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Logic Question in maths

Number 1 on both papers give me a problem help thanks
http://libraries.sta.uwi.edu/apps/index.php/PastPaperSearch/viewPaper/math1140_1_09.pdf
http://libraries.sta.uwi.edu/apps/index.php/PastPaperSearch/viewPaper/math1140_1_11.pdf

Reply 1

ghostwalker

Original post by bigmindedone

Since this is one of several questions you've posted on logic recently, what are your thoughts? What have you done so far? What are you stuck on?

Reply 2

alesssiat

I was wondering if someone could help me with a few logic questions.
For these questions, 'a' and 'b' are Real Numbers.

I have to state whether the questions are true or false (if false, why?)

1. For Every a, there exists b, such that a > b
2. There exists b such that for every a, a > b
3. For every a, there exists b such that ab = 1

Any ideas????

Reply 3

ghostwalker

Original post by alesssiat

...

Difficult to hint, without just telling you the answers.

Presumably you must have some thoughts about them....

Reply 4

alesssiat

well in my opinion its:

1.true
2. false (i dont know why)
3. true

Reply 5

ghostwalker

Original post by alesssiat

well in my opinion its:

1.true
2. false (i dont know why)
3. true

Agree with the first two.

First one:

If you put it into english:

For every number, a, there is a number ,b, that is less than it.

"for every" can also be translated as "for each" and the sentence becomes

For each number, a, there is a number ,b, that is less than it.

We can think of this as having chosen an "a", then there exists a "b" that is less than it.

For the second one:

This is saying there is a number (b) such that every number (a) is greater than it.

Try and find a counter example. Note that you'll have to find one for every "b" to show the statement is false.

One in the spoiler.

Spoiler

For the third one:

It's actually false. Try and find a counter-example.

(edited 9 years ago)

Reply 6

alesssiat

Sorry about this!!!

Second One:

For Example, when b=2 and a=-2 would show that the proposition is false?

Third One:

I don't really understand how this is false? Seeing as a real number can be anything, for example a=root(2) and b=1/root(2), then when you times them together you get 1.

Reply 7

ghostwalker

Original post by alesssiat

Sorry about this!!!

Second One:

For Example, when b=2 and a=-2 would show that the proposition is false?

Fraid not.
That shows it's not true for b=2,

The original statement says there exists a "b"; to show it's false you must show no such b exists, not just that it's not true for one particular one.

Third One:

I don't really understand how this is false? Seeing as a real number can be anything, for example a=root(2) and b=1/root(2), then when you times them together you get 1.

The third statement says "for every a".... I.e. every real number has an inverse. This is true for all bar one of them. Which one?

Edit: PS. Quote me if you want a reply as then the screen updates without having to be refreshed and I'll notice sooner.

(edited 9 years ago)

Reply 8

alesssiat

Original post by ghostwalker

Fraid not.
That shows it's not true for b=2,

The original statement says there exists a "b"; to show it's false you must show no such b exists, not just that it's not true for one particular one.

The third statement says "for every a".... I.e. every real number has an inverse. This is true for all bar one of them. Which one?

Edit: PS. Quote me if you want a reply as then the screen updates without having to be refreshed and I'll notice sooner.

So for the second one i was right? that when b=2 and a=-2, the proposition is false?

so that would mean that the third statement isnt true when one of them is zero?

Reply 9

ghostwalker

Original post by alesssiat

So for the second one i was right? that when b=2 and a=-2, the proposition is false?

The particular instance of setting b=2 and a=-2 has a truth value of "False", but that does not make the initial statement false.

The statement is "there exists a "b" such that for every "a", a>b.
To show that this is false, you need to show that it's negation is true.
The negation of that statement is:

"for every 'b', there exists an 'a', a <= b".

or

"for each 'b', there exists an 'a', a <= b.

You've only shown it for one particular b. b=2.

so that would mean that the third statement isnt true when one of them is zero?

Correct'ish

For every a, there exists b such that ab = 1

It's negation is

"There exists "a", such that for all "b", ab \= 1"

And if a=0, there is no b that would make ab=1, i.e. for all b, ab \=1. So the negation is true, and the original statement false.

Reply 10

alesssiat

Original post by ghostwalker

the particular instance of setting b=2 and a=-2 has a truth value of "false", but that does not make the initial statement false.

The statement is "there exists a "b" such that for every "a", a>b.
To show that this is false, you need to show that it's negation is true.
The negation of that statement is:

"for every 'b', there exists an 'a', a <= b".

Or

"for each 'b', there exists an 'a', a <= b.

You've only shown it for one particular b. B=2.

Correct'ish

for every a, there exists b such that ab = 1

it's negation is

"there exists "a", such that for all "b", ab \= 1"

and if a=0, there is no b that would make ab=1, i.e. For all b, ab \=1. So the negation is true, and the original statement false.