# 2 irrational number

The 2 irrational number is the important topic of the Subject Theory Of Computation.

**2 irrational number**

Suppose for the sake of contradiction that is rational. Then there are integers m^{’ }and n^{’ }with = m’/ n’.

By dividing both m’ and n’ by all the factors that are common to both, we obtain =m/n, for some integer m and n having no common factors. Since m/n= , m= n . Squaring both sides of this equation, we obtain m^{2}=2n^{2}, and therefore m^{2} is even.

If a and b are odd, then ab is odd. Since a conditional statement logically equivalent to its contrapositive. we may conclude that for any a and b, if ab not odd, then either a is not odd or b is not odd.

However, an integer is not odd if and only if it is even, and so for any a and b, if ab is even, m must be even.

This means that for some k, m=2k. Therefore, (2k) 2=2n2.

Simplifying this and canceling 2 from both sides, we obtain 2k2=n2. Therefore n2 is even.

The same argument that we have already used to show that n must be even, and so n=2j for some j.

We have shown that mand n are both divisible by 2. This contradicts the previous statement that mand n have no common factor. The assumption that is rational therefore it leads to a contradiction, and the conclusion is that is irrational: 2 irrational number

## To Prove: For every three positive integers i, j, and n, if i*j = n, then I ≤ or j ≤.

The statement we wish to prove of the general form “for every x, if p(x) then q(x).” For each x. The statement “if p(x) then q(x)” logically equivalent to “if not p(x) then not q(x).” and therefore the statement we want to prove equivalent to this: For any positive integer i, j and n, if it is not the case that i ≤ or j ≤ then i*j≠n.

if it is not true then i >and j>. A generally accepted fact from mathematical is that if a and b are the number with a> b, and c is a number >0, then ac>bc.

Applying this to the inequality i> with c=j, we obtain i*j > * j. since n>0, we know that >0, and we may apply the same fact again to the inequality j>, this time letting c=, to obtain j > =n. We now have i*j > j >n, and it follows that i*j ≠ n.

Hence, the proof is complete: 2 irrational number

**Related Terms**

Theory of Computation, Basic Terms of TOC, TOC Functions, Quantified Statement: Prime

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