-0.000 000 000 742 147 676 834 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 834(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
-0.000 000 000 742 147 676 834(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:

|-0.000 000 000 742 147 676 834| = 0.000 000 000 742 147 676 834


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: 0.000 000 000 742 147 676 834.

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) 0.000 000 000 742 147 676 834 × 2 = 0 + 0.000 000 001 484 295 353 668;
  • 2) 0.000 000 001 484 295 353 668 × 2 = 0 + 0.000 000 002 968 590 707 336;
  • 3) 0.000 000 002 968 590 707 336 × 2 = 0 + 0.000 000 005 937 181 414 672;
  • 4) 0.000 000 005 937 181 414 672 × 2 = 0 + 0.000 000 011 874 362 829 344;
  • 5) 0.000 000 011 874 362 829 344 × 2 = 0 + 0.000 000 023 748 725 658 688;
  • 6) 0.000 000 023 748 725 658 688 × 2 = 0 + 0.000 000 047 497 451 317 376;
  • 7) 0.000 000 047 497 451 317 376 × 2 = 0 + 0.000 000 094 994 902 634 752;
  • 8) 0.000 000 094 994 902 634 752 × 2 = 0 + 0.000 000 189 989 805 269 504;
  • 9) 0.000 000 189 989 805 269 504 × 2 = 0 + 0.000 000 379 979 610 539 008;
  • 10) 0.000 000 379 979 610 539 008 × 2 = 0 + 0.000 000 759 959 221 078 016;
  • 11) 0.000 000 759 959 221 078 016 × 2 = 0 + 0.000 001 519 918 442 156 032;
  • 12) 0.000 001 519 918 442 156 032 × 2 = 0 + 0.000 003 039 836 884 312 064;
  • 13) 0.000 003 039 836 884 312 064 × 2 = 0 + 0.000 006 079 673 768 624 128;
  • 14) 0.000 006 079 673 768 624 128 × 2 = 0 + 0.000 012 159 347 537 248 256;
  • 15) 0.000 012 159 347 537 248 256 × 2 = 0 + 0.000 024 318 695 074 496 512;
  • 16) 0.000 024 318 695 074 496 512 × 2 = 0 + 0.000 048 637 390 148 993 024;
  • 17) 0.000 048 637 390 148 993 024 × 2 = 0 + 0.000 097 274 780 297 986 048;
  • 18) 0.000 097 274 780 297 986 048 × 2 = 0 + 0.000 194 549 560 595 972 096;
  • 19) 0.000 194 549 560 595 972 096 × 2 = 0 + 0.000 389 099 121 191 944 192;
  • 20) 0.000 389 099 121 191 944 192 × 2 = 0 + 0.000 778 198 242 383 888 384;
  • 21) 0.000 778 198 242 383 888 384 × 2 = 0 + 0.001 556 396 484 767 776 768;
  • 22) 0.001 556 396 484 767 776 768 × 2 = 0 + 0.003 112 792 969 535 553 536;
  • 23) 0.003 112 792 969 535 553 536 × 2 = 0 + 0.006 225 585 939 071 107 072;
  • 24) 0.006 225 585 939 071 107 072 × 2 = 0 + 0.012 451 171 878 142 214 144;
  • 25) 0.012 451 171 878 142 214 144 × 2 = 0 + 0.024 902 343 756 284 428 288;
  • 26) 0.024 902 343 756 284 428 288 × 2 = 0 + 0.049 804 687 512 568 856 576;
  • 27) 0.049 804 687 512 568 856 576 × 2 = 0 + 0.099 609 375 025 137 713 152;
  • 28) 0.099 609 375 025 137 713 152 × 2 = 0 + 0.199 218 750 050 275 426 304;
  • 29) 0.199 218 750 050 275 426 304 × 2 = 0 + 0.398 437 500 100 550 852 608;
  • 30) 0.398 437 500 100 550 852 608 × 2 = 0 + 0.796 875 000 201 101 705 216;
  • 31) 0.796 875 000 201 101 705 216 × 2 = 1 + 0.593 750 000 402 203 410 432;
  • 32) 0.593 750 000 402 203 410 432 × 2 = 1 + 0.187 500 000 804 406 820 864;
  • 33) 0.187 500 000 804 406 820 864 × 2 = 0 + 0.375 000 001 608 813 641 728;
  • 34) 0.375 000 001 608 813 641 728 × 2 = 0 + 0.750 000 003 217 627 283 456;
  • 35) 0.750 000 003 217 627 283 456 × 2 = 1 + 0.500 000 006 435 254 566 912;
  • 36) 0.500 000 006 435 254 566 912 × 2 = 1 + 0.000 000 012 870 509 133 824;
  • 37) 0.000 000 012 870 509 133 824 × 2 = 0 + 0.000 000 025 741 018 267 648;
  • 38) 0.000 000 025 741 018 267 648 × 2 = 0 + 0.000 000 051 482 036 535 296;
  • 39) 0.000 000 051 482 036 535 296 × 2 = 0 + 0.000 000 102 964 073 070 592;
  • 40) 0.000 000 102 964 073 070 592 × 2 = 0 + 0.000 000 205 928 146 141 184;
  • 41) 0.000 000 205 928 146 141 184 × 2 = 0 + 0.000 000 411 856 292 282 368;
  • 42) 0.000 000 411 856 292 282 368 × 2 = 0 + 0.000 000 823 712 584 564 736;
  • 43) 0.000 000 823 712 584 564 736 × 2 = 0 + 0.000 001 647 425 169 129 472;
  • 44) 0.000 001 647 425 169 129 472 × 2 = 0 + 0.000 003 294 850 338 258 944;
  • 45) 0.000 003 294 850 338 258 944 × 2 = 0 + 0.000 006 589 700 676 517 888;
  • 46) 0.000 006 589 700 676 517 888 × 2 = 0 + 0.000 013 179 401 353 035 776;
  • 47) 0.000 013 179 401 353 035 776 × 2 = 0 + 0.000 026 358 802 706 071 552;
  • 48) 0.000 026 358 802 706 071 552 × 2 = 0 + 0.000 052 717 605 412 143 104;
  • 49) 0.000 052 717 605 412 143 104 × 2 = 0 + 0.000 105 435 210 824 286 208;
  • 50) 0.000 105 435 210 824 286 208 × 2 = 0 + 0.000 210 870 421 648 572 416;
  • 51) 0.000 210 870 421 648 572 416 × 2 = 0 + 0.000 421 740 843 297 144 832;
  • 52) 0.000 421 740 843 297 144 832 × 2 = 0 + 0.000 843 481 686 594 289 664;
  • 53) 0.000 843 481 686 594 289 664 × 2 = 0 + 0.001 686 963 373 188 579 328;
  • 54) 0.001 686 963 373 188 579 328 × 2 = 0 + 0.003 373 926 746 377 158 656;

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:


0.000 000 000 742 147 676 834(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 834(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:


0.000 000 000 742 147 676 834(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 -0.000 000 000 742 147 676 834 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