-0.000 000 000 742 147 676 566 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 566(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 147 676 566(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 742 147 676 566| = 0.000 000 000 742 147 676 566


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

Keep track of each remainder.

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


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

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 742 147 676 566.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

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


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 000 742 147 676 566 × 2 = 0 + 0.000 000 001 484 295 353 132;
  • 2) 0.000 000 001 484 295 353 132 × 2 = 0 + 0.000 000 002 968 590 706 264;
  • 3) 0.000 000 002 968 590 706 264 × 2 = 0 + 0.000 000 005 937 181 412 528;
  • 4) 0.000 000 005 937 181 412 528 × 2 = 0 + 0.000 000 011 874 362 825 056;
  • 5) 0.000 000 011 874 362 825 056 × 2 = 0 + 0.000 000 023 748 725 650 112;
  • 6) 0.000 000 023 748 725 650 112 × 2 = 0 + 0.000 000 047 497 451 300 224;
  • 7) 0.000 000 047 497 451 300 224 × 2 = 0 + 0.000 000 094 994 902 600 448;
  • 8) 0.000 000 094 994 902 600 448 × 2 = 0 + 0.000 000 189 989 805 200 896;
  • 9) 0.000 000 189 989 805 200 896 × 2 = 0 + 0.000 000 379 979 610 401 792;
  • 10) 0.000 000 379 979 610 401 792 × 2 = 0 + 0.000 000 759 959 220 803 584;
  • 11) 0.000 000 759 959 220 803 584 × 2 = 0 + 0.000 001 519 918 441 607 168;
  • 12) 0.000 001 519 918 441 607 168 × 2 = 0 + 0.000 003 039 836 883 214 336;
  • 13) 0.000 003 039 836 883 214 336 × 2 = 0 + 0.000 006 079 673 766 428 672;
  • 14) 0.000 006 079 673 766 428 672 × 2 = 0 + 0.000 012 159 347 532 857 344;
  • 15) 0.000 012 159 347 532 857 344 × 2 = 0 + 0.000 024 318 695 065 714 688;
  • 16) 0.000 024 318 695 065 714 688 × 2 = 0 + 0.000 048 637 390 131 429 376;
  • 17) 0.000 048 637 390 131 429 376 × 2 = 0 + 0.000 097 274 780 262 858 752;
  • 18) 0.000 097 274 780 262 858 752 × 2 = 0 + 0.000 194 549 560 525 717 504;
  • 19) 0.000 194 549 560 525 717 504 × 2 = 0 + 0.000 389 099 121 051 435 008;
  • 20) 0.000 389 099 121 051 435 008 × 2 = 0 + 0.000 778 198 242 102 870 016;
  • 21) 0.000 778 198 242 102 870 016 × 2 = 0 + 0.001 556 396 484 205 740 032;
  • 22) 0.001 556 396 484 205 740 032 × 2 = 0 + 0.003 112 792 968 411 480 064;
  • 23) 0.003 112 792 968 411 480 064 × 2 = 0 + 0.006 225 585 936 822 960 128;
  • 24) 0.006 225 585 936 822 960 128 × 2 = 0 + 0.012 451 171 873 645 920 256;
  • 25) 0.012 451 171 873 645 920 256 × 2 = 0 + 0.024 902 343 747 291 840 512;
  • 26) 0.024 902 343 747 291 840 512 × 2 = 0 + 0.049 804 687 494 583 681 024;
  • 27) 0.049 804 687 494 583 681 024 × 2 = 0 + 0.099 609 374 989 167 362 048;
  • 28) 0.099 609 374 989 167 362 048 × 2 = 0 + 0.199 218 749 978 334 724 096;
  • 29) 0.199 218 749 978 334 724 096 × 2 = 0 + 0.398 437 499 956 669 448 192;
  • 30) 0.398 437 499 956 669 448 192 × 2 = 0 + 0.796 874 999 913 338 896 384;
  • 31) 0.796 874 999 913 338 896 384 × 2 = 1 + 0.593 749 999 826 677 792 768;
  • 32) 0.593 749 999 826 677 792 768 × 2 = 1 + 0.187 499 999 653 355 585 536;
  • 33) 0.187 499 999 653 355 585 536 × 2 = 0 + 0.374 999 999 306 711 171 072;
  • 34) 0.374 999 999 306 711 171 072 × 2 = 0 + 0.749 999 998 613 422 342 144;
  • 35) 0.749 999 998 613 422 342 144 × 2 = 1 + 0.499 999 997 226 844 684 288;
  • 36) 0.499 999 997 226 844 684 288 × 2 = 0 + 0.999 999 994 453 689 368 576;
  • 37) 0.999 999 994 453 689 368 576 × 2 = 1 + 0.999 999 988 907 378 737 152;
  • 38) 0.999 999 988 907 378 737 152 × 2 = 1 + 0.999 999 977 814 757 474 304;
  • 39) 0.999 999 977 814 757 474 304 × 2 = 1 + 0.999 999 955 629 514 948 608;
  • 40) 0.999 999 955 629 514 948 608 × 2 = 1 + 0.999 999 911 259 029 897 216;
  • 41) 0.999 999 911 259 029 897 216 × 2 = 1 + 0.999 999 822 518 059 794 432;
  • 42) 0.999 999 822 518 059 794 432 × 2 = 1 + 0.999 999 645 036 119 588 864;
  • 43) 0.999 999 645 036 119 588 864 × 2 = 1 + 0.999 999 290 072 239 177 728;
  • 44) 0.999 999 290 072 239 177 728 × 2 = 1 + 0.999 998 580 144 478 355 456;
  • 45) 0.999 998 580 144 478 355 456 × 2 = 1 + 0.999 997 160 288 956 710 912;
  • 46) 0.999 997 160 288 956 710 912 × 2 = 1 + 0.999 994 320 577 913 421 824;
  • 47) 0.999 994 320 577 913 421 824 × 2 = 1 + 0.999 988 641 155 826 843 648;
  • 48) 0.999 988 641 155 826 843 648 × 2 = 1 + 0.999 977 282 311 653 687 296;
  • 49) 0.999 977 282 311 653 687 296 × 2 = 1 + 0.999 954 564 623 307 374 592;
  • 50) 0.999 954 564 623 307 374 592 × 2 = 1 + 0.999 909 129 246 614 749 184;
  • 51) 0.999 909 129 246 614 749 184 × 2 = 1 + 0.999 818 258 493 229 498 368;
  • 52) 0.999 818 258 493 229 498 368 × 2 = 1 + 0.999 636 516 986 458 996 736;
  • 53) 0.999 636 516 986 458 996 736 × 2 = 1 + 0.999 273 033 972 917 993 472;
  • 54) 0.999 273 033 972 917 993 472 × 2 = 1 + 0.998 546 067 945 835 986 944;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 000 742 147 676 566(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 566(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 566(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


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

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. 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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

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

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 566 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 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:

  • 1. Start with the positive version of the number:

    |-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)

    Exponent (unadjusted): 4

    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

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111