-0.000 000 000 742 147 676 646 692 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 692 3(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
-0.000 000 000 742 147 676 646 692 3(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:

|-0.000 000 000 742 147 676 646 692 3| = 0.000 000 000 742 147 676 646 692 3


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: 0.000 000 000 742 147 676 646 692 3.

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) 0.000 000 000 742 147 676 646 692 3 × 2 = 0 + 0.000 000 001 484 295 353 293 384 6;
  • 2) 0.000 000 001 484 295 353 293 384 6 × 2 = 0 + 0.000 000 002 968 590 706 586 769 2;
  • 3) 0.000 000 002 968 590 706 586 769 2 × 2 = 0 + 0.000 000 005 937 181 413 173 538 4;
  • 4) 0.000 000 005 937 181 413 173 538 4 × 2 = 0 + 0.000 000 011 874 362 826 347 076 8;
  • 5) 0.000 000 011 874 362 826 347 076 8 × 2 = 0 + 0.000 000 023 748 725 652 694 153 6;
  • 6) 0.000 000 023 748 725 652 694 153 6 × 2 = 0 + 0.000 000 047 497 451 305 388 307 2;
  • 7) 0.000 000 047 497 451 305 388 307 2 × 2 = 0 + 0.000 000 094 994 902 610 776 614 4;
  • 8) 0.000 000 094 994 902 610 776 614 4 × 2 = 0 + 0.000 000 189 989 805 221 553 228 8;
  • 9) 0.000 000 189 989 805 221 553 228 8 × 2 = 0 + 0.000 000 379 979 610 443 106 457 6;
  • 10) 0.000 000 379 979 610 443 106 457 6 × 2 = 0 + 0.000 000 759 959 220 886 212 915 2;
  • 11) 0.000 000 759 959 220 886 212 915 2 × 2 = 0 + 0.000 001 519 918 441 772 425 830 4;
  • 12) 0.000 001 519 918 441 772 425 830 4 × 2 = 0 + 0.000 003 039 836 883 544 851 660 8;
  • 13) 0.000 003 039 836 883 544 851 660 8 × 2 = 0 + 0.000 006 079 673 767 089 703 321 6;
  • 14) 0.000 006 079 673 767 089 703 321 6 × 2 = 0 + 0.000 012 159 347 534 179 406 643 2;
  • 15) 0.000 012 159 347 534 179 406 643 2 × 2 = 0 + 0.000 024 318 695 068 358 813 286 4;
  • 16) 0.000 024 318 695 068 358 813 286 4 × 2 = 0 + 0.000 048 637 390 136 717 626 572 8;
  • 17) 0.000 048 637 390 136 717 626 572 8 × 2 = 0 + 0.000 097 274 780 273 435 253 145 6;
  • 18) 0.000 097 274 780 273 435 253 145 6 × 2 = 0 + 0.000 194 549 560 546 870 506 291 2;
  • 19) 0.000 194 549 560 546 870 506 291 2 × 2 = 0 + 0.000 389 099 121 093 741 012 582 4;
  • 20) 0.000 389 099 121 093 741 012 582 4 × 2 = 0 + 0.000 778 198 242 187 482 025 164 8;
  • 21) 0.000 778 198 242 187 482 025 164 8 × 2 = 0 + 0.001 556 396 484 374 964 050 329 6;
  • 22) 0.001 556 396 484 374 964 050 329 6 × 2 = 0 + 0.003 112 792 968 749 928 100 659 2;
  • 23) 0.003 112 792 968 749 928 100 659 2 × 2 = 0 + 0.006 225 585 937 499 856 201 318 4;
  • 24) 0.006 225 585 937 499 856 201 318 4 × 2 = 0 + 0.012 451 171 874 999 712 402 636 8;
  • 25) 0.012 451 171 874 999 712 402 636 8 × 2 = 0 + 0.024 902 343 749 999 424 805 273 6;
  • 26) 0.024 902 343 749 999 424 805 273 6 × 2 = 0 + 0.049 804 687 499 998 849 610 547 2;
  • 27) 0.049 804 687 499 998 849 610 547 2 × 2 = 0 + 0.099 609 374 999 997 699 221 094 4;
  • 28) 0.099 609 374 999 997 699 221 094 4 × 2 = 0 + 0.199 218 749 999 995 398 442 188 8;
  • 29) 0.199 218 749 999 995 398 442 188 8 × 2 = 0 + 0.398 437 499 999 990 796 884 377 6;
  • 30) 0.398 437 499 999 990 796 884 377 6 × 2 = 0 + 0.796 874 999 999 981 593 768 755 2;
  • 31) 0.796 874 999 999 981 593 768 755 2 × 2 = 1 + 0.593 749 999 999 963 187 537 510 4;
  • 32) 0.593 749 999 999 963 187 537 510 4 × 2 = 1 + 0.187 499 999 999 926 375 075 020 8;
  • 33) 0.187 499 999 999 926 375 075 020 8 × 2 = 0 + 0.374 999 999 999 852 750 150 041 6;
  • 34) 0.374 999 999 999 852 750 150 041 6 × 2 = 0 + 0.749 999 999 999 705 500 300 083 2;
  • 35) 0.749 999 999 999 705 500 300 083 2 × 2 = 1 + 0.499 999 999 999 411 000 600 166 4;
  • 36) 0.499 999 999 999 411 000 600 166 4 × 2 = 0 + 0.999 999 999 998 822 001 200 332 8;
  • 37) 0.999 999 999 998 822 001 200 332 8 × 2 = 1 + 0.999 999 999 997 644 002 400 665 6;
  • 38) 0.999 999 999 997 644 002 400 665 6 × 2 = 1 + 0.999 999 999 995 288 004 801 331 2;
  • 39) 0.999 999 999 995 288 004 801 331 2 × 2 = 1 + 0.999 999 999 990 576 009 602 662 4;
  • 40) 0.999 999 999 990 576 009 602 662 4 × 2 = 1 + 0.999 999 999 981 152 019 205 324 8;
  • 41) 0.999 999 999 981 152 019 205 324 8 × 2 = 1 + 0.999 999 999 962 304 038 410 649 6;
  • 42) 0.999 999 999 962 304 038 410 649 6 × 2 = 1 + 0.999 999 999 924 608 076 821 299 2;
  • 43) 0.999 999 999 924 608 076 821 299 2 × 2 = 1 + 0.999 999 999 849 216 153 642 598 4;
  • 44) 0.999 999 999 849 216 153 642 598 4 × 2 = 1 + 0.999 999 999 698 432 307 285 196 8;
  • 45) 0.999 999 999 698 432 307 285 196 8 × 2 = 1 + 0.999 999 999 396 864 614 570 393 6;
  • 46) 0.999 999 999 396 864 614 570 393 6 × 2 = 1 + 0.999 999 998 793 729 229 140 787 2;
  • 47) 0.999 999 998 793 729 229 140 787 2 × 2 = 1 + 0.999 999 997 587 458 458 281 574 4;
  • 48) 0.999 999 997 587 458 458 281 574 4 × 2 = 1 + 0.999 999 995 174 916 916 563 148 8;
  • 49) 0.999 999 995 174 916 916 563 148 8 × 2 = 1 + 0.999 999 990 349 833 833 126 297 6;
  • 50) 0.999 999 990 349 833 833 126 297 6 × 2 = 1 + 0.999 999 980 699 667 666 252 595 2;
  • 51) 0.999 999 980 699 667 666 252 595 2 × 2 = 1 + 0.999 999 961 399 335 332 505 190 4;
  • 52) 0.999 999 961 399 335 332 505 190 4 × 2 = 1 + 0.999 999 922 798 670 665 010 380 8;
  • 53) 0.999 999 922 798 670 665 010 380 8 × 2 = 1 + 0.999 999 845 597 341 330 020 761 6;
  • 54) 0.999 999 845 597 341 330 020 761 6 × 2 = 1 + 0.999 999 691 194 682 660 041 523 2;

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:


0.000 000 000 742 147 676 646 692 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 692 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 646 692 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 0111 1111 1111 1111 111


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 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 646 692 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 691 8 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: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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