-0.000 000 000 742 147 676 632 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 632 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
-0.000 000 000 742 147 676 632 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:

|-0.000 000 000 742 147 676 632 3| = 0.000 000 000 742 147 676 632 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: 0.000 000 000 742 147 676 632 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) 0.000 000 000 742 147 676 632 3 × 2 = 0 + 0.000 000 001 484 295 353 264 6;
  • 2) 0.000 000 001 484 295 353 264 6 × 2 = 0 + 0.000 000 002 968 590 706 529 2;
  • 3) 0.000 000 002 968 590 706 529 2 × 2 = 0 + 0.000 000 005 937 181 413 058 4;
  • 4) 0.000 000 005 937 181 413 058 4 × 2 = 0 + 0.000 000 011 874 362 826 116 8;
  • 5) 0.000 000 011 874 362 826 116 8 × 2 = 0 + 0.000 000 023 748 725 652 233 6;
  • 6) 0.000 000 023 748 725 652 233 6 × 2 = 0 + 0.000 000 047 497 451 304 467 2;
  • 7) 0.000 000 047 497 451 304 467 2 × 2 = 0 + 0.000 000 094 994 902 608 934 4;
  • 8) 0.000 000 094 994 902 608 934 4 × 2 = 0 + 0.000 000 189 989 805 217 868 8;
  • 9) 0.000 000 189 989 805 217 868 8 × 2 = 0 + 0.000 000 379 979 610 435 737 6;
  • 10) 0.000 000 379 979 610 435 737 6 × 2 = 0 + 0.000 000 759 959 220 871 475 2;
  • 11) 0.000 000 759 959 220 871 475 2 × 2 = 0 + 0.000 001 519 918 441 742 950 4;
  • 12) 0.000 001 519 918 441 742 950 4 × 2 = 0 + 0.000 003 039 836 883 485 900 8;
  • 13) 0.000 003 039 836 883 485 900 8 × 2 = 0 + 0.000 006 079 673 766 971 801 6;
  • 14) 0.000 006 079 673 766 971 801 6 × 2 = 0 + 0.000 012 159 347 533 943 603 2;
  • 15) 0.000 012 159 347 533 943 603 2 × 2 = 0 + 0.000 024 318 695 067 887 206 4;
  • 16) 0.000 024 318 695 067 887 206 4 × 2 = 0 + 0.000 048 637 390 135 774 412 8;
  • 17) 0.000 048 637 390 135 774 412 8 × 2 = 0 + 0.000 097 274 780 271 548 825 6;
  • 18) 0.000 097 274 780 271 548 825 6 × 2 = 0 + 0.000 194 549 560 543 097 651 2;
  • 19) 0.000 194 549 560 543 097 651 2 × 2 = 0 + 0.000 389 099 121 086 195 302 4;
  • 20) 0.000 389 099 121 086 195 302 4 × 2 = 0 + 0.000 778 198 242 172 390 604 8;
  • 21) 0.000 778 198 242 172 390 604 8 × 2 = 0 + 0.001 556 396 484 344 781 209 6;
  • 22) 0.001 556 396 484 344 781 209 6 × 2 = 0 + 0.003 112 792 968 689 562 419 2;
  • 23) 0.003 112 792 968 689 562 419 2 × 2 = 0 + 0.006 225 585 937 379 124 838 4;
  • 24) 0.006 225 585 937 379 124 838 4 × 2 = 0 + 0.012 451 171 874 758 249 676 8;
  • 25) 0.012 451 171 874 758 249 676 8 × 2 = 0 + 0.024 902 343 749 516 499 353 6;
  • 26) 0.024 902 343 749 516 499 353 6 × 2 = 0 + 0.049 804 687 499 032 998 707 2;
  • 27) 0.049 804 687 499 032 998 707 2 × 2 = 0 + 0.099 609 374 998 065 997 414 4;
  • 28) 0.099 609 374 998 065 997 414 4 × 2 = 0 + 0.199 218 749 996 131 994 828 8;
  • 29) 0.199 218 749 996 131 994 828 8 × 2 = 0 + 0.398 437 499 992 263 989 657 6;
  • 30) 0.398 437 499 992 263 989 657 6 × 2 = 0 + 0.796 874 999 984 527 979 315 2;
  • 31) 0.796 874 999 984 527 979 315 2 × 2 = 1 + 0.593 749 999 969 055 958 630 4;
  • 32) 0.593 749 999 969 055 958 630 4 × 2 = 1 + 0.187 499 999 938 111 917 260 8;
  • 33) 0.187 499 999 938 111 917 260 8 × 2 = 0 + 0.374 999 999 876 223 834 521 6;
  • 34) 0.374 999 999 876 223 834 521 6 × 2 = 0 + 0.749 999 999 752 447 669 043 2;
  • 35) 0.749 999 999 752 447 669 043 2 × 2 = 1 + 0.499 999 999 504 895 338 086 4;
  • 36) 0.499 999 999 504 895 338 086 4 × 2 = 0 + 0.999 999 999 009 790 676 172 8;
  • 37) 0.999 999 999 009 790 676 172 8 × 2 = 1 + 0.999 999 998 019 581 352 345 6;
  • 38) 0.999 999 998 019 581 352 345 6 × 2 = 1 + 0.999 999 996 039 162 704 691 2;
  • 39) 0.999 999 996 039 162 704 691 2 × 2 = 1 + 0.999 999 992 078 325 409 382 4;
  • 40) 0.999 999 992 078 325 409 382 4 × 2 = 1 + 0.999 999 984 156 650 818 764 8;
  • 41) 0.999 999 984 156 650 818 764 8 × 2 = 1 + 0.999 999 968 313 301 637 529 6;
  • 42) 0.999 999 968 313 301 637 529 6 × 2 = 1 + 0.999 999 936 626 603 275 059 2;
  • 43) 0.999 999 936 626 603 275 059 2 × 2 = 1 + 0.999 999 873 253 206 550 118 4;
  • 44) 0.999 999 873 253 206 550 118 4 × 2 = 1 + 0.999 999 746 506 413 100 236 8;
  • 45) 0.999 999 746 506 413 100 236 8 × 2 = 1 + 0.999 999 493 012 826 200 473 6;
  • 46) 0.999 999 493 012 826 200 473 6 × 2 = 1 + 0.999 998 986 025 652 400 947 2;
  • 47) 0.999 998 986 025 652 400 947 2 × 2 = 1 + 0.999 997 972 051 304 801 894 4;
  • 48) 0.999 997 972 051 304 801 894 4 × 2 = 1 + 0.999 995 944 102 609 603 788 8;
  • 49) 0.999 995 944 102 609 603 788 8 × 2 = 1 + 0.999 991 888 205 219 207 577 6;
  • 50) 0.999 991 888 205 219 207 577 6 × 2 = 1 + 0.999 983 776 410 438 415 155 2;
  • 51) 0.999 983 776 410 438 415 155 2 × 2 = 1 + 0.999 967 552 820 876 830 310 4;
  • 52) 0.999 967 552 820 876 830 310 4 × 2 = 1 + 0.999 935 105 641 753 660 620 8;
  • 53) 0.999 935 105 641 753 660 620 8 × 2 = 1 + 0.999 870 211 283 507 321 241 6;
  • 54) 0.999 870 211 283 507 321 241 6 × 2 = 1 + 0.999 740 422 567 014 642 483 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:


0.000 000 000 742 147 676 632 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 632 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:


0.000 000 000 742 147 676 632 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 -0.000 000 000 742 147 676 632 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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