# Convert 0.000 000 000 000 076 234 564 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

## 0.000 000 000 000 076 234 564(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

### 1. First, convert to the binary (base 2) the integer part: 0. Divide the number repeatedly by 2.

#### We stop when we get a quotient that is equal to zero.

• division = quotient + remainder;
• 0 ÷ 2 = 0 + 0;

### 3. Convert to the binary (base 2) the fractional part: 0.000 000 000 000 076 234 564.

#### Stop when we get a fractional part that is equal to zero.

• #) multiplying = integer + fractional part;
• 1) 0.000 000 000 000 076 234 564 × 2 = 0 + 0.000 000 000 000 152 469 128;
• 2) 0.000 000 000 000 152 469 128 × 2 = 0 + 0.000 000 000 000 304 938 256;
• 3) 0.000 000 000 000 304 938 256 × 2 = 0 + 0.000 000 000 000 609 876 512;
• 4) 0.000 000 000 000 609 876 512 × 2 = 0 + 0.000 000 000 001 219 753 024;
• 5) 0.000 000 000 001 219 753 024 × 2 = 0 + 0.000 000 000 002 439 506 048;
• 6) 0.000 000 000 002 439 506 048 × 2 = 0 + 0.000 000 000 004 879 012 096;
• 7) 0.000 000 000 004 879 012 096 × 2 = 0 + 0.000 000 000 009 758 024 192;
• 8) 0.000 000 000 009 758 024 192 × 2 = 0 + 0.000 000 000 019 516 048 384;
• 9) 0.000 000 000 019 516 048 384 × 2 = 0 + 0.000 000 000 039 032 096 768;
• 10) 0.000 000 000 039 032 096 768 × 2 = 0 + 0.000 000 000 078 064 193 536;
• 11) 0.000 000 000 078 064 193 536 × 2 = 0 + 0.000 000 000 156 128 387 072;
• 12) 0.000 000 000 156 128 387 072 × 2 = 0 + 0.000 000 000 312 256 774 144;
• 13) 0.000 000 000 312 256 774 144 × 2 = 0 + 0.000 000 000 624 513 548 288;
• 14) 0.000 000 000 624 513 548 288 × 2 = 0 + 0.000 000 001 249 027 096 576;
• 15) 0.000 000 001 249 027 096 576 × 2 = 0 + 0.000 000 002 498 054 193 152;
• 16) 0.000 000 002 498 054 193 152 × 2 = 0 + 0.000 000 004 996 108 386 304;
• 17) 0.000 000 004 996 108 386 304 × 2 = 0 + 0.000 000 009 992 216 772 608;
• 18) 0.000 000 009 992 216 772 608 × 2 = 0 + 0.000 000 019 984 433 545 216;
• 19) 0.000 000 019 984 433 545 216 × 2 = 0 + 0.000 000 039 968 867 090 432;
• 20) 0.000 000 039 968 867 090 432 × 2 = 0 + 0.000 000 079 937 734 180 864;
• 21) 0.000 000 079 937 734 180 864 × 2 = 0 + 0.000 000 159 875 468 361 728;
• 22) 0.000 000 159 875 468 361 728 × 2 = 0 + 0.000 000 319 750 936 723 456;
• 23) 0.000 000 319 750 936 723 456 × 2 = 0 + 0.000 000 639 501 873 446 912;
• 24) 0.000 000 639 501 873 446 912 × 2 = 0 + 0.000 001 279 003 746 893 824;
• 25) 0.000 001 279 003 746 893 824 × 2 = 0 + 0.000 002 558 007 493 787 648;
• 26) 0.000 002 558 007 493 787 648 × 2 = 0 + 0.000 005 116 014 987 575 296;
• 27) 0.000 005 116 014 987 575 296 × 2 = 0 + 0.000 010 232 029 975 150 592;
• 28) 0.000 010 232 029 975 150 592 × 2 = 0 + 0.000 020 464 059 950 301 184;
• 29) 0.000 020 464 059 950 301 184 × 2 = 0 + 0.000 040 928 119 900 602 368;
• 30) 0.000 040 928 119 900 602 368 × 2 = 0 + 0.000 081 856 239 801 204 736;
• 31) 0.000 081 856 239 801 204 736 × 2 = 0 + 0.000 163 712 479 602 409 472;
• 32) 0.000 163 712 479 602 409 472 × 2 = 0 + 0.000 327 424 959 204 818 944;
• 33) 0.000 327 424 959 204 818 944 × 2 = 0 + 0.000 654 849 918 409 637 888;
• 34) 0.000 654 849 918 409 637 888 × 2 = 0 + 0.001 309 699 836 819 275 776;
• 35) 0.001 309 699 836 819 275 776 × 2 = 0 + 0.002 619 399 673 638 551 552;
• 36) 0.002 619 399 673 638 551 552 × 2 = 0 + 0.005 238 799 347 277 103 104;
• 37) 0.005 238 799 347 277 103 104 × 2 = 0 + 0.010 477 598 694 554 206 208;
• 38) 0.010 477 598 694 554 206 208 × 2 = 0 + 0.020 955 197 389 108 412 416;
• 39) 0.020 955 197 389 108 412 416 × 2 = 0 + 0.041 910 394 778 216 824 832;
• 40) 0.041 910 394 778 216 824 832 × 2 = 0 + 0.083 820 789 556 433 649 664;
• 41) 0.083 820 789 556 433 649 664 × 2 = 0 + 0.167 641 579 112 867 299 328;
• 42) 0.167 641 579 112 867 299 328 × 2 = 0 + 0.335 283 158 225 734 598 656;
• 43) 0.335 283 158 225 734 598 656 × 2 = 0 + 0.670 566 316 451 469 197 312;
• 44) 0.670 566 316 451 469 197 312 × 2 = 1 + 0.341 132 632 902 938 394 624;
• 45) 0.341 132 632 902 938 394 624 × 2 = 0 + 0.682 265 265 805 876 789 248;
• 46) 0.682 265 265 805 876 789 248 × 2 = 1 + 0.364 530 531 611 753 578 496;
• 47) 0.364 530 531 611 753 578 496 × 2 = 0 + 0.729 061 063 223 507 156 992;
• 48) 0.729 061 063 223 507 156 992 × 2 = 1 + 0.458 122 126 447 014 313 984;
• 49) 0.458 122 126 447 014 313 984 × 2 = 0 + 0.916 244 252 894 028 627 968;
• 50) 0.916 244 252 894 028 627 968 × 2 = 1 + 0.832 488 505 788 057 255 936;
• 51) 0.832 488 505 788 057 255 936 × 2 = 1 + 0.664 977 011 576 114 511 872;
• 52) 0.664 977 011 576 114 511 872 × 2 = 1 + 0.329 954 023 152 229 023 744;
• 53) 0.329 954 023 152 229 023 744 × 2 = 0 + 0.659 908 046 304 458 047 488;
• 54) 0.659 908 046 304 458 047 488 × 2 = 1 + 0.319 816 092 608 916 094 976;
• 55) 0.319 816 092 608 916 094 976 × 2 = 0 + 0.639 632 185 217 832 189 952;
• 56) 0.639 632 185 217 832 189 952 × 2 = 1 + 0.279 264 370 435 664 379 904;
• 57) 0.279 264 370 435 664 379 904 × 2 = 0 + 0.558 528 740 871 328 759 808;
• 58) 0.558 528 740 871 328 759 808 × 2 = 1 + 0.117 057 481 742 657 519 616;
• 59) 0.117 057 481 742 657 519 616 × 2 = 0 + 0.234 114 963 485 315 039 232;
• 60) 0.234 114 963 485 315 039 232 × 2 = 0 + 0.468 229 926 970 630 078 464;
• 61) 0.468 229 926 970 630 078 464 × 2 = 0 + 0.936 459 853 941 260 156 928;
• 62) 0.936 459 853 941 260 156 928 × 2 = 1 + 0.872 919 707 882 520 313 856;
• 63) 0.872 919 707 882 520 313 856 × 2 = 1 + 0.745 839 415 765 040 627 712;
• 64) 0.745 839 415 765 040 627 712 × 2 = 1 + 0.491 678 831 530 081 255 424;
• 65) 0.491 678 831 530 081 255 424 × 2 = 0 + 0.983 357 663 060 162 510 848;
• 66) 0.983 357 663 060 162 510 848 × 2 = 1 + 0.966 715 326 120 325 021 696;
• 67) 0.966 715 326 120 325 021 696 × 2 = 1 + 0.933 430 652 240 650 043 392;

### 9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

#### Use the same technique of repeatedly dividing by 2:

• division = quotient + remainder;
• 83 ÷ 2 = 41 + 1;
• 41 ÷ 2 = 20 + 1;
• 20 ÷ 2 = 10 + 0;
• 10 ÷ 2 = 5 + 0;
• 5 ÷ 2 = 2 + 1;
• 2 ÷ 2 = 1 + 0;
• 1 ÷ 2 = 0 + 1;

## Number 0.000 000 000 000 076 234 564 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point: 0 - 0101 0011 - 010 1011 1010 1010 0011 1011

(32 bits IEEE 754)

• 0

31

• 0

30
• 1

29
• 0

28
• 1

27
• 0

26
• 0

25
• 1

24
• 1

23

• 0

22
• 1

21
• 0

20
• 1

19
• 0

18
• 1

17
• 1

16
• 1

15
• 0

14
• 1

13
• 0

12
• 1

11
• 0

10
• 1

9
• 0

8
• 0

7
• 0

6
• 1

5
• 1

4
• 1

3
• 0

2
• 1

1
• 1

0

## Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

 0.000 000 000 000 076 234 564 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:42 UTC (GMT) 15 118 285 380 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:42 UTC (GMT) 0.000 003 669 45 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:42 UTC (GMT) -5.121 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:42 UTC (GMT) 12.840 800 2 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:42 UTC (GMT) 0.021 696 090 8 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:42 UTC (GMT) 13 771 013 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:42 UTC (GMT) -0.000 000 000 000 000 000 000 000 000 000 000 000 005 3 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:42 UTC (GMT) -0.122 070 1 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:41 UTC (GMT) 0.076 923 076 924 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:41 UTC (GMT) 183.2 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:41 UTC (GMT) 4.150 3 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:41 UTC (GMT) 26.541 to 32 bit single precision IEEE 754 binary floating point = ? May 12 07:41 UTC (GMT) All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

## How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

### Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

• 1. If the number to be converted is negative, start with its the positive version.
• 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
• 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
• 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
• 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
• 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
• 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
• 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
• 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

### Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

|-25.347| = 25.347

• 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
• division = quotient + remainder;
• 25 ÷ 2 = 12 + 1;
• 12 ÷ 2 = 6 + 0;
• 6 ÷ 2 = 3 + 0;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;
• We have encountered a quotient that is ZERO => FULL STOP
• 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

25(10) = 1 1001(2)

• 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
• #) multiplying = integer + fractional part;
• 1) 0.347 × 2 = 0 + 0.694;
• 2) 0.694 × 2 = 1 + 0.388;
• 3) 0.388 × 2 = 0 + 0.776;
• 4) 0.776 × 2 = 1 + 0.552;
• 5) 0.552 × 2 = 1 + 0.104;
• 6) 0.104 × 2 = 0 + 0.208;
• 7) 0.208 × 2 = 0 + 0.416;
• 8) 0.416 × 2 = 0 + 0.832;
• 9) 0.832 × 2 = 1 + 0.664;
• 10) 0.664 × 2 = 1 + 0.328;
• 11) 0.328 × 2 = 0 + 0.656;
• 12) 0.656 × 2 = 1 + 0.312;
• 13) 0.312 × 2 = 0 + 0.624;
• 14) 0.624 × 2 = 1 + 0.248;
• 15) 0.248 × 2 = 0 + 0.496;
• 16) 0.496 × 2 = 0 + 0.992;
• 17) 0.992 × 2 = 1 + 0.984;
• 18) 0.984 × 2 = 1 + 0.968;
• 19) 0.968 × 2 = 1 + 0.936;
• 20) 0.936 × 2 = 1 + 0.872;
• 21) 0.872 × 2 = 1 + 0.744;
• 22) 0.744 × 2 = 1 + 0.488;
• 23) 0.488 × 2 = 0 + 0.976;
• 24) 0.976 × 2 = 1 + 0.952;
• We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
• 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

• 6. Summarizing - the positive number before normalization:

25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

• 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

25.347(10) =
1 1001.0101 1000 1101 0100 1111 1101(2) =
1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
1.1001 0101 1000 1101 0100 1111 1101(2) × 24

• 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

Sign: 1 (a negative number)

Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

• 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
1000 0011(2)

• 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

Mantissa (normalized): 100 1010 1100 0110 1010 0111

• Conclusion:

Sign (1 bit) = 1 (a negative number)

Exponent (8 bits) = 1000 0011

Mantissa (23 bits) = 100 1010 1100 0110 1010 0111