-0.000 000 000 742 147 676 646 719 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 719(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
-0.000 000 000 742 147 676 646 719(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:

|-0.000 000 000 742 147 676 646 719| = 0.000 000 000 742 147 676 646 719


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: 0.000 000 000 742 147 676 646 719.

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) 0.000 000 000 742 147 676 646 719 × 2 = 0 + 0.000 000 001 484 295 353 293 438;
  • 2) 0.000 000 001 484 295 353 293 438 × 2 = 0 + 0.000 000 002 968 590 706 586 876;
  • 3) 0.000 000 002 968 590 706 586 876 × 2 = 0 + 0.000 000 005 937 181 413 173 752;
  • 4) 0.000 000 005 937 181 413 173 752 × 2 = 0 + 0.000 000 011 874 362 826 347 504;
  • 5) 0.000 000 011 874 362 826 347 504 × 2 = 0 + 0.000 000 023 748 725 652 695 008;
  • 6) 0.000 000 023 748 725 652 695 008 × 2 = 0 + 0.000 000 047 497 451 305 390 016;
  • 7) 0.000 000 047 497 451 305 390 016 × 2 = 0 + 0.000 000 094 994 902 610 780 032;
  • 8) 0.000 000 094 994 902 610 780 032 × 2 = 0 + 0.000 000 189 989 805 221 560 064;
  • 9) 0.000 000 189 989 805 221 560 064 × 2 = 0 + 0.000 000 379 979 610 443 120 128;
  • 10) 0.000 000 379 979 610 443 120 128 × 2 = 0 + 0.000 000 759 959 220 886 240 256;
  • 11) 0.000 000 759 959 220 886 240 256 × 2 = 0 + 0.000 001 519 918 441 772 480 512;
  • 12) 0.000 001 519 918 441 772 480 512 × 2 = 0 + 0.000 003 039 836 883 544 961 024;
  • 13) 0.000 003 039 836 883 544 961 024 × 2 = 0 + 0.000 006 079 673 767 089 922 048;
  • 14) 0.000 006 079 673 767 089 922 048 × 2 = 0 + 0.000 012 159 347 534 179 844 096;
  • 15) 0.000 012 159 347 534 179 844 096 × 2 = 0 + 0.000 024 318 695 068 359 688 192;
  • 16) 0.000 024 318 695 068 359 688 192 × 2 = 0 + 0.000 048 637 390 136 719 376 384;
  • 17) 0.000 048 637 390 136 719 376 384 × 2 = 0 + 0.000 097 274 780 273 438 752 768;
  • 18) 0.000 097 274 780 273 438 752 768 × 2 = 0 + 0.000 194 549 560 546 877 505 536;
  • 19) 0.000 194 549 560 546 877 505 536 × 2 = 0 + 0.000 389 099 121 093 755 011 072;
  • 20) 0.000 389 099 121 093 755 011 072 × 2 = 0 + 0.000 778 198 242 187 510 022 144;
  • 21) 0.000 778 198 242 187 510 022 144 × 2 = 0 + 0.001 556 396 484 375 020 044 288;
  • 22) 0.001 556 396 484 375 020 044 288 × 2 = 0 + 0.003 112 792 968 750 040 088 576;
  • 23) 0.003 112 792 968 750 040 088 576 × 2 = 0 + 0.006 225 585 937 500 080 177 152;
  • 24) 0.006 225 585 937 500 080 177 152 × 2 = 0 + 0.012 451 171 875 000 160 354 304;
  • 25) 0.012 451 171 875 000 160 354 304 × 2 = 0 + 0.024 902 343 750 000 320 708 608;
  • 26) 0.024 902 343 750 000 320 708 608 × 2 = 0 + 0.049 804 687 500 000 641 417 216;
  • 27) 0.049 804 687 500 000 641 417 216 × 2 = 0 + 0.099 609 375 000 001 282 834 432;
  • 28) 0.099 609 375 000 001 282 834 432 × 2 = 0 + 0.199 218 750 000 002 565 668 864;
  • 29) 0.199 218 750 000 002 565 668 864 × 2 = 0 + 0.398 437 500 000 005 131 337 728;
  • 30) 0.398 437 500 000 005 131 337 728 × 2 = 0 + 0.796 875 000 000 010 262 675 456;
  • 31) 0.796 875 000 000 010 262 675 456 × 2 = 1 + 0.593 750 000 000 020 525 350 912;
  • 32) 0.593 750 000 000 020 525 350 912 × 2 = 1 + 0.187 500 000 000 041 050 701 824;
  • 33) 0.187 500 000 000 041 050 701 824 × 2 = 0 + 0.375 000 000 000 082 101 403 648;
  • 34) 0.375 000 000 000 082 101 403 648 × 2 = 0 + 0.750 000 000 000 164 202 807 296;
  • 35) 0.750 000 000 000 164 202 807 296 × 2 = 1 + 0.500 000 000 000 328 405 614 592;
  • 36) 0.500 000 000 000 328 405 614 592 × 2 = 1 + 0.000 000 000 000 656 811 229 184;
  • 37) 0.000 000 000 000 656 811 229 184 × 2 = 0 + 0.000 000 000 001 313 622 458 368;
  • 38) 0.000 000 000 001 313 622 458 368 × 2 = 0 + 0.000 000 000 002 627 244 916 736;
  • 39) 0.000 000 000 002 627 244 916 736 × 2 = 0 + 0.000 000 000 005 254 489 833 472;
  • 40) 0.000 000 000 005 254 489 833 472 × 2 = 0 + 0.000 000 000 010 508 979 666 944;
  • 41) 0.000 000 000 010 508 979 666 944 × 2 = 0 + 0.000 000 000 021 017 959 333 888;
  • 42) 0.000 000 000 021 017 959 333 888 × 2 = 0 + 0.000 000 000 042 035 918 667 776;
  • 43) 0.000 000 000 042 035 918 667 776 × 2 = 0 + 0.000 000 000 084 071 837 335 552;
  • 44) 0.000 000 000 084 071 837 335 552 × 2 = 0 + 0.000 000 000 168 143 674 671 104;
  • 45) 0.000 000 000 168 143 674 671 104 × 2 = 0 + 0.000 000 000 336 287 349 342 208;
  • 46) 0.000 000 000 336 287 349 342 208 × 2 = 0 + 0.000 000 000 672 574 698 684 416;
  • 47) 0.000 000 000 672 574 698 684 416 × 2 = 0 + 0.000 000 001 345 149 397 368 832;
  • 48) 0.000 000 001 345 149 397 368 832 × 2 = 0 + 0.000 000 002 690 298 794 737 664;
  • 49) 0.000 000 002 690 298 794 737 664 × 2 = 0 + 0.000 000 005 380 597 589 475 328;
  • 50) 0.000 000 005 380 597 589 475 328 × 2 = 0 + 0.000 000 010 761 195 178 950 656;
  • 51) 0.000 000 010 761 195 178 950 656 × 2 = 0 + 0.000 000 021 522 390 357 901 312;
  • 52) 0.000 000 021 522 390 357 901 312 × 2 = 0 + 0.000 000 043 044 780 715 802 624;
  • 53) 0.000 000 043 044 780 715 802 624 × 2 = 0 + 0.000 000 086 089 561 431 605 248;
  • 54) 0.000 000 086 089 561 431 605 248 × 2 = 0 + 0.000 000 172 179 122 863 210 496;

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:


0.000 000 000 742 147 676 646 719(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 719(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:


0.000 000 000 742 147 676 646 719(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 -0.000 000 000 742 147 676 646 719 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 646 76 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 minutes

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