-0.000 000 000 742 147 676 206 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 206(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
-0.000 000 000 742 147 676 206(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:

|-0.000 000 000 742 147 676 206| = 0.000 000 000 742 147 676 206


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: 0.000 000 000 742 147 676 206.

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) 0.000 000 000 742 147 676 206 × 2 = 0 + 0.000 000 001 484 295 352 412;
  • 2) 0.000 000 001 484 295 352 412 × 2 = 0 + 0.000 000 002 968 590 704 824;
  • 3) 0.000 000 002 968 590 704 824 × 2 = 0 + 0.000 000 005 937 181 409 648;
  • 4) 0.000 000 005 937 181 409 648 × 2 = 0 + 0.000 000 011 874 362 819 296;
  • 5) 0.000 000 011 874 362 819 296 × 2 = 0 + 0.000 000 023 748 725 638 592;
  • 6) 0.000 000 023 748 725 638 592 × 2 = 0 + 0.000 000 047 497 451 277 184;
  • 7) 0.000 000 047 497 451 277 184 × 2 = 0 + 0.000 000 094 994 902 554 368;
  • 8) 0.000 000 094 994 902 554 368 × 2 = 0 + 0.000 000 189 989 805 108 736;
  • 9) 0.000 000 189 989 805 108 736 × 2 = 0 + 0.000 000 379 979 610 217 472;
  • 10) 0.000 000 379 979 610 217 472 × 2 = 0 + 0.000 000 759 959 220 434 944;
  • 11) 0.000 000 759 959 220 434 944 × 2 = 0 + 0.000 001 519 918 440 869 888;
  • 12) 0.000 001 519 918 440 869 888 × 2 = 0 + 0.000 003 039 836 881 739 776;
  • 13) 0.000 003 039 836 881 739 776 × 2 = 0 + 0.000 006 079 673 763 479 552;
  • 14) 0.000 006 079 673 763 479 552 × 2 = 0 + 0.000 012 159 347 526 959 104;
  • 15) 0.000 012 159 347 526 959 104 × 2 = 0 + 0.000 024 318 695 053 918 208;
  • 16) 0.000 024 318 695 053 918 208 × 2 = 0 + 0.000 048 637 390 107 836 416;
  • 17) 0.000 048 637 390 107 836 416 × 2 = 0 + 0.000 097 274 780 215 672 832;
  • 18) 0.000 097 274 780 215 672 832 × 2 = 0 + 0.000 194 549 560 431 345 664;
  • 19) 0.000 194 549 560 431 345 664 × 2 = 0 + 0.000 389 099 120 862 691 328;
  • 20) 0.000 389 099 120 862 691 328 × 2 = 0 + 0.000 778 198 241 725 382 656;
  • 21) 0.000 778 198 241 725 382 656 × 2 = 0 + 0.001 556 396 483 450 765 312;
  • 22) 0.001 556 396 483 450 765 312 × 2 = 0 + 0.003 112 792 966 901 530 624;
  • 23) 0.003 112 792 966 901 530 624 × 2 = 0 + 0.006 225 585 933 803 061 248;
  • 24) 0.006 225 585 933 803 061 248 × 2 = 0 + 0.012 451 171 867 606 122 496;
  • 25) 0.012 451 171 867 606 122 496 × 2 = 0 + 0.024 902 343 735 212 244 992;
  • 26) 0.024 902 343 735 212 244 992 × 2 = 0 + 0.049 804 687 470 424 489 984;
  • 27) 0.049 804 687 470 424 489 984 × 2 = 0 + 0.099 609 374 940 848 979 968;
  • 28) 0.099 609 374 940 848 979 968 × 2 = 0 + 0.199 218 749 881 697 959 936;
  • 29) 0.199 218 749 881 697 959 936 × 2 = 0 + 0.398 437 499 763 395 919 872;
  • 30) 0.398 437 499 763 395 919 872 × 2 = 0 + 0.796 874 999 526 791 839 744;
  • 31) 0.796 874 999 526 791 839 744 × 2 = 1 + 0.593 749 999 053 583 679 488;
  • 32) 0.593 749 999 053 583 679 488 × 2 = 1 + 0.187 499 998 107 167 358 976;
  • 33) 0.187 499 998 107 167 358 976 × 2 = 0 + 0.374 999 996 214 334 717 952;
  • 34) 0.374 999 996 214 334 717 952 × 2 = 0 + 0.749 999 992 428 669 435 904;
  • 35) 0.749 999 992 428 669 435 904 × 2 = 1 + 0.499 999 984 857 338 871 808;
  • 36) 0.499 999 984 857 338 871 808 × 2 = 0 + 0.999 999 969 714 677 743 616;
  • 37) 0.999 999 969 714 677 743 616 × 2 = 1 + 0.999 999 939 429 355 487 232;
  • 38) 0.999 999 939 429 355 487 232 × 2 = 1 + 0.999 999 878 858 710 974 464;
  • 39) 0.999 999 878 858 710 974 464 × 2 = 1 + 0.999 999 757 717 421 948 928;
  • 40) 0.999 999 757 717 421 948 928 × 2 = 1 + 0.999 999 515 434 843 897 856;
  • 41) 0.999 999 515 434 843 897 856 × 2 = 1 + 0.999 999 030 869 687 795 712;
  • 42) 0.999 999 030 869 687 795 712 × 2 = 1 + 0.999 998 061 739 375 591 424;
  • 43) 0.999 998 061 739 375 591 424 × 2 = 1 + 0.999 996 123 478 751 182 848;
  • 44) 0.999 996 123 478 751 182 848 × 2 = 1 + 0.999 992 246 957 502 365 696;
  • 45) 0.999 992 246 957 502 365 696 × 2 = 1 + 0.999 984 493 915 004 731 392;
  • 46) 0.999 984 493 915 004 731 392 × 2 = 1 + 0.999 968 987 830 009 462 784;
  • 47) 0.999 968 987 830 009 462 784 × 2 = 1 + 0.999 937 975 660 018 925 568;
  • 48) 0.999 937 975 660 018 925 568 × 2 = 1 + 0.999 875 951 320 037 851 136;
  • 49) 0.999 875 951 320 037 851 136 × 2 = 1 + 0.999 751 902 640 075 702 272;
  • 50) 0.999 751 902 640 075 702 272 × 2 = 1 + 0.999 503 805 280 151 404 544;
  • 51) 0.999 503 805 280 151 404 544 × 2 = 1 + 0.999 007 610 560 302 809 088;
  • 52) 0.999 007 610 560 302 809 088 × 2 = 1 + 0.998 015 221 120 605 618 176;
  • 53) 0.998 015 221 120 605 618 176 × 2 = 1 + 0.996 030 442 241 211 236 352;
  • 54) 0.996 030 442 241 211 236 352 × 2 = 1 + 0.992 060 884 482 422 472 704;

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:


0.000 000 000 742 147 676 206(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 206(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:


0.000 000 000 742 147 676 206(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 -0.000 000 000 742 147 676 206 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 147 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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