Math 230 Exam

Problem

A)

i

Our question ∀x ∃y (x + y ≥ ) means for every real number x, there’s "some" real number y
where x + y is greater than or equal to

this is true because the freedom of the allows the user to pick any y vaule to make this true for all instances in which you may want to pick an X vaule.

if x = y =

+ = ≥

for all states of X let say it was for x
i can be some number for Y like and thus make it true as this is

ii

you have to really read what it is saying, the argument implies

there is some number out there , that for all vaules of Y in a real number set

output to be >

this is NOT the same as for all x there is some y that makes it true to be greater than

our argument is dependent on one single x vaule holding true to makes all instances of Y vaules true. not possible .

B)

i

the question says "There are two numbers whose ratio is less than "

This means is "some" or orther words E.... two numbers x and y
when divided by each other they are less than <

I can write this as logical expression

∃x ∃y (x/y < )

ii

the question says

"The reciprocal of every positive number is also positive."

Logical expression is ∀x (x > → ( / x) > )

This means for all in other words every x as postive , then its reciprocal /x is positive. we use the arrow as it jumps to the conclusion

Problem

i

Let x and y be real numbers such that x + y is rational. Prove by contrapositive that if x is irrational, then x - y is irrational.

this can be expressed as if p then q.

p

the contrapostive of this is
- q → - p

this can be worded in context of a contrapostive proof where rational is in the arugment so

if q not rational than p not rational

ill use x = √

y = - √

to prove the proof

x + y =

√ + ( - √) = (rational)

x - y =

√ - ( - √) = √ - (irrational)

same vaules for x and y

and solve the problem

ii

Prove by contradiction that for any positive two real numbers, x and y, if x * y ≤ , then either x < or y <

ok so we will assume the opposite in that
it’s not true that either x < or y <

so we can use

x ≥ and y ≥ this sets the case that

* at the lowest inputs which is NOT mathematically true

proving that our orginal statment of x < and y < when mutiplied is less than to be true

problem

Let n ≥ 1, x be a real number, and x - Prove the following statement using mathematical induction:

( + x)^n ≥ + nx

For

n=1

( + x) ^1 + * x

our left side is + x after distrubting a in the paranthesis

our right side well you might think to add together but actually you times it by x to get just x leaving now just x +

these are x + x + which is true

we tested the base which ensures us we heading in the right dieriction and this argument works its almost like getting on a ladder but now we need to take our steps up soo

we now test if the argument works with n being k +

The use of

k+1

ensures that if the statement is true for ( k ), it must also be true for the next number, creating a chain of truth

In mathematical induction,

k +

is used in the inductive step to extend the proof from one case to the next like the ladder we are climbing haha. ok so we test our arguement and see if it holds

to prove: ( + x)^(k+) >= + (k+)x for all k >= and x >= -

Start with the expression ( + x)^(k+) = ( + x) * ( + x)^k.

By the inductive hypothesis, we assume ( + x)^k >= + kx holds. Therefore:

( + x)^(k+) >= ( + x) * ( + kx).

Expand the right-hand side:

( + x)( + kx) = + kx + x + kx^ = + (k + )x + kx^

Since k >= and x^ >= , it follows that kx^ >= Thus:

+ (k + )x + kx^ >= + (k + )x.

Also, since x >= -, we have + x >= , so multiplying both sides of the inductive hypothesis ( + x)^k >= + kx by the non-negative term ( + x) preserves the inequality.

Therefore, ( + x)^(k+) >= + (k + )x.

By mathematical induction, the inequality ( + x)^n >= + nx holds for all n >= and x >= -

Problem

a)