-0.000 000 000 742 147 676 647 43 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 647 43(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 43(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 43| = 0.000 000 000 742 147 676 647 43


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 43.

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 43 × 2 = 0 + 0.000 000 001 484 295 353 294 86;
  • 2) 0.000 000 001 484 295 353 294 86 × 2 = 0 + 0.000 000 002 968 590 706 589 72;
  • 3) 0.000 000 002 968 590 706 589 72 × 2 = 0 + 0.000 000 005 937 181 413 179 44;
  • 4) 0.000 000 005 937 181 413 179 44 × 2 = 0 + 0.000 000 011 874 362 826 358 88;
  • 5) 0.000 000 011 874 362 826 358 88 × 2 = 0 + 0.000 000 023 748 725 652 717 76;
  • 6) 0.000 000 023 748 725 652 717 76 × 2 = 0 + 0.000 000 047 497 451 305 435 52;
  • 7) 0.000 000 047 497 451 305 435 52 × 2 = 0 + 0.000 000 094 994 902 610 871 04;
  • 8) 0.000 000 094 994 902 610 871 04 × 2 = 0 + 0.000 000 189 989 805 221 742 08;
  • 9) 0.000 000 189 989 805 221 742 08 × 2 = 0 + 0.000 000 379 979 610 443 484 16;
  • 10) 0.000 000 379 979 610 443 484 16 × 2 = 0 + 0.000 000 759 959 220 886 968 32;
  • 11) 0.000 000 759 959 220 886 968 32 × 2 = 0 + 0.000 001 519 918 441 773 936 64;
  • 12) 0.000 001 519 918 441 773 936 64 × 2 = 0 + 0.000 003 039 836 883 547 873 28;
  • 13) 0.000 003 039 836 883 547 873 28 × 2 = 0 + 0.000 006 079 673 767 095 746 56;
  • 14) 0.000 006 079 673 767 095 746 56 × 2 = 0 + 0.000 012 159 347 534 191 493 12;
  • 15) 0.000 012 159 347 534 191 493 12 × 2 = 0 + 0.000 024 318 695 068 382 986 24;
  • 16) 0.000 024 318 695 068 382 986 24 × 2 = 0 + 0.000 048 637 390 136 765 972 48;
  • 17) 0.000 048 637 390 136 765 972 48 × 2 = 0 + 0.000 097 274 780 273 531 944 96;
  • 18) 0.000 097 274 780 273 531 944 96 × 2 = 0 + 0.000 194 549 560 547 063 889 92;
  • 19) 0.000 194 549 560 547 063 889 92 × 2 = 0 + 0.000 389 099 121 094 127 779 84;
  • 20) 0.000 389 099 121 094 127 779 84 × 2 = 0 + 0.000 778 198 242 188 255 559 68;
  • 21) 0.000 778 198 242 188 255 559 68 × 2 = 0 + 0.001 556 396 484 376 511 119 36;
  • 22) 0.001 556 396 484 376 511 119 36 × 2 = 0 + 0.003 112 792 968 753 022 238 72;
  • 23) 0.003 112 792 968 753 022 238 72 × 2 = 0 + 0.006 225 585 937 506 044 477 44;
  • 24) 0.006 225 585 937 506 044 477 44 × 2 = 0 + 0.012 451 171 875 012 088 954 88;
  • 25) 0.012 451 171 875 012 088 954 88 × 2 = 0 + 0.024 902 343 750 024 177 909 76;
  • 26) 0.024 902 343 750 024 177 909 76 × 2 = 0 + 0.049 804 687 500 048 355 819 52;
  • 27) 0.049 804 687 500 048 355 819 52 × 2 = 0 + 0.099 609 375 000 096 711 639 04;
  • 28) 0.099 609 375 000 096 711 639 04 × 2 = 0 + 0.199 218 750 000 193 423 278 08;
  • 29) 0.199 218 750 000 193 423 278 08 × 2 = 0 + 0.398 437 500 000 386 846 556 16;
  • 30) 0.398 437 500 000 386 846 556 16 × 2 = 0 + 0.796 875 000 000 773 693 112 32;
  • 31) 0.796 875 000 000 773 693 112 32 × 2 = 1 + 0.593 750 000 001 547 386 224 64;
  • 32) 0.593 750 000 001 547 386 224 64 × 2 = 1 + 0.187 500 000 003 094 772 449 28;
  • 33) 0.187 500 000 003 094 772 449 28 × 2 = 0 + 0.375 000 000 006 189 544 898 56;
  • 34) 0.375 000 000 006 189 544 898 56 × 2 = 0 + 0.750 000 000 012 379 089 797 12;
  • 35) 0.750 000 000 012 379 089 797 12 × 2 = 1 + 0.500 000 000 024 758 179 594 24;
  • 36) 0.500 000 000 024 758 179 594 24 × 2 = 1 + 0.000 000 000 049 516 359 188 48;
  • 37) 0.000 000 000 049 516 359 188 48 × 2 = 0 + 0.000 000 000 099 032 718 376 96;
  • 38) 0.000 000 000 099 032 718 376 96 × 2 = 0 + 0.000 000 000 198 065 436 753 92;
  • 39) 0.000 000 000 198 065 436 753 92 × 2 = 0 + 0.000 000 000 396 130 873 507 84;
  • 40) 0.000 000 000 396 130 873 507 84 × 2 = 0 + 0.000 000 000 792 261 747 015 68;
  • 41) 0.000 000 000 792 261 747 015 68 × 2 = 0 + 0.000 000 001 584 523 494 031 36;
  • 42) 0.000 000 001 584 523 494 031 36 × 2 = 0 + 0.000 000 003 169 046 988 062 72;
  • 43) 0.000 000 003 169 046 988 062 72 × 2 = 0 + 0.000 000 006 338 093 976 125 44;
  • 44) 0.000 000 006 338 093 976 125 44 × 2 = 0 + 0.000 000 012 676 187 952 250 88;
  • 45) 0.000 000 012 676 187 952 250 88 × 2 = 0 + 0.000 000 025 352 375 904 501 76;
  • 46) 0.000 000 025 352 375 904 501 76 × 2 = 0 + 0.000 000 050 704 751 809 003 52;
  • 47) 0.000 000 050 704 751 809 003 52 × 2 = 0 + 0.000 000 101 409 503 618 007 04;
  • 48) 0.000 000 101 409 503 618 007 04 × 2 = 0 + 0.000 000 202 819 007 236 014 08;
  • 49) 0.000 000 202 819 007 236 014 08 × 2 = 0 + 0.000 000 405 638 014 472 028 16;
  • 50) 0.000 000 405 638 014 472 028 16 × 2 = 0 + 0.000 000 811 276 028 944 056 32;
  • 51) 0.000 000 811 276 028 944 056 32 × 2 = 0 + 0.000 001 622 552 057 888 112 64;
  • 52) 0.000 001 622 552 057 888 112 64 × 2 = 0 + 0.000 003 245 104 115 776 225 28;
  • 53) 0.000 003 245 104 115 776 225 28 × 2 = 0 + 0.000 006 490 208 231 552 450 56;
  • 54) 0.000 006 490 208 231 552 450 56 × 2 = 0 + 0.000 012 980 416 463 104 901 12;

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 43(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 43(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 43(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 43 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