-0.000 000 349 613 904 800 000 238 907 549 058 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 349 613 904 800 000 238 907 549 058(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
-0.000 000 349 613 904 800 000 238 907 549 058(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:

|-0.000 000 349 613 904 800 000 238 907 549 058| = 0.000 000 349 613 904 800 000 238 907 549 058


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: 0.000 000 349 613 904 800 000 238 907 549 058.

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) 0.000 000 349 613 904 800 000 238 907 549 058 × 2 = 0 + 0.000 000 699 227 809 600 000 477 815 098 116;
  • 2) 0.000 000 699 227 809 600 000 477 815 098 116 × 2 = 0 + 0.000 001 398 455 619 200 000 955 630 196 232;
  • 3) 0.000 001 398 455 619 200 000 955 630 196 232 × 2 = 0 + 0.000 002 796 911 238 400 001 911 260 392 464;
  • 4) 0.000 002 796 911 238 400 001 911 260 392 464 × 2 = 0 + 0.000 005 593 822 476 800 003 822 520 784 928;
  • 5) 0.000 005 593 822 476 800 003 822 520 784 928 × 2 = 0 + 0.000 011 187 644 953 600 007 645 041 569 856;
  • 6) 0.000 011 187 644 953 600 007 645 041 569 856 × 2 = 0 + 0.000 022 375 289 907 200 015 290 083 139 712;
  • 7) 0.000 022 375 289 907 200 015 290 083 139 712 × 2 = 0 + 0.000 044 750 579 814 400 030 580 166 279 424;
  • 8) 0.000 044 750 579 814 400 030 580 166 279 424 × 2 = 0 + 0.000 089 501 159 628 800 061 160 332 558 848;
  • 9) 0.000 089 501 159 628 800 061 160 332 558 848 × 2 = 0 + 0.000 179 002 319 257 600 122 320 665 117 696;
  • 10) 0.000 179 002 319 257 600 122 320 665 117 696 × 2 = 0 + 0.000 358 004 638 515 200 244 641 330 235 392;
  • 11) 0.000 358 004 638 515 200 244 641 330 235 392 × 2 = 0 + 0.000 716 009 277 030 400 489 282 660 470 784;
  • 12) 0.000 716 009 277 030 400 489 282 660 470 784 × 2 = 0 + 0.001 432 018 554 060 800 978 565 320 941 568;
  • 13) 0.001 432 018 554 060 800 978 565 320 941 568 × 2 = 0 + 0.002 864 037 108 121 601 957 130 641 883 136;
  • 14) 0.002 864 037 108 121 601 957 130 641 883 136 × 2 = 0 + 0.005 728 074 216 243 203 914 261 283 766 272;
  • 15) 0.005 728 074 216 243 203 914 261 283 766 272 × 2 = 0 + 0.011 456 148 432 486 407 828 522 567 532 544;
  • 16) 0.011 456 148 432 486 407 828 522 567 532 544 × 2 = 0 + 0.022 912 296 864 972 815 657 045 135 065 088;
  • 17) 0.022 912 296 864 972 815 657 045 135 065 088 × 2 = 0 + 0.045 824 593 729 945 631 314 090 270 130 176;
  • 18) 0.045 824 593 729 945 631 314 090 270 130 176 × 2 = 0 + 0.091 649 187 459 891 262 628 180 540 260 352;
  • 19) 0.091 649 187 459 891 262 628 180 540 260 352 × 2 = 0 + 0.183 298 374 919 782 525 256 361 080 520 704;
  • 20) 0.183 298 374 919 782 525 256 361 080 520 704 × 2 = 0 + 0.366 596 749 839 565 050 512 722 161 041 408;
  • 21) 0.366 596 749 839 565 050 512 722 161 041 408 × 2 = 0 + 0.733 193 499 679 130 101 025 444 322 082 816;
  • 22) 0.733 193 499 679 130 101 025 444 322 082 816 × 2 = 1 + 0.466 386 999 358 260 202 050 888 644 165 632;
  • 23) 0.466 386 999 358 260 202 050 888 644 165 632 × 2 = 0 + 0.932 773 998 716 520 404 101 777 288 331 264;
  • 24) 0.932 773 998 716 520 404 101 777 288 331 264 × 2 = 1 + 0.865 547 997 433 040 808 203 554 576 662 528;
  • 25) 0.865 547 997 433 040 808 203 554 576 662 528 × 2 = 1 + 0.731 095 994 866 081 616 407 109 153 325 056;
  • 26) 0.731 095 994 866 081 616 407 109 153 325 056 × 2 = 1 + 0.462 191 989 732 163 232 814 218 306 650 112;
  • 27) 0.462 191 989 732 163 232 814 218 306 650 112 × 2 = 0 + 0.924 383 979 464 326 465 628 436 613 300 224;
  • 28) 0.924 383 979 464 326 465 628 436 613 300 224 × 2 = 1 + 0.848 767 958 928 652 931 256 873 226 600 448;
  • 29) 0.848 767 958 928 652 931 256 873 226 600 448 × 2 = 1 + 0.697 535 917 857 305 862 513 746 453 200 896;
  • 30) 0.697 535 917 857 305 862 513 746 453 200 896 × 2 = 1 + 0.395 071 835 714 611 725 027 492 906 401 792;
  • 31) 0.395 071 835 714 611 725 027 492 906 401 792 × 2 = 0 + 0.790 143 671 429 223 450 054 985 812 803 584;
  • 32) 0.790 143 671 429 223 450 054 985 812 803 584 × 2 = 1 + 0.580 287 342 858 446 900 109 971 625 607 168;
  • 33) 0.580 287 342 858 446 900 109 971 625 607 168 × 2 = 1 + 0.160 574 685 716 893 800 219 943 251 214 336;
  • 34) 0.160 574 685 716 893 800 219 943 251 214 336 × 2 = 0 + 0.321 149 371 433 787 600 439 886 502 428 672;
  • 35) 0.321 149 371 433 787 600 439 886 502 428 672 × 2 = 0 + 0.642 298 742 867 575 200 879 773 004 857 344;
  • 36) 0.642 298 742 867 575 200 879 773 004 857 344 × 2 = 1 + 0.284 597 485 735 150 401 759 546 009 714 688;
  • 37) 0.284 597 485 735 150 401 759 546 009 714 688 × 2 = 0 + 0.569 194 971 470 300 803 519 092 019 429 376;
  • 38) 0.569 194 971 470 300 803 519 092 019 429 376 × 2 = 1 + 0.138 389 942 940 601 607 038 184 038 858 752;
  • 39) 0.138 389 942 940 601 607 038 184 038 858 752 × 2 = 0 + 0.276 779 885 881 203 214 076 368 077 717 504;
  • 40) 0.276 779 885 881 203 214 076 368 077 717 504 × 2 = 0 + 0.553 559 771 762 406 428 152 736 155 435 008;
  • 41) 0.553 559 771 762 406 428 152 736 155 435 008 × 2 = 1 + 0.107 119 543 524 812 856 305 472 310 870 016;
  • 42) 0.107 119 543 524 812 856 305 472 310 870 016 × 2 = 0 + 0.214 239 087 049 625 712 610 944 621 740 032;
  • 43) 0.214 239 087 049 625 712 610 944 621 740 032 × 2 = 0 + 0.428 478 174 099 251 425 221 889 243 480 064;
  • 44) 0.428 478 174 099 251 425 221 889 243 480 064 × 2 = 0 + 0.856 956 348 198 502 850 443 778 486 960 128;
  • 45) 0.856 956 348 198 502 850 443 778 486 960 128 × 2 = 1 + 0.713 912 696 397 005 700 887 556 973 920 256;

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:


0.000 000 349 613 904 800 000 238 907 549 058(10) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1(2)

6. Positive number before normalization:

0.000 000 349 613 904 800 000 238 907 549 058(10) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 349 613 904 800 000 238 907 549 058(10) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1(2) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1(2) × 20 =

1.0111 0111 0110 0101 0010 001(2) × 2-22


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): -22

Mantissa (not normalized):
1.0111 0111 0110 0101 0010 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(8-1) - 1 =

(-22 + 127)(10) =

105(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;
  • 105 ÷ 2 = 52 + 1;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

105(10) =

0110 1001(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. 011 1011 1011 0010 1001 0001 =

011 1011 1011 0010 1001 0001


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 1001

Mantissa (23 bits) =
011 1011 1011 0010 1001 0001


Decimal number -0.000 000 349 613 904 800 000 238 907 549 058 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1001 - 011 1011 1011 0010 1001 0001

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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