-0.000 000 000 742 147 676 336 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 336(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
-0.000 000 000 742 147 676 336(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:

|-0.000 000 000 742 147 676 336| = 0.000 000 000 742 147 676 336


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: 0.000 000 000 742 147 676 336.

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) 0.000 000 000 742 147 676 336 × 2 = 0 + 0.000 000 001 484 295 352 672;
  • 2) 0.000 000 001 484 295 352 672 × 2 = 0 + 0.000 000 002 968 590 705 344;
  • 3) 0.000 000 002 968 590 705 344 × 2 = 0 + 0.000 000 005 937 181 410 688;
  • 4) 0.000 000 005 937 181 410 688 × 2 = 0 + 0.000 000 011 874 362 821 376;
  • 5) 0.000 000 011 874 362 821 376 × 2 = 0 + 0.000 000 023 748 725 642 752;
  • 6) 0.000 000 023 748 725 642 752 × 2 = 0 + 0.000 000 047 497 451 285 504;
  • 7) 0.000 000 047 497 451 285 504 × 2 = 0 + 0.000 000 094 994 902 571 008;
  • 8) 0.000 000 094 994 902 571 008 × 2 = 0 + 0.000 000 189 989 805 142 016;
  • 9) 0.000 000 189 989 805 142 016 × 2 = 0 + 0.000 000 379 979 610 284 032;
  • 10) 0.000 000 379 979 610 284 032 × 2 = 0 + 0.000 000 759 959 220 568 064;
  • 11) 0.000 000 759 959 220 568 064 × 2 = 0 + 0.000 001 519 918 441 136 128;
  • 12) 0.000 001 519 918 441 136 128 × 2 = 0 + 0.000 003 039 836 882 272 256;
  • 13) 0.000 003 039 836 882 272 256 × 2 = 0 + 0.000 006 079 673 764 544 512;
  • 14) 0.000 006 079 673 764 544 512 × 2 = 0 + 0.000 012 159 347 529 089 024;
  • 15) 0.000 012 159 347 529 089 024 × 2 = 0 + 0.000 024 318 695 058 178 048;
  • 16) 0.000 024 318 695 058 178 048 × 2 = 0 + 0.000 048 637 390 116 356 096;
  • 17) 0.000 048 637 390 116 356 096 × 2 = 0 + 0.000 097 274 780 232 712 192;
  • 18) 0.000 097 274 780 232 712 192 × 2 = 0 + 0.000 194 549 560 465 424 384;
  • 19) 0.000 194 549 560 465 424 384 × 2 = 0 + 0.000 389 099 120 930 848 768;
  • 20) 0.000 389 099 120 930 848 768 × 2 = 0 + 0.000 778 198 241 861 697 536;
  • 21) 0.000 778 198 241 861 697 536 × 2 = 0 + 0.001 556 396 483 723 395 072;
  • 22) 0.001 556 396 483 723 395 072 × 2 = 0 + 0.003 112 792 967 446 790 144;
  • 23) 0.003 112 792 967 446 790 144 × 2 = 0 + 0.006 225 585 934 893 580 288;
  • 24) 0.006 225 585 934 893 580 288 × 2 = 0 + 0.012 451 171 869 787 160 576;
  • 25) 0.012 451 171 869 787 160 576 × 2 = 0 + 0.024 902 343 739 574 321 152;
  • 26) 0.024 902 343 739 574 321 152 × 2 = 0 + 0.049 804 687 479 148 642 304;
  • 27) 0.049 804 687 479 148 642 304 × 2 = 0 + 0.099 609 374 958 297 284 608;
  • 28) 0.099 609 374 958 297 284 608 × 2 = 0 + 0.199 218 749 916 594 569 216;
  • 29) 0.199 218 749 916 594 569 216 × 2 = 0 + 0.398 437 499 833 189 138 432;
  • 30) 0.398 437 499 833 189 138 432 × 2 = 0 + 0.796 874 999 666 378 276 864;
  • 31) 0.796 874 999 666 378 276 864 × 2 = 1 + 0.593 749 999 332 756 553 728;
  • 32) 0.593 749 999 332 756 553 728 × 2 = 1 + 0.187 499 998 665 513 107 456;
  • 33) 0.187 499 998 665 513 107 456 × 2 = 0 + 0.374 999 997 331 026 214 912;
  • 34) 0.374 999 997 331 026 214 912 × 2 = 0 + 0.749 999 994 662 052 429 824;
  • 35) 0.749 999 994 662 052 429 824 × 2 = 1 + 0.499 999 989 324 104 859 648;
  • 36) 0.499 999 989 324 104 859 648 × 2 = 0 + 0.999 999 978 648 209 719 296;
  • 37) 0.999 999 978 648 209 719 296 × 2 = 1 + 0.999 999 957 296 419 438 592;
  • 38) 0.999 999 957 296 419 438 592 × 2 = 1 + 0.999 999 914 592 838 877 184;
  • 39) 0.999 999 914 592 838 877 184 × 2 = 1 + 0.999 999 829 185 677 754 368;
  • 40) 0.999 999 829 185 677 754 368 × 2 = 1 + 0.999 999 658 371 355 508 736;
  • 41) 0.999 999 658 371 355 508 736 × 2 = 1 + 0.999 999 316 742 711 017 472;
  • 42) 0.999 999 316 742 711 017 472 × 2 = 1 + 0.999 998 633 485 422 034 944;
  • 43) 0.999 998 633 485 422 034 944 × 2 = 1 + 0.999 997 266 970 844 069 888;
  • 44) 0.999 997 266 970 844 069 888 × 2 = 1 + 0.999 994 533 941 688 139 776;
  • 45) 0.999 994 533 941 688 139 776 × 2 = 1 + 0.999 989 067 883 376 279 552;
  • 46) 0.999 989 067 883 376 279 552 × 2 = 1 + 0.999 978 135 766 752 559 104;
  • 47) 0.999 978 135 766 752 559 104 × 2 = 1 + 0.999 956 271 533 505 118 208;
  • 48) 0.999 956 271 533 505 118 208 × 2 = 1 + 0.999 912 543 067 010 236 416;
  • 49) 0.999 912 543 067 010 236 416 × 2 = 1 + 0.999 825 086 134 020 472 832;
  • 50) 0.999 825 086 134 020 472 832 × 2 = 1 + 0.999 650 172 268 040 945 664;
  • 51) 0.999 650 172 268 040 945 664 × 2 = 1 + 0.999 300 344 536 081 891 328;
  • 52) 0.999 300 344 536 081 891 328 × 2 = 1 + 0.998 600 689 072 163 782 656;
  • 53) 0.998 600 689 072 163 782 656 × 2 = 1 + 0.997 201 378 144 327 565 312;
  • 54) 0.997 201 378 144 327 565 312 × 2 = 1 + 0.994 402 756 288 655 130 624;

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:


0.000 000 000 742 147 676 336(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 336(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:


0.000 000 000 742 147 676 336(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 -0.000 000 000 742 147 676 336 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