-0.000 000 000 742 147 676 648 95 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 648 95(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
-0.000 000 000 742 147 676 648 95(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:

|-0.000 000 000 742 147 676 648 95| = 0.000 000 000 742 147 676 648 95


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: 0.000 000 000 742 147 676 648 95.

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) 0.000 000 000 742 147 676 648 95 × 2 = 0 + 0.000 000 001 484 295 353 297 9;
  • 2) 0.000 000 001 484 295 353 297 9 × 2 = 0 + 0.000 000 002 968 590 706 595 8;
  • 3) 0.000 000 002 968 590 706 595 8 × 2 = 0 + 0.000 000 005 937 181 413 191 6;
  • 4) 0.000 000 005 937 181 413 191 6 × 2 = 0 + 0.000 000 011 874 362 826 383 2;
  • 5) 0.000 000 011 874 362 826 383 2 × 2 = 0 + 0.000 000 023 748 725 652 766 4;
  • 6) 0.000 000 023 748 725 652 766 4 × 2 = 0 + 0.000 000 047 497 451 305 532 8;
  • 7) 0.000 000 047 497 451 305 532 8 × 2 = 0 + 0.000 000 094 994 902 611 065 6;
  • 8) 0.000 000 094 994 902 611 065 6 × 2 = 0 + 0.000 000 189 989 805 222 131 2;
  • 9) 0.000 000 189 989 805 222 131 2 × 2 = 0 + 0.000 000 379 979 610 444 262 4;
  • 10) 0.000 000 379 979 610 444 262 4 × 2 = 0 + 0.000 000 759 959 220 888 524 8;
  • 11) 0.000 000 759 959 220 888 524 8 × 2 = 0 + 0.000 001 519 918 441 777 049 6;
  • 12) 0.000 001 519 918 441 777 049 6 × 2 = 0 + 0.000 003 039 836 883 554 099 2;
  • 13) 0.000 003 039 836 883 554 099 2 × 2 = 0 + 0.000 006 079 673 767 108 198 4;
  • 14) 0.000 006 079 673 767 108 198 4 × 2 = 0 + 0.000 012 159 347 534 216 396 8;
  • 15) 0.000 012 159 347 534 216 396 8 × 2 = 0 + 0.000 024 318 695 068 432 793 6;
  • 16) 0.000 024 318 695 068 432 793 6 × 2 = 0 + 0.000 048 637 390 136 865 587 2;
  • 17) 0.000 048 637 390 136 865 587 2 × 2 = 0 + 0.000 097 274 780 273 731 174 4;
  • 18) 0.000 097 274 780 273 731 174 4 × 2 = 0 + 0.000 194 549 560 547 462 348 8;
  • 19) 0.000 194 549 560 547 462 348 8 × 2 = 0 + 0.000 389 099 121 094 924 697 6;
  • 20) 0.000 389 099 121 094 924 697 6 × 2 = 0 + 0.000 778 198 242 189 849 395 2;
  • 21) 0.000 778 198 242 189 849 395 2 × 2 = 0 + 0.001 556 396 484 379 698 790 4;
  • 22) 0.001 556 396 484 379 698 790 4 × 2 = 0 + 0.003 112 792 968 759 397 580 8;
  • 23) 0.003 112 792 968 759 397 580 8 × 2 = 0 + 0.006 225 585 937 518 795 161 6;
  • 24) 0.006 225 585 937 518 795 161 6 × 2 = 0 + 0.012 451 171 875 037 590 323 2;
  • 25) 0.012 451 171 875 037 590 323 2 × 2 = 0 + 0.024 902 343 750 075 180 646 4;
  • 26) 0.024 902 343 750 075 180 646 4 × 2 = 0 + 0.049 804 687 500 150 361 292 8;
  • 27) 0.049 804 687 500 150 361 292 8 × 2 = 0 + 0.099 609 375 000 300 722 585 6;
  • 28) 0.099 609 375 000 300 722 585 6 × 2 = 0 + 0.199 218 750 000 601 445 171 2;
  • 29) 0.199 218 750 000 601 445 171 2 × 2 = 0 + 0.398 437 500 001 202 890 342 4;
  • 30) 0.398 437 500 001 202 890 342 4 × 2 = 0 + 0.796 875 000 002 405 780 684 8;
  • 31) 0.796 875 000 002 405 780 684 8 × 2 = 1 + 0.593 750 000 004 811 561 369 6;
  • 32) 0.593 750 000 004 811 561 369 6 × 2 = 1 + 0.187 500 000 009 623 122 739 2;
  • 33) 0.187 500 000 009 623 122 739 2 × 2 = 0 + 0.375 000 000 019 246 245 478 4;
  • 34) 0.375 000 000 019 246 245 478 4 × 2 = 0 + 0.750 000 000 038 492 490 956 8;
  • 35) 0.750 000 000 038 492 490 956 8 × 2 = 1 + 0.500 000 000 076 984 981 913 6;
  • 36) 0.500 000 000 076 984 981 913 6 × 2 = 1 + 0.000 000 000 153 969 963 827 2;
  • 37) 0.000 000 000 153 969 963 827 2 × 2 = 0 + 0.000 000 000 307 939 927 654 4;
  • 38) 0.000 000 000 307 939 927 654 4 × 2 = 0 + 0.000 000 000 615 879 855 308 8;
  • 39) 0.000 000 000 615 879 855 308 8 × 2 = 0 + 0.000 000 001 231 759 710 617 6;
  • 40) 0.000 000 001 231 759 710 617 6 × 2 = 0 + 0.000 000 002 463 519 421 235 2;
  • 41) 0.000 000 002 463 519 421 235 2 × 2 = 0 + 0.000 000 004 927 038 842 470 4;
  • 42) 0.000 000 004 927 038 842 470 4 × 2 = 0 + 0.000 000 009 854 077 684 940 8;
  • 43) 0.000 000 009 854 077 684 940 8 × 2 = 0 + 0.000 000 019 708 155 369 881 6;
  • 44) 0.000 000 019 708 155 369 881 6 × 2 = 0 + 0.000 000 039 416 310 739 763 2;
  • 45) 0.000 000 039 416 310 739 763 2 × 2 = 0 + 0.000 000 078 832 621 479 526 4;
  • 46) 0.000 000 078 832 621 479 526 4 × 2 = 0 + 0.000 000 157 665 242 959 052 8;
  • 47) 0.000 000 157 665 242 959 052 8 × 2 = 0 + 0.000 000 315 330 485 918 105 6;
  • 48) 0.000 000 315 330 485 918 105 6 × 2 = 0 + 0.000 000 630 660 971 836 211 2;
  • 49) 0.000 000 630 660 971 836 211 2 × 2 = 0 + 0.000 001 261 321 943 672 422 4;
  • 50) 0.000 001 261 321 943 672 422 4 × 2 = 0 + 0.000 002 522 643 887 344 844 8;
  • 51) 0.000 002 522 643 887 344 844 8 × 2 = 0 + 0.000 005 045 287 774 689 689 6;
  • 52) 0.000 005 045 287 774 689 689 6 × 2 = 0 + 0.000 010 090 575 549 379 379 2;
  • 53) 0.000 010 090 575 549 379 379 2 × 2 = 0 + 0.000 020 181 151 098 758 758 4;
  • 54) 0.000 020 181 151 098 758 758 4 × 2 = 0 + 0.000 040 362 302 197 517 516 8;

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:


0.000 000 000 742 147 676 648 95(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 648 95(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:


0.000 000 000 742 147 676 648 95(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 -0.000 000 000 742 147 676 648 95 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