-0.000 000 000 742 147 676 646 84 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 84(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
-0.000 000 000 742 147 676 646 84(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:

|-0.000 000 000 742 147 676 646 84| = 0.000 000 000 742 147 676 646 84


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: 0.000 000 000 742 147 676 646 84.

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) 0.000 000 000 742 147 676 646 84 × 2 = 0 + 0.000 000 001 484 295 353 293 68;
  • 2) 0.000 000 001 484 295 353 293 68 × 2 = 0 + 0.000 000 002 968 590 706 587 36;
  • 3) 0.000 000 002 968 590 706 587 36 × 2 = 0 + 0.000 000 005 937 181 413 174 72;
  • 4) 0.000 000 005 937 181 413 174 72 × 2 = 0 + 0.000 000 011 874 362 826 349 44;
  • 5) 0.000 000 011 874 362 826 349 44 × 2 = 0 + 0.000 000 023 748 725 652 698 88;
  • 6) 0.000 000 023 748 725 652 698 88 × 2 = 0 + 0.000 000 047 497 451 305 397 76;
  • 7) 0.000 000 047 497 451 305 397 76 × 2 = 0 + 0.000 000 094 994 902 610 795 52;
  • 8) 0.000 000 094 994 902 610 795 52 × 2 = 0 + 0.000 000 189 989 805 221 591 04;
  • 9) 0.000 000 189 989 805 221 591 04 × 2 = 0 + 0.000 000 379 979 610 443 182 08;
  • 10) 0.000 000 379 979 610 443 182 08 × 2 = 0 + 0.000 000 759 959 220 886 364 16;
  • 11) 0.000 000 759 959 220 886 364 16 × 2 = 0 + 0.000 001 519 918 441 772 728 32;
  • 12) 0.000 001 519 918 441 772 728 32 × 2 = 0 + 0.000 003 039 836 883 545 456 64;
  • 13) 0.000 003 039 836 883 545 456 64 × 2 = 0 + 0.000 006 079 673 767 090 913 28;
  • 14) 0.000 006 079 673 767 090 913 28 × 2 = 0 + 0.000 012 159 347 534 181 826 56;
  • 15) 0.000 012 159 347 534 181 826 56 × 2 = 0 + 0.000 024 318 695 068 363 653 12;
  • 16) 0.000 024 318 695 068 363 653 12 × 2 = 0 + 0.000 048 637 390 136 727 306 24;
  • 17) 0.000 048 637 390 136 727 306 24 × 2 = 0 + 0.000 097 274 780 273 454 612 48;
  • 18) 0.000 097 274 780 273 454 612 48 × 2 = 0 + 0.000 194 549 560 546 909 224 96;
  • 19) 0.000 194 549 560 546 909 224 96 × 2 = 0 + 0.000 389 099 121 093 818 449 92;
  • 20) 0.000 389 099 121 093 818 449 92 × 2 = 0 + 0.000 778 198 242 187 636 899 84;
  • 21) 0.000 778 198 242 187 636 899 84 × 2 = 0 + 0.001 556 396 484 375 273 799 68;
  • 22) 0.001 556 396 484 375 273 799 68 × 2 = 0 + 0.003 112 792 968 750 547 599 36;
  • 23) 0.003 112 792 968 750 547 599 36 × 2 = 0 + 0.006 225 585 937 501 095 198 72;
  • 24) 0.006 225 585 937 501 095 198 72 × 2 = 0 + 0.012 451 171 875 002 190 397 44;
  • 25) 0.012 451 171 875 002 190 397 44 × 2 = 0 + 0.024 902 343 750 004 380 794 88;
  • 26) 0.024 902 343 750 004 380 794 88 × 2 = 0 + 0.049 804 687 500 008 761 589 76;
  • 27) 0.049 804 687 500 008 761 589 76 × 2 = 0 + 0.099 609 375 000 017 523 179 52;
  • 28) 0.099 609 375 000 017 523 179 52 × 2 = 0 + 0.199 218 750 000 035 046 359 04;
  • 29) 0.199 218 750 000 035 046 359 04 × 2 = 0 + 0.398 437 500 000 070 092 718 08;
  • 30) 0.398 437 500 000 070 092 718 08 × 2 = 0 + 0.796 875 000 000 140 185 436 16;
  • 31) 0.796 875 000 000 140 185 436 16 × 2 = 1 + 0.593 750 000 000 280 370 872 32;
  • 32) 0.593 750 000 000 280 370 872 32 × 2 = 1 + 0.187 500 000 000 560 741 744 64;
  • 33) 0.187 500 000 000 560 741 744 64 × 2 = 0 + 0.375 000 000 001 121 483 489 28;
  • 34) 0.375 000 000 001 121 483 489 28 × 2 = 0 + 0.750 000 000 002 242 966 978 56;
  • 35) 0.750 000 000 002 242 966 978 56 × 2 = 1 + 0.500 000 000 004 485 933 957 12;
  • 36) 0.500 000 000 004 485 933 957 12 × 2 = 1 + 0.000 000 000 008 971 867 914 24;
  • 37) 0.000 000 000 008 971 867 914 24 × 2 = 0 + 0.000 000 000 017 943 735 828 48;
  • 38) 0.000 000 000 017 943 735 828 48 × 2 = 0 + 0.000 000 000 035 887 471 656 96;
  • 39) 0.000 000 000 035 887 471 656 96 × 2 = 0 + 0.000 000 000 071 774 943 313 92;
  • 40) 0.000 000 000 071 774 943 313 92 × 2 = 0 + 0.000 000 000 143 549 886 627 84;
  • 41) 0.000 000 000 143 549 886 627 84 × 2 = 0 + 0.000 000 000 287 099 773 255 68;
  • 42) 0.000 000 000 287 099 773 255 68 × 2 = 0 + 0.000 000 000 574 199 546 511 36;
  • 43) 0.000 000 000 574 199 546 511 36 × 2 = 0 + 0.000 000 001 148 399 093 022 72;
  • 44) 0.000 000 001 148 399 093 022 72 × 2 = 0 + 0.000 000 002 296 798 186 045 44;
  • 45) 0.000 000 002 296 798 186 045 44 × 2 = 0 + 0.000 000 004 593 596 372 090 88;
  • 46) 0.000 000 004 593 596 372 090 88 × 2 = 0 + 0.000 000 009 187 192 744 181 76;
  • 47) 0.000 000 009 187 192 744 181 76 × 2 = 0 + 0.000 000 018 374 385 488 363 52;
  • 48) 0.000 000 018 374 385 488 363 52 × 2 = 0 + 0.000 000 036 748 770 976 727 04;
  • 49) 0.000 000 036 748 770 976 727 04 × 2 = 0 + 0.000 000 073 497 541 953 454 08;
  • 50) 0.000 000 073 497 541 953 454 08 × 2 = 0 + 0.000 000 146 995 083 906 908 16;
  • 51) 0.000 000 146 995 083 906 908 16 × 2 = 0 + 0.000 000 293 990 167 813 816 32;
  • 52) 0.000 000 293 990 167 813 816 32 × 2 = 0 + 0.000 000 587 980 335 627 632 64;
  • 53) 0.000 000 587 980 335 627 632 64 × 2 = 0 + 0.000 001 175 960 671 255 265 28;
  • 54) 0.000 001 175 960 671 255 265 28 × 2 = 0 + 0.000 002 351 921 342 510 530 56;

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:


0.000 000 000 742 147 676 646 84(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 84(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:


0.000 000 000 742 147 676 646 84(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 -0.000 000 000 742 147 676 646 84 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