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So having $1$'s and $0$'s can we combine them in a way starting with a $1$ that will have only finitely repeating patterns in any sub-row of $1$'s and $0$'s of the number? A sub-row being a sequence of $1$'s and $0$'s taken from the whole number. Clarifying: one $0$ or one $1$ is also a pattern. So infinitely many repeating $0$'s counts.

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    By example if we define strictly increasing gaps of zeros separated by $1$'s then there is no pattern repetition. – 2017-01-24
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    Combine them in what manner? What does "only finitely repeating pattern" mean? What is a "sub-row"? Clarify these points and probably you will find the answer is known. As it currently stands your question is too vague to be answered. – 2017-01-24
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    it is not too vague. see first comment. – 2017-01-24
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    If you mean an infinite binary *sequence*, we should be careful about calling this a *number*. In general the fractional part of a *binary representation* is not unique, though if we restrict ourselves to expansions that are not *periodic*, then the representation is unique and we could identify such a sequence with a "number" in this fashion. – 2017-01-24
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    I have to agree with @ErickWong; as the question stands, it's vague. – 2017-01-24
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    @Masacroso There is *no* pattern repetition? But the pattern of "00000000" repeats an awful lot of times in that sequence. What counts as repetition? This is evidence supporting my claim that the question cannot be answered without clarification. – 2017-01-24
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    Guys, it's his first question, we shouldn't doggy-pile downvotes. – 2017-01-24
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    Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. – 2017-01-24
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    What you are looking for is called a "normal number." https://en.wikipedia.org/wiki/Normal_number – 2017-01-24
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    $1110111$ has six repetitions of the pattern $1$. The repetitions are not _consecutive,_ but you have not said anything about whether the repetitions are consecutive. That is one of the reasons we have so much difficulty understanding what you are trying to ask. – 2017-01-24

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Your sequence either has infinitely many $1$s, or infinitely many $0$s (or both). Either way, we get a "subrow" (subsequence) consisting entirely of the same digit. So the answer is no (unless I've misunderstood the question).

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    This seems to be the most reasonable interpretation of the question as it is literally written. I doubt the OP intended to express this interpretation, though :). – 2017-01-24
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    So there will always atleast be an infinite row of 1's or 0's? Because that is what I meant... Where can I learn this fact and the proof? – 2017-01-24
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    Not sure if I need to tag you @Noah Schweber – 2017-01-24
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    @XJ We actually don't know what you mean by an "infinite row" (this is not standard terminology)! Can you give some examples? – 2017-01-24
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Sure, assuming you accept that the phrase "infinite number" has meaning.

How about $N = \sum_0^{i=\infty} 2^{2^i}$ ? This ends with: $\dots10001011$ The number of zeroes increases going to the left, but there's no repeating pattern.

If you want a finite number, then how about: $N = \sum_0^{i=\infty} 2^{-2^{i}}$? That comes to: $0.11010001\dots$

Again, no repetition.

Another number would be $\pi$ represented in binary; that has no infinitely repeating patterns.

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    $\pi$ is generally believed to be normal, but it is unproven. – 2017-01-24
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    @DougM The fact that $\pi$ is irrational is enough to exclude eventually periodic digits. That's one interpretation of "infinitely repeating". On the other hand, it is known that $\pi$ contains infinitely many instances of the substring "01". Is that what "infinitely repeating" means? Who knows (until the OP clarifies)... – 2017-01-24
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The Thue-Morse sequence has a very low amount of repetition in the following precise sense:

For any binary string $X$ of length $\ge 1$, the sequence never contains three consecutive copies of $X$, e.g. if $X$ is $011$ then the sequence completely avoids the string $011011011$. So not only are there no consecutive repeating patterns of infinite length (which would be true for any irrational number), but in fact the number of consecutive repetitions of any pattern is uniformly bounded (by $2$).