c)=c then the left side of the equality is c. This means c≤a and at the same time c≤b. Then not knowing what is min(a,b) (it may be either a or b) we can conclude that c≤min(a,b), thus min(min(a,b),c)=c.

So we’ve checked the equality for each case and thus min(a,b,c)=min(min(a,b),c) is true for any real numbers a,b,c ∈ R.

## Comments

Assignment Expert08.10.15, 17:54Inequality a≥b means that a is greater than or equal to b.

alexandra molina07.10.15, 23:52what does a≥b mean

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