-0.000 000 000 742 147 676 646 729 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 729 8(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
-0.000 000 000 742 147 676 646 729 8(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:

|-0.000 000 000 742 147 676 646 729 8| = 0.000 000 000 742 147 676 646 729 8


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: 0.000 000 000 742 147 676 646 729 8.

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) 0.000 000 000 742 147 676 646 729 8 × 2 = 0 + 0.000 000 001 484 295 353 293 459 6;
  • 2) 0.000 000 001 484 295 353 293 459 6 × 2 = 0 + 0.000 000 002 968 590 706 586 919 2;
  • 3) 0.000 000 002 968 590 706 586 919 2 × 2 = 0 + 0.000 000 005 937 181 413 173 838 4;
  • 4) 0.000 000 005 937 181 413 173 838 4 × 2 = 0 + 0.000 000 011 874 362 826 347 676 8;
  • 5) 0.000 000 011 874 362 826 347 676 8 × 2 = 0 + 0.000 000 023 748 725 652 695 353 6;
  • 6) 0.000 000 023 748 725 652 695 353 6 × 2 = 0 + 0.000 000 047 497 451 305 390 707 2;
  • 7) 0.000 000 047 497 451 305 390 707 2 × 2 = 0 + 0.000 000 094 994 902 610 781 414 4;
  • 8) 0.000 000 094 994 902 610 781 414 4 × 2 = 0 + 0.000 000 189 989 805 221 562 828 8;
  • 9) 0.000 000 189 989 805 221 562 828 8 × 2 = 0 + 0.000 000 379 979 610 443 125 657 6;
  • 10) 0.000 000 379 979 610 443 125 657 6 × 2 = 0 + 0.000 000 759 959 220 886 251 315 2;
  • 11) 0.000 000 759 959 220 886 251 315 2 × 2 = 0 + 0.000 001 519 918 441 772 502 630 4;
  • 12) 0.000 001 519 918 441 772 502 630 4 × 2 = 0 + 0.000 003 039 836 883 545 005 260 8;
  • 13) 0.000 003 039 836 883 545 005 260 8 × 2 = 0 + 0.000 006 079 673 767 090 010 521 6;
  • 14) 0.000 006 079 673 767 090 010 521 6 × 2 = 0 + 0.000 012 159 347 534 180 021 043 2;
  • 15) 0.000 012 159 347 534 180 021 043 2 × 2 = 0 + 0.000 024 318 695 068 360 042 086 4;
  • 16) 0.000 024 318 695 068 360 042 086 4 × 2 = 0 + 0.000 048 637 390 136 720 084 172 8;
  • 17) 0.000 048 637 390 136 720 084 172 8 × 2 = 0 + 0.000 097 274 780 273 440 168 345 6;
  • 18) 0.000 097 274 780 273 440 168 345 6 × 2 = 0 + 0.000 194 549 560 546 880 336 691 2;
  • 19) 0.000 194 549 560 546 880 336 691 2 × 2 = 0 + 0.000 389 099 121 093 760 673 382 4;
  • 20) 0.000 389 099 121 093 760 673 382 4 × 2 = 0 + 0.000 778 198 242 187 521 346 764 8;
  • 21) 0.000 778 198 242 187 521 346 764 8 × 2 = 0 + 0.001 556 396 484 375 042 693 529 6;
  • 22) 0.001 556 396 484 375 042 693 529 6 × 2 = 0 + 0.003 112 792 968 750 085 387 059 2;
  • 23) 0.003 112 792 968 750 085 387 059 2 × 2 = 0 + 0.006 225 585 937 500 170 774 118 4;
  • 24) 0.006 225 585 937 500 170 774 118 4 × 2 = 0 + 0.012 451 171 875 000 341 548 236 8;
  • 25) 0.012 451 171 875 000 341 548 236 8 × 2 = 0 + 0.024 902 343 750 000 683 096 473 6;
  • 26) 0.024 902 343 750 000 683 096 473 6 × 2 = 0 + 0.049 804 687 500 001 366 192 947 2;
  • 27) 0.049 804 687 500 001 366 192 947 2 × 2 = 0 + 0.099 609 375 000 002 732 385 894 4;
  • 28) 0.099 609 375 000 002 732 385 894 4 × 2 = 0 + 0.199 218 750 000 005 464 771 788 8;
  • 29) 0.199 218 750 000 005 464 771 788 8 × 2 = 0 + 0.398 437 500 000 010 929 543 577 6;
  • 30) 0.398 437 500 000 010 929 543 577 6 × 2 = 0 + 0.796 875 000 000 021 859 087 155 2;
  • 31) 0.796 875 000 000 021 859 087 155 2 × 2 = 1 + 0.593 750 000 000 043 718 174 310 4;
  • 32) 0.593 750 000 000 043 718 174 310 4 × 2 = 1 + 0.187 500 000 000 087 436 348 620 8;
  • 33) 0.187 500 000 000 087 436 348 620 8 × 2 = 0 + 0.375 000 000 000 174 872 697 241 6;
  • 34) 0.375 000 000 000 174 872 697 241 6 × 2 = 0 + 0.750 000 000 000 349 745 394 483 2;
  • 35) 0.750 000 000 000 349 745 394 483 2 × 2 = 1 + 0.500 000 000 000 699 490 788 966 4;
  • 36) 0.500 000 000 000 699 490 788 966 4 × 2 = 1 + 0.000 000 000 001 398 981 577 932 8;
  • 37) 0.000 000 000 001 398 981 577 932 8 × 2 = 0 + 0.000 000 000 002 797 963 155 865 6;
  • 38) 0.000 000 000 002 797 963 155 865 6 × 2 = 0 + 0.000 000 000 005 595 926 311 731 2;
  • 39) 0.000 000 000 005 595 926 311 731 2 × 2 = 0 + 0.000 000 000 011 191 852 623 462 4;
  • 40) 0.000 000 000 011 191 852 623 462 4 × 2 = 0 + 0.000 000 000 022 383 705 246 924 8;
  • 41) 0.000 000 000 022 383 705 246 924 8 × 2 = 0 + 0.000 000 000 044 767 410 493 849 6;
  • 42) 0.000 000 000 044 767 410 493 849 6 × 2 = 0 + 0.000 000 000 089 534 820 987 699 2;
  • 43) 0.000 000 000 089 534 820 987 699 2 × 2 = 0 + 0.000 000 000 179 069 641 975 398 4;
  • 44) 0.000 000 000 179 069 641 975 398 4 × 2 = 0 + 0.000 000 000 358 139 283 950 796 8;
  • 45) 0.000 000 000 358 139 283 950 796 8 × 2 = 0 + 0.000 000 000 716 278 567 901 593 6;
  • 46) 0.000 000 000 716 278 567 901 593 6 × 2 = 0 + 0.000 000 001 432 557 135 803 187 2;
  • 47) 0.000 000 001 432 557 135 803 187 2 × 2 = 0 + 0.000 000 002 865 114 271 606 374 4;
  • 48) 0.000 000 002 865 114 271 606 374 4 × 2 = 0 + 0.000 000 005 730 228 543 212 748 8;
  • 49) 0.000 000 005 730 228 543 212 748 8 × 2 = 0 + 0.000 000 011 460 457 086 425 497 6;
  • 50) 0.000 000 011 460 457 086 425 497 6 × 2 = 0 + 0.000 000 022 920 914 172 850 995 2;
  • 51) 0.000 000 022 920 914 172 850 995 2 × 2 = 0 + 0.000 000 045 841 828 345 701 990 4;
  • 52) 0.000 000 045 841 828 345 701 990 4 × 2 = 0 + 0.000 000 091 683 656 691 403 980 8;
  • 53) 0.000 000 091 683 656 691 403 980 8 × 2 = 0 + 0.000 000 183 367 313 382 807 961 6;
  • 54) 0.000 000 183 367 313 382 807 961 6 × 2 = 0 + 0.000 000 366 734 626 765 615 923 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:


0.000 000 000 742 147 676 646 729 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 729 8(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:


0.000 000 000 742 147 676 646 729 8(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 -0.000 000 000 742 147 676 646 729 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 646 739 1 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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