-0.000 000 000 742 147 676 647 58 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 647 58(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
-0.000 000 000 742 147 676 647 58(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:

|-0.000 000 000 742 147 676 647 58| = 0.000 000 000 742 147 676 647 58


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: 0.000 000 000 742 147 676 647 58.

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) 0.000 000 000 742 147 676 647 58 × 2 = 0 + 0.000 000 001 484 295 353 295 16;
  • 2) 0.000 000 001 484 295 353 295 16 × 2 = 0 + 0.000 000 002 968 590 706 590 32;
  • 3) 0.000 000 002 968 590 706 590 32 × 2 = 0 + 0.000 000 005 937 181 413 180 64;
  • 4) 0.000 000 005 937 181 413 180 64 × 2 = 0 + 0.000 000 011 874 362 826 361 28;
  • 5) 0.000 000 011 874 362 826 361 28 × 2 = 0 + 0.000 000 023 748 725 652 722 56;
  • 6) 0.000 000 023 748 725 652 722 56 × 2 = 0 + 0.000 000 047 497 451 305 445 12;
  • 7) 0.000 000 047 497 451 305 445 12 × 2 = 0 + 0.000 000 094 994 902 610 890 24;
  • 8) 0.000 000 094 994 902 610 890 24 × 2 = 0 + 0.000 000 189 989 805 221 780 48;
  • 9) 0.000 000 189 989 805 221 780 48 × 2 = 0 + 0.000 000 379 979 610 443 560 96;
  • 10) 0.000 000 379 979 610 443 560 96 × 2 = 0 + 0.000 000 759 959 220 887 121 92;
  • 11) 0.000 000 759 959 220 887 121 92 × 2 = 0 + 0.000 001 519 918 441 774 243 84;
  • 12) 0.000 001 519 918 441 774 243 84 × 2 = 0 + 0.000 003 039 836 883 548 487 68;
  • 13) 0.000 003 039 836 883 548 487 68 × 2 = 0 + 0.000 006 079 673 767 096 975 36;
  • 14) 0.000 006 079 673 767 096 975 36 × 2 = 0 + 0.000 012 159 347 534 193 950 72;
  • 15) 0.000 012 159 347 534 193 950 72 × 2 = 0 + 0.000 024 318 695 068 387 901 44;
  • 16) 0.000 024 318 695 068 387 901 44 × 2 = 0 + 0.000 048 637 390 136 775 802 88;
  • 17) 0.000 048 637 390 136 775 802 88 × 2 = 0 + 0.000 097 274 780 273 551 605 76;
  • 18) 0.000 097 274 780 273 551 605 76 × 2 = 0 + 0.000 194 549 560 547 103 211 52;
  • 19) 0.000 194 549 560 547 103 211 52 × 2 = 0 + 0.000 389 099 121 094 206 423 04;
  • 20) 0.000 389 099 121 094 206 423 04 × 2 = 0 + 0.000 778 198 242 188 412 846 08;
  • 21) 0.000 778 198 242 188 412 846 08 × 2 = 0 + 0.001 556 396 484 376 825 692 16;
  • 22) 0.001 556 396 484 376 825 692 16 × 2 = 0 + 0.003 112 792 968 753 651 384 32;
  • 23) 0.003 112 792 968 753 651 384 32 × 2 = 0 + 0.006 225 585 937 507 302 768 64;
  • 24) 0.006 225 585 937 507 302 768 64 × 2 = 0 + 0.012 451 171 875 014 605 537 28;
  • 25) 0.012 451 171 875 014 605 537 28 × 2 = 0 + 0.024 902 343 750 029 211 074 56;
  • 26) 0.024 902 343 750 029 211 074 56 × 2 = 0 + 0.049 804 687 500 058 422 149 12;
  • 27) 0.049 804 687 500 058 422 149 12 × 2 = 0 + 0.099 609 375 000 116 844 298 24;
  • 28) 0.099 609 375 000 116 844 298 24 × 2 = 0 + 0.199 218 750 000 233 688 596 48;
  • 29) 0.199 218 750 000 233 688 596 48 × 2 = 0 + 0.398 437 500 000 467 377 192 96;
  • 30) 0.398 437 500 000 467 377 192 96 × 2 = 0 + 0.796 875 000 000 934 754 385 92;
  • 31) 0.796 875 000 000 934 754 385 92 × 2 = 1 + 0.593 750 000 001 869 508 771 84;
  • 32) 0.593 750 000 001 869 508 771 84 × 2 = 1 + 0.187 500 000 003 739 017 543 68;
  • 33) 0.187 500 000 003 739 017 543 68 × 2 = 0 + 0.375 000 000 007 478 035 087 36;
  • 34) 0.375 000 000 007 478 035 087 36 × 2 = 0 + 0.750 000 000 014 956 070 174 72;
  • 35) 0.750 000 000 014 956 070 174 72 × 2 = 1 + 0.500 000 000 029 912 140 349 44;
  • 36) 0.500 000 000 029 912 140 349 44 × 2 = 1 + 0.000 000 000 059 824 280 698 88;
  • 37) 0.000 000 000 059 824 280 698 88 × 2 = 0 + 0.000 000 000 119 648 561 397 76;
  • 38) 0.000 000 000 119 648 561 397 76 × 2 = 0 + 0.000 000 000 239 297 122 795 52;
  • 39) 0.000 000 000 239 297 122 795 52 × 2 = 0 + 0.000 000 000 478 594 245 591 04;
  • 40) 0.000 000 000 478 594 245 591 04 × 2 = 0 + 0.000 000 000 957 188 491 182 08;
  • 41) 0.000 000 000 957 188 491 182 08 × 2 = 0 + 0.000 000 001 914 376 982 364 16;
  • 42) 0.000 000 001 914 376 982 364 16 × 2 = 0 + 0.000 000 003 828 753 964 728 32;
  • 43) 0.000 000 003 828 753 964 728 32 × 2 = 0 + 0.000 000 007 657 507 929 456 64;
  • 44) 0.000 000 007 657 507 929 456 64 × 2 = 0 + 0.000 000 015 315 015 858 913 28;
  • 45) 0.000 000 015 315 015 858 913 28 × 2 = 0 + 0.000 000 030 630 031 717 826 56;
  • 46) 0.000 000 030 630 031 717 826 56 × 2 = 0 + 0.000 000 061 260 063 435 653 12;
  • 47) 0.000 000 061 260 063 435 653 12 × 2 = 0 + 0.000 000 122 520 126 871 306 24;
  • 48) 0.000 000 122 520 126 871 306 24 × 2 = 0 + 0.000 000 245 040 253 742 612 48;
  • 49) 0.000 000 245 040 253 742 612 48 × 2 = 0 + 0.000 000 490 080 507 485 224 96;
  • 50) 0.000 000 490 080 507 485 224 96 × 2 = 0 + 0.000 000 980 161 014 970 449 92;
  • 51) 0.000 000 980 161 014 970 449 92 × 2 = 0 + 0.000 001 960 322 029 940 899 84;
  • 52) 0.000 001 960 322 029 940 899 84 × 2 = 0 + 0.000 003 920 644 059 881 799 68;
  • 53) 0.000 003 920 644 059 881 799 68 × 2 = 0 + 0.000 007 841 288 119 763 599 36;
  • 54) 0.000 007 841 288 119 763 599 36 × 2 = 0 + 0.000 015 682 576 239 527 198 72;

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:


0.000 000 000 742 147 676 647 58(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 647 58(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 647 58(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 647 58 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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