-0.000 000 000 742 147 676 646 786 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 786(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
-0.000 000 000 742 147 676 646 786(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:

|-0.000 000 000 742 147 676 646 786| = 0.000 000 000 742 147 676 646 786


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: 0.000 000 000 742 147 676 646 786.

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) 0.000 000 000 742 147 676 646 786 × 2 = 0 + 0.000 000 001 484 295 353 293 572;
  • 2) 0.000 000 001 484 295 353 293 572 × 2 = 0 + 0.000 000 002 968 590 706 587 144;
  • 3) 0.000 000 002 968 590 706 587 144 × 2 = 0 + 0.000 000 005 937 181 413 174 288;
  • 4) 0.000 000 005 937 181 413 174 288 × 2 = 0 + 0.000 000 011 874 362 826 348 576;
  • 5) 0.000 000 011 874 362 826 348 576 × 2 = 0 + 0.000 000 023 748 725 652 697 152;
  • 6) 0.000 000 023 748 725 652 697 152 × 2 = 0 + 0.000 000 047 497 451 305 394 304;
  • 7) 0.000 000 047 497 451 305 394 304 × 2 = 0 + 0.000 000 094 994 902 610 788 608;
  • 8) 0.000 000 094 994 902 610 788 608 × 2 = 0 + 0.000 000 189 989 805 221 577 216;
  • 9) 0.000 000 189 989 805 221 577 216 × 2 = 0 + 0.000 000 379 979 610 443 154 432;
  • 10) 0.000 000 379 979 610 443 154 432 × 2 = 0 + 0.000 000 759 959 220 886 308 864;
  • 11) 0.000 000 759 959 220 886 308 864 × 2 = 0 + 0.000 001 519 918 441 772 617 728;
  • 12) 0.000 001 519 918 441 772 617 728 × 2 = 0 + 0.000 003 039 836 883 545 235 456;
  • 13) 0.000 003 039 836 883 545 235 456 × 2 = 0 + 0.000 006 079 673 767 090 470 912;
  • 14) 0.000 006 079 673 767 090 470 912 × 2 = 0 + 0.000 012 159 347 534 180 941 824;
  • 15) 0.000 012 159 347 534 180 941 824 × 2 = 0 + 0.000 024 318 695 068 361 883 648;
  • 16) 0.000 024 318 695 068 361 883 648 × 2 = 0 + 0.000 048 637 390 136 723 767 296;
  • 17) 0.000 048 637 390 136 723 767 296 × 2 = 0 + 0.000 097 274 780 273 447 534 592;
  • 18) 0.000 097 274 780 273 447 534 592 × 2 = 0 + 0.000 194 549 560 546 895 069 184;
  • 19) 0.000 194 549 560 546 895 069 184 × 2 = 0 + 0.000 389 099 121 093 790 138 368;
  • 20) 0.000 389 099 121 093 790 138 368 × 2 = 0 + 0.000 778 198 242 187 580 276 736;
  • 21) 0.000 778 198 242 187 580 276 736 × 2 = 0 + 0.001 556 396 484 375 160 553 472;
  • 22) 0.001 556 396 484 375 160 553 472 × 2 = 0 + 0.003 112 792 968 750 321 106 944;
  • 23) 0.003 112 792 968 750 321 106 944 × 2 = 0 + 0.006 225 585 937 500 642 213 888;
  • 24) 0.006 225 585 937 500 642 213 888 × 2 = 0 + 0.012 451 171 875 001 284 427 776;
  • 25) 0.012 451 171 875 001 284 427 776 × 2 = 0 + 0.024 902 343 750 002 568 855 552;
  • 26) 0.024 902 343 750 002 568 855 552 × 2 = 0 + 0.049 804 687 500 005 137 711 104;
  • 27) 0.049 804 687 500 005 137 711 104 × 2 = 0 + 0.099 609 375 000 010 275 422 208;
  • 28) 0.099 609 375 000 010 275 422 208 × 2 = 0 + 0.199 218 750 000 020 550 844 416;
  • 29) 0.199 218 750 000 020 550 844 416 × 2 = 0 + 0.398 437 500 000 041 101 688 832;
  • 30) 0.398 437 500 000 041 101 688 832 × 2 = 0 + 0.796 875 000 000 082 203 377 664;
  • 31) 0.796 875 000 000 082 203 377 664 × 2 = 1 + 0.593 750 000 000 164 406 755 328;
  • 32) 0.593 750 000 000 164 406 755 328 × 2 = 1 + 0.187 500 000 000 328 813 510 656;
  • 33) 0.187 500 000 000 328 813 510 656 × 2 = 0 + 0.375 000 000 000 657 627 021 312;
  • 34) 0.375 000 000 000 657 627 021 312 × 2 = 0 + 0.750 000 000 001 315 254 042 624;
  • 35) 0.750 000 000 001 315 254 042 624 × 2 = 1 + 0.500 000 000 002 630 508 085 248;
  • 36) 0.500 000 000 002 630 508 085 248 × 2 = 1 + 0.000 000 000 005 261 016 170 496;
  • 37) 0.000 000 000 005 261 016 170 496 × 2 = 0 + 0.000 000 000 010 522 032 340 992;
  • 38) 0.000 000 000 010 522 032 340 992 × 2 = 0 + 0.000 000 000 021 044 064 681 984;
  • 39) 0.000 000 000 021 044 064 681 984 × 2 = 0 + 0.000 000 000 042 088 129 363 968;
  • 40) 0.000 000 000 042 088 129 363 968 × 2 = 0 + 0.000 000 000 084 176 258 727 936;
  • 41) 0.000 000 000 084 176 258 727 936 × 2 = 0 + 0.000 000 000 168 352 517 455 872;
  • 42) 0.000 000 000 168 352 517 455 872 × 2 = 0 + 0.000 000 000 336 705 034 911 744;
  • 43) 0.000 000 000 336 705 034 911 744 × 2 = 0 + 0.000 000 000 673 410 069 823 488;
  • 44) 0.000 000 000 673 410 069 823 488 × 2 = 0 + 0.000 000 001 346 820 139 646 976;
  • 45) 0.000 000 001 346 820 139 646 976 × 2 = 0 + 0.000 000 002 693 640 279 293 952;
  • 46) 0.000 000 002 693 640 279 293 952 × 2 = 0 + 0.000 000 005 387 280 558 587 904;
  • 47) 0.000 000 005 387 280 558 587 904 × 2 = 0 + 0.000 000 010 774 561 117 175 808;
  • 48) 0.000 000 010 774 561 117 175 808 × 2 = 0 + 0.000 000 021 549 122 234 351 616;
  • 49) 0.000 000 021 549 122 234 351 616 × 2 = 0 + 0.000 000 043 098 244 468 703 232;
  • 50) 0.000 000 043 098 244 468 703 232 × 2 = 0 + 0.000 000 086 196 488 937 406 464;
  • 51) 0.000 000 086 196 488 937 406 464 × 2 = 0 + 0.000 000 172 392 977 874 812 928;
  • 52) 0.000 000 172 392 977 874 812 928 × 2 = 0 + 0.000 000 344 785 955 749 625 856;
  • 53) 0.000 000 344 785 955 749 625 856 × 2 = 0 + 0.000 000 689 571 911 499 251 712;
  • 54) 0.000 000 689 571 911 499 251 712 × 2 = 0 + 0.000 001 379 143 822 998 503 424;

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:


0.000 000 000 742 147 676 646 786(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 786(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:


0.000 000 000 742 147 676 646 786(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 -0.000 000 000 742 147 676 646 786 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