-0.000 000 000 000 151 576 98 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 151 576 98(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
-0.000 000 000 000 151 576 98(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:

|-0.000 000 000 000 151 576 98| = 0.000 000 000 000 151 576 98


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: 0.000 000 000 000 151 576 98.

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) 0.000 000 000 000 151 576 98 × 2 = 0 + 0.000 000 000 000 303 153 96;
  • 2) 0.000 000 000 000 303 153 96 × 2 = 0 + 0.000 000 000 000 606 307 92;
  • 3) 0.000 000 000 000 606 307 92 × 2 = 0 + 0.000 000 000 001 212 615 84;
  • 4) 0.000 000 000 001 212 615 84 × 2 = 0 + 0.000 000 000 002 425 231 68;
  • 5) 0.000 000 000 002 425 231 68 × 2 = 0 + 0.000 000 000 004 850 463 36;
  • 6) 0.000 000 000 004 850 463 36 × 2 = 0 + 0.000 000 000 009 700 926 72;
  • 7) 0.000 000 000 009 700 926 72 × 2 = 0 + 0.000 000 000 019 401 853 44;
  • 8) 0.000 000 000 019 401 853 44 × 2 = 0 + 0.000 000 000 038 803 706 88;
  • 9) 0.000 000 000 038 803 706 88 × 2 = 0 + 0.000 000 000 077 607 413 76;
  • 10) 0.000 000 000 077 607 413 76 × 2 = 0 + 0.000 000 000 155 214 827 52;
  • 11) 0.000 000 000 155 214 827 52 × 2 = 0 + 0.000 000 000 310 429 655 04;
  • 12) 0.000 000 000 310 429 655 04 × 2 = 0 + 0.000 000 000 620 859 310 08;
  • 13) 0.000 000 000 620 859 310 08 × 2 = 0 + 0.000 000 001 241 718 620 16;
  • 14) 0.000 000 001 241 718 620 16 × 2 = 0 + 0.000 000 002 483 437 240 32;
  • 15) 0.000 000 002 483 437 240 32 × 2 = 0 + 0.000 000 004 966 874 480 64;
  • 16) 0.000 000 004 966 874 480 64 × 2 = 0 + 0.000 000 009 933 748 961 28;
  • 17) 0.000 000 009 933 748 961 28 × 2 = 0 + 0.000 000 019 867 497 922 56;
  • 18) 0.000 000 019 867 497 922 56 × 2 = 0 + 0.000 000 039 734 995 845 12;
  • 19) 0.000 000 039 734 995 845 12 × 2 = 0 + 0.000 000 079 469 991 690 24;
  • 20) 0.000 000 079 469 991 690 24 × 2 = 0 + 0.000 000 158 939 983 380 48;
  • 21) 0.000 000 158 939 983 380 48 × 2 = 0 + 0.000 000 317 879 966 760 96;
  • 22) 0.000 000 317 879 966 760 96 × 2 = 0 + 0.000 000 635 759 933 521 92;
  • 23) 0.000 000 635 759 933 521 92 × 2 = 0 + 0.000 001 271 519 867 043 84;
  • 24) 0.000 001 271 519 867 043 84 × 2 = 0 + 0.000 002 543 039 734 087 68;
  • 25) 0.000 002 543 039 734 087 68 × 2 = 0 + 0.000 005 086 079 468 175 36;
  • 26) 0.000 005 086 079 468 175 36 × 2 = 0 + 0.000 010 172 158 936 350 72;
  • 27) 0.000 010 172 158 936 350 72 × 2 = 0 + 0.000 020 344 317 872 701 44;
  • 28) 0.000 020 344 317 872 701 44 × 2 = 0 + 0.000 040 688 635 745 402 88;
  • 29) 0.000 040 688 635 745 402 88 × 2 = 0 + 0.000 081 377 271 490 805 76;
  • 30) 0.000 081 377 271 490 805 76 × 2 = 0 + 0.000 162 754 542 981 611 52;
  • 31) 0.000 162 754 542 981 611 52 × 2 = 0 + 0.000 325 509 085 963 223 04;
  • 32) 0.000 325 509 085 963 223 04 × 2 = 0 + 0.000 651 018 171 926 446 08;
  • 33) 0.000 651 018 171 926 446 08 × 2 = 0 + 0.001 302 036 343 852 892 16;
  • 34) 0.001 302 036 343 852 892 16 × 2 = 0 + 0.002 604 072 687 705 784 32;
  • 35) 0.002 604 072 687 705 784 32 × 2 = 0 + 0.005 208 145 375 411 568 64;
  • 36) 0.005 208 145 375 411 568 64 × 2 = 0 + 0.010 416 290 750 823 137 28;
  • 37) 0.010 416 290 750 823 137 28 × 2 = 0 + 0.020 832 581 501 646 274 56;
  • 38) 0.020 832 581 501 646 274 56 × 2 = 0 + 0.041 665 163 003 292 549 12;
  • 39) 0.041 665 163 003 292 549 12 × 2 = 0 + 0.083 330 326 006 585 098 24;
  • 40) 0.083 330 326 006 585 098 24 × 2 = 0 + 0.166 660 652 013 170 196 48;
  • 41) 0.166 660 652 013 170 196 48 × 2 = 0 + 0.333 321 304 026 340 392 96;
  • 42) 0.333 321 304 026 340 392 96 × 2 = 0 + 0.666 642 608 052 680 785 92;
  • 43) 0.666 642 608 052 680 785 92 × 2 = 1 + 0.333 285 216 105 361 571 84;
  • 44) 0.333 285 216 105 361 571 84 × 2 = 0 + 0.666 570 432 210 723 143 68;
  • 45) 0.666 570 432 210 723 143 68 × 2 = 1 + 0.333 140 864 421 446 287 36;
  • 46) 0.333 140 864 421 446 287 36 × 2 = 0 + 0.666 281 728 842 892 574 72;
  • 47) 0.666 281 728 842 892 574 72 × 2 = 1 + 0.332 563 457 685 785 149 44;
  • 48) 0.332 563 457 685 785 149 44 × 2 = 0 + 0.665 126 915 371 570 298 88;
  • 49) 0.665 126 915 371 570 298 88 × 2 = 1 + 0.330 253 830 743 140 597 76;
  • 50) 0.330 253 830 743 140 597 76 × 2 = 0 + 0.660 507 661 486 281 195 52;
  • 51) 0.660 507 661 486 281 195 52 × 2 = 1 + 0.321 015 322 972 562 391 04;
  • 52) 0.321 015 322 972 562 391 04 × 2 = 0 + 0.642 030 645 945 124 782 08;
  • 53) 0.642 030 645 945 124 782 08 × 2 = 1 + 0.284 061 291 890 249 564 16;
  • 54) 0.284 061 291 890 249 564 16 × 2 = 0 + 0.568 122 583 780 499 128 32;
  • 55) 0.568 122 583 780 499 128 32 × 2 = 1 + 0.136 245 167 560 998 256 64;
  • 56) 0.136 245 167 560 998 256 64 × 2 = 0 + 0.272 490 335 121 996 513 28;
  • 57) 0.272 490 335 121 996 513 28 × 2 = 0 + 0.544 980 670 243 993 026 56;
  • 58) 0.544 980 670 243 993 026 56 × 2 = 1 + 0.089 961 340 487 986 053 12;
  • 59) 0.089 961 340 487 986 053 12 × 2 = 0 + 0.179 922 680 975 972 106 24;
  • 60) 0.179 922 680 975 972 106 24 × 2 = 0 + 0.359 845 361 951 944 212 48;
  • 61) 0.359 845 361 951 944 212 48 × 2 = 0 + 0.719 690 723 903 888 424 96;
  • 62) 0.719 690 723 903 888 424 96 × 2 = 1 + 0.439 381 447 807 776 849 92;
  • 63) 0.439 381 447 807 776 849 92 × 2 = 0 + 0.878 762 895 615 553 699 84;
  • 64) 0.878 762 895 615 553 699 84 × 2 = 1 + 0.757 525 791 231 107 399 68;
  • 65) 0.757 525 791 231 107 399 68 × 2 = 1 + 0.515 051 582 462 214 799 36;
  • 66) 0.515 051 582 462 214 799 36 × 2 = 1 + 0.030 103 164 924 429 598 72;

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:


0.000 000 000 000 151 576 98(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 0100 0101 11(2)

6. Positive number before normalization:

0.000 000 000 000 151 576 98(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 0100 0101 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 151 576 98(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 0100 0101 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 1010 1010 1010 0100 0101 11(2) × 20 =

1.0101 0101 0101 0010 0010 111(2) × 2-43


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): -43

Mantissa (not normalized):
1.0101 0101 0101 0010 0010 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-43 + 2(8-1) - 1 =

(-43 + 127)(10) =

84(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;
  • 84 ÷ 2 = 42 + 0;
  • 42 ÷ 2 = 21 + 0;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

84(10) =

0101 0100(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. 010 1010 1010 1001 0001 0111 =

010 1010 1010 1001 0001 0111


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) =
0101 0100

Mantissa (23 bits) =
010 1010 1010 1001 0001 0111


Decimal number -0.000 000 000 000 151 576 98 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 0100 - 010 1010 1010 1001 0001 0111

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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